Delftse foundations of computation

2018

TU Delft Open

Delftse Foundations of Computation is a textbook for a one quarter introductory course in theoretical computer science. It includes topics from propositional and predicate logic, proof techniques, set theory and the theory of computation, along with practical applications to computer science. It has no prerequisites other than a general familiarity with computer programming.

Computer Science & Information Systems

English

9789463660839

Syntax Warning: Invalid Font Weight
Stefan Hugtenburg and Neil Yorke-Smith

Delftse Foundations of
Computation

Delftse Foundations of Computation
First TU Delft Edition (TUD-1.0.2, November 2018)

Stefan Hugtenburg and Neil Yorke-Smith

This work is licensed under CC BY-NC-SA 4.0

Delftse Foundations of Computation
First TU Delft Edition (TUD-1.0.2, November 2018)

Stefan Hugtenburg and Neil Yorke-Smith
Faculty of Electrical Engineering, Mathematics and Computer Science
Delft University of Technology
Delft, The Netherlands
{s.hugtenburg,n.yorke-smith}@tudelft.nl
———
Derived from Foundations of Computation by Carol Critchlow and David Eck
This work is licensed under CC BY-NC-SA 4.0

©2011 Carol Critchlow and David Eck
©2018 Stefan Hugtenburg and Neil Yorke-Smith
©2018 TU Delft Open
Delftse Foundations of Computation is a textbook for a one quarter introductory course in theoretical computer science. It includes topics from propositional
and predicate logic, proof techniques, set theory and the theory of computation,
along with practical applications to computer science. It has no prerequisites
other than a general familiarity with computer programming.
This book is derived from Foundations of Computation by Carol Critchlow and
David Eck, Version 2.3 (Summer 2011), which is licensed under CC BY-NC-SA
4.0. Critchlow and Eck are not associated with the TU Delft editions. This book
also uses some material from Wikipedia (English) (en.wikipedia.org), which is
licensed under CC BY-SA 3.0. The authors of the TU Delft editions are responsible
for any errors, and welcome bug reports and suggestions by email or in person.
Thanks to M. de Jong, T. Klos, F. Mulder, H. Tonino and E. Walraven.
This work can be redistributed in unmodified form, or in modified form with proper attribution and under the same license as the original, for non-commercial uses only, as specified by the Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License
(creativecommons.org/licenses/by-nc-sa/4.0/).
The latest edition of this book is available for online use and for free download from
the TU Delft Open Textbook repository at textbooks.open.tudelft.nl. The original
(non-TU Delft) version contains additional material, and is available at math.hws.edu/
FoundationsOfComputation/.
Typeset in TEX Gyre Pagella
Published by Delft University of Technology, Delft, The Netherlands, November 2018
Version TUD-1.0.2
ISBN 978-94-6366-083-9 | DOI 10.5074/t.isbn.9789463660839

iv

To the students of CSE1300

v

Contents
1 Introduction

1

2 Logic
2.1 Propositional Logic . . . . . . . . . . . . . .
2.1.1 Propositions . . . . . . . . . . . . . .
2.1.2 Logical operators . . . . . . . . . . .
2.1.3 Precedence rules . . . . . . . . . . .
2.1.4 Logical equivalence . . . . . . . . . .
2.1.5 More logical operaters . . . . . . . .
2.1.6 Implications in English . . . . . . . .
2.1.7 More forms of implication . . . . . .
2.1.8 Exclusive or . . . . . . . . . . . . . .
2.1.9 Universal operators . . . . . . . . . .
2.1.10 Classifying propositions . . . . . . .
2.2 Boolean Algebra . . . . . . . . . . . . . . . .
2.2.1 Basics of Boolean Algebra . . . . . .
2.2.2 Substitution laws . . . . . . . . . . .
2.2.3 Simplifications . . . . . . . . . . . .
2.2.4 More rules of Boolean algebra . . . .
2.3 Application: Logic Circuits . . . . . . . . .
2.3.1 Logic gates * . . . . . . . . . . . . . .
2.3.2 Combining gates to create circuits *
2.3.3 From circuits to propositions * . . .
2.3.4 Disjunctive Normal Form . . . . . .
2.3.5 Binary addition * . . . . . . . . . . .
2.4 Predicate Logic . . . . . . . . . . . . . . . .
2.4.1 Predicates . . . . . . . . . . . . . . .
2.4.2 Quantifiers . . . . . . . . . . . . . . .
2.4.3 Operators . . . . . . . . . . . . . . .
2.4.4 Tarski’s world and formal structures
vii

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3
5
5
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6
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27
28
32
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42
46
46
48
51

Proof
3.1 A Little Historical Background . . . . . .
3.2 Mathematical Proof . . . . . . . . . . . . .
3.2.1 How to write a proof . . . . . . . .
3.2.2 Some terminology . . . . . . . . .
3.2.3 Examples . . . . . . . . . . . . . .
3.3 Proof by Contradiction . . . . . . . . . . .
3.4 Mathematical Induction . . . . . . . . . .
3.4.1 How to write a proof by induction
3.4.2 Examples . . . . . . . . . . . . . .
3.4.3 More examples . . . . . . . . . . .
3.5 Strong Mathematical Induction . . . . . .
3.6 Application: Recursion and Induction . .
3.6.1 Recursive factorials . . . . . . . . .
3.6.2 Towers of Hanoi . . . . . . . . . .
3.6.3 Binary trees * . . . . . . . . . . . .
3.7 Recursive Definitions . . . . . . . . . . . .
3.8 Invariants . . . . . . . . . . . . . . . . . . .

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55
55
57
60
62
63
67
70
70
71
73
75
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78
79
81
85
87

Sets, Functions, and Relations
4.1 Basic Concepts . . . . . . . . . . . . . . . . .
4.1.1 Elements of sets . . . . . . . . . . . .
4.1.2 Set-builder notation . . . . . . . . .
4.1.3 Operations on sets . . . . . . . . . .
4.1.4 Visualising sets . . . . . . . . . . . .
4.1.5 Sets of sets . . . . . . . . . . . . . . .
4.1.6 Mathematical induction revisited . .
4.1.7 Structural Induction . . . . . . . . .
4.2 The Boolean Algebra of Sets . . . . . . . . .
4.2.1 Set complement . . . . . . . . . . . .
4.2.2 Link between logic and set theory .
4.3 Application: Programming with Sets * . . .
4.3.1 Representing sets . . . . . . . . . . .
4.3.2 Computing with sets . . . . . . . . .
4.4 Functions . . . . . . . . . . . . . . . . . . . .
4.4.1 Formalising the notion of functions

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89
89
91
92
93
95
96
99
100
102
103
106
110
111
113
117
117

2.5

3

4

2.4.5 Logical equivalence . . . . .
Deduction . . . . . . . . . . . . . .
2.5.1 Arguments . . . . . . . . .
2.5.2 Valid arguments and proofs
2.5.3 Proofs in predicate logic . .

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viii

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4.5
4.6

4.7

4.8

4.4.2 Operations on functions . . . . . . . .
4.4.3 Properties of functions . . . . . . . . .
4.4.4 First-class objects . . . . . . . . . . . .
Application: Programming with Functions *
4.5.1 Functions as first-class objects . . . . .
Counting Past Infinity . . . . . . . . . . . . .
4.6.1 Cardinality . . . . . . . . . . . . . . .
4.6.2 Counting to infinity . . . . . . . . . .
4.6.3 Uncountable sets * . . . . . . . . . . .
4.6.4 A final note on infinities * . . . . . . .
Relations . . . . . . . . . . . . . . . . . . . . .
4.7.1 Properties of relations . . . . . . . . .
4.7.2 Equivalence relations . . . . . . . . . .
Application: Relational Databases * . . . . . .

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118
121
123
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126
129
129
133
134
137
139
140
141
145

5 Looking Beyond *

153

Further Reading

157

Index

159

ix

Chapter 1

Introduction
ogic originally meant ‘the word’ or ‘what is spoken’ in Ancient Greece, and today

L means ‘thought’ or ‘reason’. As a subject, logic is concerned with the most general

laws of truth. Why study this kind of reasoning in computer science?
Logic is important because digital computers work with precision, and because designing algorithms requires precision, and because comparing algorithms requires precision.
Even when when a computer is, seemingly, computing with vague or imprecise
quantities, the underlying computation is precise.1 For example, when a deep neural
network is being trained to recognise cats, the algorithm being used to train the network
is specified precisely. More than this, the criteria we use to assess whether the network
has learned well enough are also specified precisely. And any theoretical properties
about the algorithm have been proven precisely.
Reasoning and logic, and related mathematical concepts such as sets, are foundational for computer science. A third of your TUDelft CSE curriculum is mathematics for
the first year: Reasoning & Logic, Calculus, Linear Algebra and Probability Theory & Statistics.
As a computer scientist, you have to be capable of solving complex problems. One
important aspect is to be able to come to the right conclusions. On the basis of theorems
and partial observations you can acquire more knowledge and evidence to help prove
that a specific conclusion is mathematically and logically correct. You learn how to do
this with Reasoning & Logic.
The foundational mathematical skills you learn in Reasoning & Logic are used in all
the other mathematics courses, and in Computer Organisation, Algorithms & Data Structures, Information & Data Management, Machine Learning, and many other courses. Logic
is studied not only in mathematics and computer science, but also in philosophy (since
Ancient Greece) and today in fields such as linguistics and psychology.
This book is designed to help you achieve the learning goals of Reasoning & Logic:
1 You

can take a course on quantum computing to learn whether they are an exception.

1

2

1

Chapter 1. Introduction
1. Translate a logically precise claim to and from natural language.
2. Describe the operation of logical connectors and quantifiers.
3. Describe the notion of logical validity.
4. Explain and apply basic set operations.
5. Define and perform computations with functions, relations and equivalence classes.
6. Construct a direct or indirect proof—by contradiction, division into cases, generalisation, or (structural) induction—or counterexample for (in)valid arguments or
logical equivalence in propositional or predicate logic and in set theory.
7. Construct and interpret recursive definitions.
8. Identify what type of proof is appropriate for a given claim.
9. Construct the right function or relation given a description (in natural language or
formal notation).
10. Develop specifications for verification tools like SAT or SMT solvers.
11. Interpret the response of verification tools like SAT or SMT solvers.



We do not cover every topic at the same level of detail and some topics (SAT
and SMT solvers) we do not cover at all, and the lectures will not cover
everything in the book. Some topics in the lectures which you will prepare
for using other materials: these will be announced.



Starred sections in the contents of this book are not included in the syllabus
for Reasoning & Logic.

The theme of the book is about coming to the right conclusion: proving the logical
validity of arguments. What is a valid argument? When is an argument logically valid
and when is it not? How can we determine whether an argument is logically valid? How
can we derive a logically valid conclusion from the premises? Or how can we prove that
a conclusion is not a logical consequence of the premises?

Chapter 2

Logic
n a sense, we know a lot more than we realise, because everything that we know has

I consequences—logical consequences—that follow automatically. If you know that all

humans are mortal, and you know that Socrates is human, then in a sense you know
that Socrates is mortal, whether or not you have ever considered or wanted to consider
that fact. This is an example of logical deduction: From the premises that “All humans
are mortal” and “Socrates is human”, the conclusion that “Socrates is mortal” can be
deduced by logic.



Socrates was a Greek philosopher who suffered a most
unfortunate fate. In one of the most famous arguments—
the one outlined above—he is the man that is destined
to die. History has since taught us we were right as Socrates died after a long life (71 years), sentenced to death
for corrupting the minds of the youths of Athens. His
disciple Plato wrote many Socratic dialogues which give
an insight into the philosophy of Socrates often summarised as: “I know that I know nothing”. The last words
of this mortal man were (according to Plato): “Crito, we
owe a cock to Asclepius, Pay it and do not neglect it.”
Source: en.wikipedia.org/wiki/Socrates.

Logical deduction is a kind of computation. By applying rules of logic to a given set of
premises, conclusions that follow from those premises can be generated automatically.
This computational process could for instance be carried out by a computer. Once you
3

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2

Chapter 2. Logic

know the premises, or are willing to accept them for the sake of argument, you are forced
by logic to accept the conclusions. Still, to say that you ‘know’ those conclusions would
be misleading. The problem is that there are too many of them (infinitely many), and,
in general, most of them are not particularly interesting. Until you have actually made
the deduction, you don’t really know the conclusion, and knowing which of the possible
chains of deduction to follow is not easy. The art of logic is to find an interesting conclusion and a chain of logical deductions that leads from the premises to that conclusion.
Checking that the deductions are valid is the mechanical, computational side of logic.



Later in Reasoning & Logic, you will see some automated computational techniques that can help us checking the deductions. We don’t cover these in this
edition of this book. In fact, there are proof assistants that can even help us
finding interesting conclusions.

This chapter is mostly about the mechanics of logic. We will investigate logic as a
branch of mathematics, with its own symbols, formulas, and rules of computation. Your
objective is to learn the rules of logic, to understand why they are valid, and to develop
skill in applying them. As with any branch of mathematics, there is a certain beauty to
the symbols and formulas themselves. But it is the applications that bring the subject
to life for most people. We will, of course, cover some applications as we go along. In
a sense, though, the real applications of logic include much of computer science and of
mathematics itself.
Among the fundamental elements of thought, and therefore of logic, are propositions.
A proposition is a statement that has a truth value: It is either true or false. “Delft is a
city” and “2 + 2 = 42” are propositions. In the first part of this chapter, we will study
propositional logic, which takes propositions and considers how they can be combined
and manipulated. This branch of logic has surprising application to the design of the
electronic circuits that make up computers. This ties closely to the digital and boolean
logic you will study in your course Computer Organisation.
Logic gets more interesting when we consider the internal structure of propositions.
In English, a proposition is expressed as a sentence, and, as you know from studying
grammar, sentences have parts. A simple sentence like “Delft is a city” has a subject and
a predicate. The sentence says something about its subject. The subject of “Delft is a city”
is Delft. The sentence says something about Delft. The something that the sentence says
about its subject is the predicate. In the example, the predicate is the phrase “is a city”.
Once we start working with predicates, we can create propositions using quantifiers
like ‘all’, ‘some’, and ‘no’. For example, working with the predicate “has a university”
we can move from simple propositions like “Delft has a university” to “All cities have a
university” or to “No city has a university” or to the rather more realistic “Some cities
have a university”.
Logical deduction usually deals with quantified statements, as shown by the basic

2.1. Propositional Logic

5

example of human mortality with which we began this chapter. Logical deduction will
be a major topic of this chapter; and under the name of proof , it will be the topic of the
next chapter and a major tool for the rest of this book.

2.1 Propositional Logic
To start modeling the ambigous and often vague natural languages we first consider
propositional logic. This form of logic is arguably the easiest to work with, but also
has limited expressive power. However even with this form we can already encapsulate
many arguments and power a number of applications, for instance digital logic in chip
design.

2.1.1 Propositions
A proposition is a statement which is either true or false. In propositional logic, we
reason only about propositions and see what we can do with them. Since this is mathematics, we need to be able to talk about propositions without saying which particular
propositions we are talking about, so we use symbolic names to represent them. We will
always use lowercase letters such as p, q, and r to represent propositions. A letter used
in this way is called a propositional variable. Remember that when I say something
like “Let p be a proposition”, I mean “For the rest of this discussion, let the symbol p
stand for some particular statement, which is either true or false (although I am not at
the moment making any assumption about which it is).” The discussion has mathematical generality in that p can represent any statement, and the discussion will be valid no
matter which statement it represents.



Propositional variables are a little bit like variables in a programming language like Java. A basic Java variable such as int x can take any integer
value. I say there is “a little bit” of similarity between the two notions of
variables—don’t take the analogy too far at this point in your learning!

2.1.2 Logical operators
What we do with propositions is combine them with logical operators, also referred to
as logical connectives. A logical operator can be applied to one or more propositions
to produce a new proposition. The truth value of the new proposition is completely
determined by the operator and by the truth values of the propositions to which it is

2

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Chapter 2. Logic

applied.1 In English, logical operators are represented by words such as ‘and’, ‘or’, and
‘not’. For example, the proposition “I wanted to leave and I left” is formed from two
simpler propositions joined by the word ‘and’. Adding the word ‘not’ to the proposition
“I left” gives “I did not leave” (after a bit of necessary grammatical adjustment).
But English is a little too rich for mathematical logic. When you read the sentence
“I wanted to leave and I left”, you probably see a connotation of causality: I left because
I wanted to leave. This implication does not follow from the logical combination of the
truth values of the two propositions “I wanted to leave” and “I left”. Or consider the
proposition “I wanted to leave but I did not leave”. Here, the word ‘but’ has the same
logical meaning as the word ‘and’, but the connotation is very different. So, in mathematical logic, we use symbols to represent logical operators. These symbols do not carry any
connotation beyond their defined logical meaning. The logical operators corresponding
to the English words ‘and’, ‘or’, and ‘not’ are ∧, ∨, and ¬.2
Definition 2.1. Let p and q be propositions. Then p ∨ q, p ∧ q, and ¬ p are propositions,
whose truth values are given by the rules:
• p ∧ q is true when both p is true and q is true, and in no other case.
• p ∨ q is true when either p is true, or q is true, or both p and q are true, and in no
other case.
• ¬ p is true when p is false, and in no other case.
The operators ∧, ∨, and ¬ are referred to as conjunction, disjunction, and negation,
respectively. (Note that p ∧ q is read as ‘p and q’, p ∨ q is read as ‘p or q’, and ¬ p is
read as ‘not p’.)


2.1.3

Consider the statement “I am a CSE student or I am not a TPM student.”
Taking p to mean “I am a CSE student” and q to mean “I am a TPM student”,
you can write this as p ∨ ¬q.

Precedence rules

These operators can be used in more complicated expressions, such as p ∧ ¬q or ( p ∨
q) ∧ (q ∨ r ). A proposition made up of simpler propositions and logical operators is
1 It is not always true that the truth value of a sentence can be determined from the truth values of its
component parts. For example, if p is a proposition, then “Johan Cruyff believes p” is also a proposition, so
“Cruyff believes” is some kind of operator. However, it does not count as a logical operator because just from
knowing whether or not p is true, we get no information at all about whether “Johan Cruyff believes p” is true.
2 Other textbooks might use different notations to represent a negation. For instance a bar over the variable
x̄ or a ∼ symbol. In Boolean algebra (and thus in your Computer Organisation course) you will also often find
the + symbol to represent an ‘or’ and a · (dot) symbol to represent an ‘and’.

2.1. Propositional Logic

7

called a compound proposition. Just like in mathematics, parentheses can be used in
compound expressions to indicate the order in which the operators are to be evaluated.
In the absence of parentheses, the order of evaluation is determined by precedence rules.
For the logical operators defined above, the rules are that ¬ has higher precedence than
∧, and ∧ has precedence over ∨. This means that in the absence of parentheses, any ¬
operators are evaluated first, followed by any ∧ operators, followed by any ∨ operators.
For example, the expression ¬ p ∨ q ∧ r is equivalent to the expression (¬ p) ∨ (q ∧ r ),
while p ∨ q ∧ q ∨ r is equivalent to p ∨ (q ∧ q) ∨ r.
This still leaves open the question of which of the ∧ operators in the expression p ∧
q ∧ r is evaluated first. This is settled by the following rule: When several operators of
equal precedence occur in the absence of parentheses, they are evaluated from left to
right. Thus, the expression p ∧ q ∧ r is equivalent to ( p ∧ q) ∧ r rather than to p ∧ (q ∧ r ).
In this particular case, as a matter of fact, it doesn’t really matter which ∧ operator is
evaluated first, since the two compound propositions ( p ∧ q) ∧ r and p ∧ (q ∧ r ) always
have the same value, no matter what logical values the component propositions p, q, and
r have. We say that ∧ is an associative operation. We’ll see more about associativity and
other properties of operations in the next section.



In practice however you should always add parentheses in places where ambiguity may arise. In fact some text books even add them to single operators
as well, e.g., writing ( p ∧ q) instead of p ∧ q. Although for this course we do
not require them around single operators, we should never need the precedence rules outlined above. Your parentheses should make clear the order of
operations!

Every compound proposition has a main connective. The main connective is the connective that is evaluated last, according to the precedence rules and parentheses. There
should be no ambiguity over which is the main connective in a compound proposition.

2.1.4 Logical equivalence
Suppose we want to verify that, in fact, ( p ∧ q) ∧ r and p ∧ (q ∧ r ) do always have the
same value. To do so, we have to consider all possible combinations of values of p, q, and
r, and check that for all such combinations, the two compound expressions do indeed
have the same value. It is convenient to organize this computation into a truth table. A
truth table is a table that shows the value of one or more compound propositions for
each possible combination of values of the propositional variables that they contain. We
call each such combination a situation. Figure 2.1 is a truth table that compares the
value of ( p ∧ q) ∧ r to the value of p ∧ (q ∧ r ) for all possible values of p, q, and r. There
are eight rows in the table because there are exactly eight different ways in which truth

2

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Chapter 2. Logic

2

p

q

r

p∧q

q∧r

( p ∧ q) ∧ r

p ∧ (q ∧ r )

0
0
0
0
1
1
1
1

0
0
1
1
0
0
1
1

0
1
0
1
0
1
0
1

0
0
0
0
0
0
1
1

0
0
0
1
0
0
0
1

0
0
0
0
0
0
0
1

0
0
0
0
0
0
0
1

Figure 2.1: A truth table that demonstrates the logical equivalence of ( p ∧ q) ∧ r and
p ∧ (q ∧ r ). The fact that the last two columns of this table are identical shows that
these two expressions have the same value for all eight possible combinations of values
of p, q, and r.

values can be assigned to p, q, and r.3 In this table, we see that the last two columns,
representing the values of ( p ∧ q) ∧ r and p ∧ (q ∧ r ), are identical.



We discuss the creation of truth tables for statements written in propositional logic in more detail in one of the pencasts of this course: youtu.be/
SOZqDm0bDuc. We also discuss using truth tables to test for equivalence in
another pencast youtu.be/sWu0fUu7s5c.



You can write the rows in a truth table in any order you like. I suggest you
write them in a sorted order, as in Table 2.1. This helps you to be systematic
in writing out the table.

More generally, we say that two compound propositions are logically equivalent if
they always have the same value, no matter what truth values are assigned to the propositional variables that they contain. If the number of propositional variables is small,
it is easy to use a truth table to check whether or not two propositions are logically equivalent.
3 In general, if there are n variables, then there are 2n different ways to assign truth values to the variables,
i.e., 2n situations. This might become clear to you if you try to come up with a scheme for systematically listing
all possible sets of values. As this should not satisfy you, you’ll find a rigorous proof of the fact later in this
chapter.

2.1. Propositional Logic



9

When writing a piece of code you will often have your code make decisions.
For instance in a bit of Java code—such as in your Object-Oriented Programming course—you might encounter an if-statement to check if the user has
inputted the right type of data. Since the input you expect can be rather
difficult, the if-statement is a complex combination of many simple checked
chained together by &&’s and ||’s. After taking a look at the code, you believe it can be simplified to a much smaller expression. Using a truth table
you can prove that your simplified version is equivalent to the original.

2.1.5 More logical operaters
There are other logical operators besides ∧, ∨, and ¬. We will consider the conditional
operator, →, the biconditional operator, ↔, and the exclusive or operator, ⊕.4 These
operators can be completely defined by a truth table that shows their values for the four
possible combinations of truth values of p and q.
Definition 2.2. For any propositions p and q, we define the propositions p → q, p ↔ q,
and p ⊕ q according to the truth table:



p

q

p→q

p↔q

p⊕q

0
0
1
1

0
1
0
1

1
1
0
1

1
0
0
1

0
1
1
0

When these operators are used in expressions, in the absence of parentheses
to indicate order of evaluation, we use the following precedence rules: The
exclusive or operator, ⊕, has the same precedence as ∨. The conditional
operator, →, has lower precedence than ∧, ∨, ¬, and ⊕, and is therefore
evaluated after them. Finally, the biconditional operator, ↔, has the lowest
precedence and is therefore evaluated last. For example, the expression p →
q ∧ r ↔ ¬ p ⊕ s is evaluated as if it were written ( p → (q ∧ r )) ↔ ((¬ p) ⊕ s).
But again you should always include the parentheses!

In order to work effectively with the logical operators, you need to know more about
4 Note that the symbols for these operations also differ from textbook to textbook. While → is fairly standard,
↔ is sometimes represented by ≡ or ⇔. There is even less standardization of the exclusive or operator, but
that operator is generally not so important as the others.

2

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Chapter 2. Logic

their meaning and how they relate to ordinary English expressions. To that end we first
consider the conditional operator in more detail in the next section.

2

2.1.6

Implications in English

The proposition p → q is called an implication or a conditional. It is usually read as “p
implies q”. In such an implication p and q also get special names of their own. p is called
the hypothesis or antecedent and q is called the conclusion or consequent.
Furthermore we say that if the implication p → q holds, then p is sufficient for q. That
is if p is true that is sufficient to also make q true. Conversely we say that q is necessary
for p. Without q being true, it is impossible for p to be true. That is if q is false, then p
also has to be false.
In English, p → q is often expressed as ‘if p then q’. For example, if p represents
the proposition “Karel Luyben is Rector Magnificus of TU Delft” and q represents “Prometheus is blessed by the gods”, then p → q could be expressed in English as “If Karel
Luyben is Rector Magnificus of TU Delft, then Prometheus is blessed by the gods.” In
this example, p is false and q is also false. Checking the definition of p → q, we see that
p → q is a true statement. Most people would agree with this, even though it is not
immediately obvious.



The letter ‘T’ in the TUDelft logo bears a stylized flame on
top, referring to the flame that Prometheus brought from
Mount Olympus to the people, against the will of Zeus. Because of this, Prometheus is sometimes considered as the
first engineer, and he is an important symbol for the university. His bronze statue stands in the Mekelpark at the centre
of campus.
Source: en.wikipedia.org/wiki/Delft_University_of_
Technology. Image: weblog.library.tudelft.nl/2016/
01/04/english-prometheus-is-back/.

It is worth looking at a similar example in more detail. Suppose that I assert that “If
Feyenoord is a great team, then I’m the King of the Netherlands”. This statement has
the form m → k where m is the proposition “Feyenoord is a great team” and k is the
proposition “I’m the king of the Netherlands”. Now, demonstrably I am not the king of
the Netherlands, so k is false. Since k is false, the only way for m → k to be true is for m
to be false as well. (Check the definition of → in the table, if you are not convinced!) So,

2.1. Propositional Logic

11

by asserting m → k, I am really asserting that the Feyenoord is not a great team.
Or consider the statement, “If the party is on Tuesday, then I’ll be there.” What am
I trying to say if I assert this statement? I am asserting that p → q is true, where p
represents “The party is on Tuesday” and q represents “I will be at the party”. Suppose
that p is true, that is, the party does in fact take place on Tuesday. Checking the definition
of →, we see that in the only case where p is true and p → q is true, q is also true. So from
the truth of “If the party is on Tuesday, then I will be at the party” and “The party is in
fact on Tuesday”, you can deduce that “I will be at the party” is also true. But suppose,
on the other hand, that the party is actually on Wednesday. Then p is false. When p is
false and p → q is true, the definition of p → q allows q to be either true or false. So, in
this case, you can’t make any deduction about whether or not I will be at the party. The
statement “If the party is on Tuesday, then I’ll be there” doesn’t assert anything about
what will happen if the party is on some other day than Tuesday.

2.1.7 More forms of implication
The implication ¬q → ¬ p is called the contrapositive of p → q. An implication is logically equivalent to its contrapositive. The contrapositive of “If this is Tuesday, then we are
in Belgium” is “If we aren’t in Belgium, then this isn’t Tuesday”. These two sentences
assert exactly the same thing.
Note that p → q is not logically equivalent to q → p. The implication q → p is called
the converse of p → q. The converse of “If this is Tuesday, then we are in Belgium” is “If
we are in Belgium, then this is Tuesday”. Note that it is possible for either one of these
statements to be true while the other is false. In English, I might express the fact that both
statements are true by saying “If this is Tuesday, then we are in Belgium, and conversely”.
In logic, this would be expressed with a proposition of the form ( p → q) ∧ (q → p).
Similarly p → q is not logically equivalent to ¬ p → ¬q. The implication ¬ p → ¬q
is called the inverse of p → q. Although this mistake is commonly made in English, for
instance people often assume that when I say: “If it is morning, I drink some coffee”, I
also mean that when it is not morning I do not drink coffee. But my original statement
does not tell you anything about what I do when it is not morning.
The biconditional operator is closely related to the conditional operator. In fact, p ↔
q is logically equivalent to ( p → q) ∧ (q → p). The proposition p ↔ q is usually read as
“p if and only if q”. (The “p if q” part represents q → p, while “p only if q” is another
way of asserting that p → q.) It could also be expressed as “if p then q, and conversely”.
Occasionally in English, ‘if… then’ is used when what is really meant is ‘if and only if’.
For example, if a parent tells a child, “If you are good, Sinterklaas will bring you toys”,
the parent probably really means to say “Sinterklaas will bring you toys if and only if you
are good”. (The parent would probably not respond well to the child’s perfectly logical
plea “But you never said what would happen if I wasn’t good!”)

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2.1.8

2

Chapter 2. Logic

Exclusive or

Finally, we turn to the exclusive or operator. The English word ‘or’ is actually somewhat
ambiguous. The two operators ⊕ and ∨ express the two possible meanings of this word.
The proposition p ∨ q can be expressed unambiguously as “p or q, or both”, while p ⊕ q
stands for “p or q, but not both”. If a menu says that you can choose soup or salad, it
doesn’t mean that you can have both. In this case, ‘or’ is an exclusive or. On the other
hand, in “You are at risk of heart disease if you smoke or drink”, the or is inclusive since
you certainly don’t get off the hook if you both smoke and drink. In theoretical computer
science and mathematics, the word ‘or’ is always taken in the inclusive sense of p ∨ q.

2.1.9

Universal operators

Now, any compound proposition that uses any of the operators →, ↔, and ⊕ can be
rewritten as a logically equivalent proposition that uses only ∧, ∨, and ¬. It is easy to
check that p → q is logically equivalent to ¬ p ∨ q. (Just make a truth table for ¬ p ∨ q.)
Similarly, p ↔ q can be expressed as (¬ p ∨ q) ∧ (¬q ∨ p), So, in a strict logical sense, →,
↔, and ⊕ are unnecessary. (Nevertheless, they are useful and important, and we won’t
give them up.)
Even more is true: In a strict logical sense, we could do without the conjunction
operator ∧. It is easy to check that p ∧ q is logically equivalent to ¬(¬ p ∨ ¬q), so any
expression that uses ∧ can be rewritten as one that uses only ¬ and ∨. Alternatively, we
could do without ∨ and write everything in terms of ¬ and ∧. We shall study some of
these rewrite rules in more detail in Section 2.2.
We call a set of operators that can express all operations: functionally complete. More
formally we would state the following:
Definition 2.3. A set of logical operators is functionally complete if and only if all formulas in propositional logic can be rewritten to an equivalent form that uses only operators
from the set.
Consider for instance the set {¬, ∨}. As shown above the ∧, → and ↔-operators can
be expressed using only these operators. In fact all possible operations can be expressed
using only {¬, ∨}. To prove this you will show in one of the exercises that all possible
formulas in propositional logic can be expressed using {¬, ∨, ∧, →, ↔}. So by showing
that we do not need ∧, →, and ↔ we can prove that {¬, ∨} is also functionally complete.

2.1.10

Classifying propositions

Certain types of proposition will play a special role in our further work with logic. In
particular, we define tautologies, contradictions, and contingencies as follows:
Definition 2.4. A compound proposition is said to be a tautology if and only if it is
true for all possible combinations of truth values of the propositional variables which it

2.1. Propositional Logic

13

contains. A compound proposition is said to be a contradiction if and only if it is false for
all possible combinations of truth values of the propositional variables which it contains.
A compound proposition is said to be a contingency if and only if it is neither a tautology
nor a contradiction.
For example, the proposition (( p ∨ q) ∧ ¬q) → p is a tautology. This can be checked
with a truth table:
p

q

p∨q

¬q

( p ∨ q) ∧ ¬q

(( p ∨ q) ∧ ¬q) → p

0
0
1
1

0
1
0
1

0
1
1
1

1
0
1
0

0
0
1
0

1
1
1
1

The fact that all entries in the last column are true tells us that this expression is a
tautology. Note that for any compound proposition P, P is a tautology if and only if ¬ P
is a contradiction. (Here and in the future, I use uppercase letters to represent compound
propositions. P stands for any formula made up of simple propositions, propositional
variables, and logical operators.)
Logical equivalence can be defined in terms of tautology:
Definition 2.5. Two compound propositions, P and Q, are said to be logically equivalent
if and only if the proposition P ↔ Q is a tautology.
The assertion that P is logically equivalent to Q will be expressed symbolically as
“P ≡ Q”. For example, ( p → q) ≡ (¬ p ∨ q), and p ⊕ q ≡ ( p ∨ q) ∧ ¬( p ∧ q).



What if P → Q and P is false? From a false premise we can derive any
conclusion (check the truth table of →). So if k stands for “I’m the King of the
Netherlands”, then k → Q is true for any compound proposition Q. You can
substitute anything for Q, and the implication k → Q will hold. For example,
it a logically valid deduction that: If I’m the King of the Netherlands, then
unicorns exist. Taking this further, from a contradiction we can derive any
conclusion. This is called the Principle of Explosion.

Exercises
1. Give the three truth tables that define the logical operators ∧, ∨, and ¬.
2. Some of the following compound propositions are tautologies, some are contradictions, and
some are neither (i.e.,, so are contingencies). In each case, use a truth table to decide to which
of these categories the proposition belongs:

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Chapter 2. Logic
a) ( p ∧ ( p → q)) → q
c) p ∧ ¬ p
e) p ∨ ¬ p

2

b) (( p → q) ∧ (q → r )) → ( p → r )
d) ( p ∨ q) → ( p ∧ q)
f) ( p ∧ q) → ( p ∨ q)

3. Use truth tables to show that each of the following propositions is logically equivalent to p ↔ q.
a) ( p → q) ∧ (q → p)
b) ¬ p ↔ ¬q
c) ( p → q) ∧ (¬ p → ¬q)
d) ¬( p ⊕ q)
4. Is → an associative operation? This is, is ( p → q) → r logically equivalent to p → (q → r )? Is
↔ associative?
5. Let p represent the proposition “You leave” and let q represent the proposition “I leave”. Express the following sentences as compound propositions using p and q, and show that they
are logically equivalent:
a) Either you leave or I do. (Or both!)
b) If you don’t leave, I will.
6. Suppose that m represents the proposition “The Earth moves”, c represents “The Earth is the
centre of the universe”, and g represents “Galileo was falsely accused”. Translate each of the
following compound propositions into English:
a) ¬ g ∧ c
b) m → ¬c
c) m ↔ ¬c
d) (m → g) ∧ (c → ¬ g)
7. Give the converse and the contrapositive of each of the following English sentences:
a) If you are good, Sinterklaas brings you toys.
b) If the package weighs more than one kilo, then you need extra postage.
c) If I have a choice, I don’t eat courgette.
8. In an ordinary deck of fifty-two playing cards, for how many cards is it true
a) that “This card is a ten and this card is a heart”?
b) that “This card is a ten or this card is a heart”?
c) that “If this card is a ten, then this card is a heart”?
d) that “This card is a ten if and only if this card is a heart”?
9. Define a logical operator ↓ so that p ↓ q is logically equivalent to ¬( p ∨ q). (This operator is
usually referred to as ‘nor’, short for ‘not or’.) Show that each of the propositions ¬ p, p ∧ q,
p ∨ q, p → q, p ↔ q, and p ⊕ q can be rewritten as a logically equivalent proposition that uses
↓ as its only operator.
10. For our proof that {¬, ∨} is functionally complete, we need to show that all formulas in propositional logic can be expressed in an equivalent form using only {¬, ∧, ∨, →, ↔}.
a) How many unique truth tables exist for formulas containing two atoms?
b) Create a function for each of the possible truth tables that uses only the 5 operators listed
above.
c) Give an (informal) argument why this means all formulas in propositional logic can be
expressed using only these five operators.

2.2

Boolean Algebra

So far we have discussed how to write and interpret propositions. This section deals
with manipulating them. For this, we need algebra. Ordinary algebra, of the sort taught

2.2. Boolean Algebra

15

in high school, is about manipulating numbers, variables that represent numbers, and
operators such as + and × that apply to numbers. Now, we need an algebra that applies
to logical values, propositional variables, and logical operators. The first person to think
of logic in terms of algebra was the mathematician, George Boole, who introduced the
idea in a book that he published in 1854. The algebra of logic is now called Boolean
algebra in his honour.



George Boole (1815–1864) was a largely self-taught British mathematician, philosopher and logician, most of
whose short career was spent as the first professor of
mathematics at Queen’s College, Cork in Ireland. He
worked in the fields of differential equations and algebraic logic, and is best known as the author of The Laws
of Thought (1854). Among TU Delft students he is best
known for the room named after him in the EEMCS
building 28.
Boolean logic is credited with laying the foundations for
the information age: essentially, computer science. Boole maintained that:
“No general method for the solution of questions in the theory of probabilities can be established which does not explicitly recognise, not only the special numerical bases of the science, but also those universal laws of thought
which are the basis of all reasoning, and which, whatever they may be as to
their essence, are at least mathematical as to their form.”
Source: en.wikipedia.org/wiki/George_Boole.

The algebra of numbers includes a large number of rules for manipulating expressions. The distributive law, for example, says that x (y + z) = xy + xz, where x, y, and z
are variables that stand for any numbers or numerical expressions. This law means that
whenever you see something of the form xy + xz in a numerical expression, you can
substitute x (y + z) without changing the value of the expression, and vice versa. Note
that the equals sign in x (y + z) = xy + xz means “has the same value as, no matter what
numerical values x, y, and z have”.
In Boolean algebra, we work with logical values instead of numerical values. There
are only two logical values, true and false. We will write these values as T and F or 1 and
0. The symbols T and F play a similar role in Boolean algebra to the role that constant
numbers such as 1 and 3.14159 play in ordinary algebra. Instead of the equals sign,
Boolean algebra uses logical equivalence, ≡, which has essentially the same meaning.5
5 In ordinary algebra, it is easy to be confused by the equals sign, because it has two very different roles. In
an identity such as the distributive law, it means “is always equal to”. On the other hand, an equation such

2

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Chapter 2. Logic

2

Double negation

¬(¬ p) ≡ p

Excluded middle

p ∨ ¬p ≡ T

Contradiction

p ∧ ¬p ≡ F

Identity laws

T∧ p ≡ p
F∨ p ≡ p

Idempotent laws

p∧p ≡ p
p∨p ≡ p

Commutative laws

p∧q ≡ q∧p
p∨q ≡ q∨p

Associative laws

( p ∧ q) ∧ r ≡ p ∧ (q ∧ r )
( p ∨ q) ∨ r ≡ p ∨ (q ∨ r )

Distributive laws

p ∧ (q ∨ r ) ≡ ( p ∧ q) ∨ ( p ∧ r )
p ∨ (q ∧ r ) ≡ ( p ∨ q) ∧ ( p ∨ r )

DeMorgan’s laws

¬( p ∧ q) ≡ (¬ p) ∨ (¬q)

¬( p ∨ q) ≡ (¬ p) ∧ (¬q)
Figure 2.2: Laws of Boolean Algebra. These laws hold for any propositions p, q, and r.

For example, for propositions p, q, and r, the ≡ operator in p ∧ (q ∧ r ) ≡ ( p ∧ q) ∧ r
means “has the same value as, no matter what logical values p, q, and r have”.

2.2.1

Basics of Boolean Algebra

Many of the rules of Boolean algebra are fairly obvious, if you think a bit about what
they mean. Even those that are not obvious can be verified easily by using a truth table.
Figure 2.2 lists the most important of these laws. You will notice that all these laws,
except the first, come in pairs: Each law in the pair can be obtained from the other by
interchanging ∧ with ∨ and T with F. This cuts down on the number of facts you have
to remember.6
as x2 + 3x = 4 is a statement that might or might not be true, depending on the value of x. Boolean algebra
has two operators, ≡ and ↔, that play roles similar to the two roles of the equals sign. ≡ is used for identity,
whereas ↔ is used in equations that may or may not be true.
6 It is also an example of a more general fact known as duality, which asserts that given any tautology that
uses only the operators ∧, ∨, and ¬, another tautology can be obtained from it by interchanging ∧ with ∨ and
T with F. We won’t attempt to prove this here, but I encourage you to try it!

2.2. Boolean Algebra

17

Just as an example, let’s verify the first rule in the table, the Law of Double Negation.
This law is just the old, basic grammar rule that two negatives make a positive. Although
the way this rule applies to English is questionable, if you look at how it is used—no
matter what the grammarian says, “I can’t get no satisfaction” doesn’t really mean “I
can get satisfaction”—the validity of the rule in logic can be verified just by computing
the two possible cases: when p is true and when p is false. When p is true, then by the
definition of the ¬ operator, ¬ p is false. But then, again by the definition of ¬, the value
of ¬(¬ p) is true, which is the same as the value of p. Similarly, in the case where p is
false, ¬(¬ p) is also false. Organized into a truth table, this argument takes the rather
simple form
p

¬p

¬(¬ p)

0
1

1
0

0
1

The fact that the first and last columns are identical shows the logical equivalence
of p and ¬(¬ p). The point here is not just that ¬(¬ p) ≡ p, but also that this logical
equivalence is valid because it can be verified computationally based just on the relevant
definitions. Its validity does not follow from the fact that “it’s obvious” or “it’s a wellknown rule of grammar”.



Students often ask “Why do I have to prove something when it’s obvious?”
The point is that logic—and mathematics more generally—is its own little
world with its own set of rules. Although this world is related somehow to
the real world, when you say that something is obvious (in the real world),
you aren’t playing by the rules of the world of logic. The real magic of mathematics is that by playing by its rules, you can come up with things that are
decidedly not obvious, but that still say something about the real world—
often, something interesting and important.

Each of the rules in Figure 2.2 can be verified in the same way, by making a truth table
to check all the possible cases. In one of the pencasts of this course we further discuss
how to check the equivalence of two propositions using truth tables.

2.2.2 Substitution laws
It’s important to understand that the propositional variables in the laws of Boolean algebra can stand for any propositions, including compound propositions. It is not just
true, as the Double Negation Law states, that ¬(¬ p) ≡ p. It is also true that ¬(¬q) ≡ q,
that ¬(¬( p ∧ q)) ≡ ( p ∧ q), that ¬(¬( p → (q ∧ ¬ p))) ≡ ( p → (q ∧ ¬ p)), and an infinite

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Chapter 2. Logic

number of other statements of the same form. Here, a “statement of the same form” is
one that can be obtained by substituting something for p in both places where it occurs
in ¬(¬ p) ≡ p. How can I be sure that all these infinitely many statements are valid when
all that I’ve checked is one little two-line truth table? The answer is that any given proposition, Q, no matter how complicated, has a particular truth value, either true or false.
So, the question of the validity of ¬(¬ Q) ≡ Q always reduces to one of the two cases I
already checked in the truth table. (Note that for this argument to be valid, the same Q
must be substituted for p in every position where it occurs.) While this argument may
be ‘obvious’, it is not exactly a proof, but for now we will just accept the validity of the
following theorem:
Theorem 2.1 (First Substitution Law). Suppose that Q is any proposition, and that p is a
propositional variable. Consider any tautology. If ( Q) is substituted for p in all places where p
occurs in the tautology, then the result is also a tautology.
Since logical equivalence is defined in terms of tautology, it is also true that when ( Q)
is substituted for p in a logical equivalence, the result is again a logical equivalence.7
The First Substitution Law lets you do algebra! For example, you can substitute p →
q for p in the law of double negation, ¬(¬ p) ≡ p. This allows you to ‘simplify’ the
expression ¬(¬(r → q)) to r → q with confidence that the resulting expression has
the same logical value as the expression you started with. (That’s what it means for
¬(¬(r → q)) and r → q to be logically equivalent.) You can play similar tricks with all
the laws in Figure 2.2. Even more important is the Second Substitution Law, which says
that you can substitute an expression for a logically equivalent expression, wherever it
occurs. Once again, we will accept this as a theorem without trying to prove it here. It
is surprisingly hard to put this law into words:
Theorem 2.2 (Second Substitution Law). Suppose that P and Q are any propositions such
that P ≡ Q. Suppose that R is any compound proposition in which ( P) occurs as a subproposition. Let R′ be the proposition that is obtained by substituting ( Q) for that occurrence of ( P) in
R. Then R ≡ R′ .
Note that in this case, the theorem does not require ( Q) to be substituted for every
occurrence of ( P) in R. You are free to substitute for one, two, or as many occurrences of
( P) as you like, and the result is still logically equivalent to R.
The Second Substitution Law allows us to use the logical equivalence ¬(¬ p) ≡ p
to ‘simplify’ the expression q → (¬(¬ p)) by substituting ¬(¬ p) for p. The resulting
expression, q → p, is logically equivalent to the original q → (¬(¬ p)). Once again, we
have to be careful about parentheses: The fact that p ∨ p ≡ p does not allow us to rewrite
q ∧ p ∨ p ∧ r as q ∧ p ∧ r. The problem is that q ∧ p ∨ p ∧ r means (q ∧ p) ∨ ( p ∧ r ), so that
7 I’ve added parentheses around Q here for technical reasons. Sometimes, the parentheses are necessary to
make sure that Q is evaluated as a whole, so that its final value is used in place of p. As an example of what
can go wrong, consider q ∧ r. If this is substituted literally for p in ¬(¬ p), without parentheses, the result is
¬(¬q ∧ r ). But this expression means ¬((¬q) ∧ r ), which is not equivalent to q ∧ r.

2.2. Boolean Algebra

19

( p ∨ p) is not a subexpression. This again underlines the importance of always writing
parantheses in your propositional formulas.

2
2.2.3 Simplifications
The final piece of algebra in Boolean algebra is the observation that we can chain logical
equivalences together. That is, from P ≡ Q and Q ≡ R, it follows that P ≡ R. This
is really just a consequence of the Second Substitution Law: The equivalence Q ≡ R
allows us to substitute R for Q in the statement P ≡ Q, giving P ≡ R. (Remember that,
by Definition 2.5, logical equivalence is defined in terms of a proposition.) This means
that we can show that two compound propositions are logically equivalent by finding a
chain of logical equivalences that lead from one to the other. For example:

p ∧ ( p → q) ≡ p ∧ (¬ p ∨ q)

≡ ( p ∧ ¬ p) ∨ ( p ∧ q)
≡ F ∨ ( p ∧ q)
≡ ( p ∧ q)

definition of p → q, Theorem 2.2
Distributive Law
Law of Contradiction, Theorem 2.2
Identity Law

Each step in the chain has its own justification. In several cases, a substitution law is
used without stating as much. In the first line, for example, the definition of p → q is
that p → q ≡ ¬ p ∨ q. The Second Substitution Law allows us to substitute (¬ p ∨ q) for
( p → q). In the last line, we implicitly applied the First Substitution Law to the Identity
Law, F ∨ p ≡ p, to obtain F ∨ ( p ∧ q) ≡ ( p ∧ q).
The chain of equivalences in the above example allows us to conclude that p ∧ ( p → q)
is logically equivalent to p ∧ q. This means that if you were to make a truth table for these
two expressions, the truth values in the column for p ∧ ( p → q) would be identical to
those in the column for p ∧ q. We know this without actually making the table. In this
case, the table would only be four lines long and easy enough to make. But Boolean
algebra can be applied in cases where the number of propositional variables is too large
for a truth table to be practical.

20


2

Chapter 2. Logic
Let’s do another example. Recall that a compound proposition is a tautology
if it is true for all possible combinations of truth values of the propositional
variables that it contains. But another way of saying the same thing is that
P is a tautology if P ≡ T. So, we can prove that a compound proposition, P,
is a tautology by finding a chain of logical equivalences leading from P to T.
For example:

(( p ∨ q) ∧ ¬ p) → q
≡ (¬(( p ∨ q) ∧ ¬ p)) ∨ q
≡ (¬( p ∨ q) ∨ ¬(¬ p)) ∨ q
≡ (¬( p ∨ q) ∨ p) ∨ q
≡ (¬( p ∨ q)) ∨ ( p ∨ q)
≡T

definition of →
DeMorgan’s Law, Theorem 2.2
Double Negation, Theorem 2.2
Associative Law for ∨
Law of Excluded Middle

From this chain of equivalences, we can conclude that (( p ∨ q) ∧ ¬ p) → q is
a tautology.

Now, it takes some practice to look at an expression and see which rules can be applied to it; to see (¬( p ∨ q)) ∨ ( p ∨ q) as an application of the law of the excluded middle
for example, you need to mentally substitute ( p ∨ q) for p in the law as it is stated in
Figure 2.2. Often, there are several rules that apply, and there are no definite guidelines
about which one you should try. This is what makes algebra something of an art.

2.2.4

More rules of Boolean algebra

It is certainly not true that all possible rules of Boolean algebra are given in Figure 2.2.
For one thing, there are many rules that are easy consequences of the rules that are listed
there. For example, although the table asserts only that F ∨ p ≡ p, it is also true that
p ∨ F ≡ p. This can be checked directly or by a simple calculation:
p∨F ≡ F∨ p

≡p

Commutative Law
Identity Law as given in the table

Additional rules can be obtained by applying the Commutative Law to other rules in the
table, and we will use such rules freely in the future.
Another sort of easy extension can be applied to the Associative Law, ( p ∨ q) ∨ r ≡
p ∨ (q ∨ r ). The law is stated for the ∨ operator applied to three terms, but it generalizes

2.2. Boolean Algebra

21

to four or more terms. For example

(( p ∨ q) ∨ r ) ∨ s
≡ ( p ∨ q ) ∨ (r ∨ s )
≡ p ∨ (q ∨ (r ∨ s))

2
by the Associative Law for three terms
by the Associative Law for three terms

We will, of course, often write this expression as p ∨ q ∨ r ∨ s, with no parentheses at all,
knowing that wherever we put the parentheses the value is the same.



One other thing that you should keep in mind is that rules can be applied in
either direction. The Distributive Law, for example, allows you to distribute
the p in p ∨ (q ∧ ¬ p) to get ( p ∨ q) ∧ ( p ∨ ¬ p). But it can also be used in
reverse to ‘factor out’ a term, as when you start with (q ∨ ( p → q)) ∧ (q ∨
(q → p)) and factor out the q to get q ∨ (( p → q) ∧ (q → p)).

So far in this section, we have been working with the laws of Boolean algebra without
saying much about what they mean or why they are reasonable. Of course, you can
apply the laws in calculations without understanding them. But if you want to figure
out which calculations to do, you need some understanding. Most of the laws are clear
enough with a little thought. For example, if we already know that q is false, then p ∨ q
will be true when p is true and false when p is false. That is, p ∨ F has the same logical
value as p. But that’s just what the Identity Law for ∨ says. A few of the laws need more
discussion.
The Law of the Excluded Middle, p ∨ ¬ p ≡ T, says that given any proposition p, at
least one of p or ¬ p must be true. Since ¬ p is true exactly when p is false, this is the
same as saying that p must be either true or false. There is no middle ground. The Law
of Contradiction, p ∧ ¬ p ≡ F, says that it is not possible for both p and ¬ p to be true.
Both of these rules are obvious.

22


2

Chapter 2. Logic
There are some who set out to question the law of there being no middle
ground. Already in the 1920’s people like Tarski (who we will meet
later) talked about other forms of logic where another value representing
‘unknown’ or ‘not proven’ also exists. You can also see this in some programming languages where they are referred to as ‘tri-state booleans’.
These so-called non-standard logics have been developed and have also lead to things like ‘fuzzy logic’,
which some consider quite controversial. Lotfi Zadeh is
credited as the first person to refer to this type of logic
as fuzzy logic in his work on ‘fuzzy sets’ in 1965. Zadeh
was later quoted as saying: “Not being afraid to get embroiled in controversy. … That’s part of my character, too.
I can be very stubborn. That’s probably been beneficial
for the development of Fuzzy Logic.”
Source: en.wikipedia.org/wiki/Lotfi_A._Zadeh

The Distributive Laws cannot be called obvious, but a few examples can show that
they are reasonable. Consider the statement, “This card is the ace of spades or clubs.”
Clearly, this is equivalent to “This card is the ace of spaces or this card is the ace of clubs.”
But this is just an example of the first distributive law! For, let a represent the proposition
“This card is an ace”, let s represent “This card is a spade” and let c represent “This card
is a club”. Then “This card is the ace of spades or clubs” can be translated into logic as
a ∧ (s ∨ c), while “This card is the ace of spades or this card is the ace of clubs” becomes
( a ∧ s) ∨ ( a ∧ c). And the distributive law assures us that a ∧ (s ∨ c) ≡ ( a ∧ s) ∨ ( a ∧ c).
The second distributive law tells us, for example, that “This card is either a joker or is the
ten of diamonds” is logically equivalent to “This card is either a joker or a ten, and it is
either a joker or a diamond”. That is, j ∨ (t ∧ d) ≡ ( j ∨ t) ∧ ( j ∨ d). The distributive laws
are powerful tools and you should keep them in mind whenever you are faced with a
mixture of ∧ and ∨ operators.
DeMorgan’s Laws must also be less than obvious, since people often get them wrong.
Fortunately you get to practice them both in Reasoning & Logic, as well as in Computer
Organisation, so you will soon get them right. More importantly perhaps they do also
make sense. When considering ¬( p ∧ q), you should ask yourself, how can “p and q” fail
to be true. It will fail to be true if either p is false or if q is false (or both). That is, ¬( p ∧ q)
is equivalent to (¬ p) ∨ (¬q). Consider the sentence “A raven is large and black.” If a
bird is not large and black, then it is not a raven. But what exactly does it mean to be “not
(large and black)”? How can you tell whether the assertion “not (large and black)” is true
of something? This will be true if it is either not large or not black. (It doesn’t have to be
both—it could be large and white, it could be small and black.) Similarly, for “p or q” to

2.2. Boolean Algebra

23

fail to be true, both p and q must be false. That is, ¬( p ∨ q) is equivalent to (¬ p) ∧ (¬q).
This is DeMorgan’s second law.
Recalling that p → q is equivalent to (¬ p) ∨ q, we can apply DeMorgan’s law to obtain
a formula for the negation an implication:

¬( p → q) ≡ ¬((¬ p) ∨ q)
≡ (¬(¬ p)) ∧ (¬q)
≡ p ∧ ¬q
That is, p → q is false exactly when both p is true and q is false. For example, the negation
of “If you have an ace, you win” is “You have an ace, and you don’t win”. Think of it this
way: if you had an ace and you didn’t win, then the statement “If you have an ace, you
win” was not true.

Exercises
1. Construct truth tables to demonstrate the validity of each of the distributive laws.
2. Construct the following truth tables:
a) Construct truth tables to demonstrate that ¬( p ∧ q) is not logically equivalent to (¬ p) ∧
(¬q).
b) Construct truth tables to demonstrate that ¬( p ∨ q) is not logically equivalent to (¬ p) ∨
(¬q).
c) Construct truth tables to demonstrate the validity of both DeMorgan’s Laws.
3. Construct truth tables to demonstrate that ¬( p → q) is not logically equivalent to any of the
following.
a) (¬ p) → (¬q)
b) (¬ p) → q
c) p → (¬q)
Refer back to this section for a formula that is logically equivalent to ¬( p → q).
4. Is ¬( p ↔ q) logically equivalent to (¬ p) ↔ (¬q)?
5. In the algebra of numbers, there is a distributive law of multiplication over addition: x (y + z) =
xy + xz. What would a distributive law of addition over multiplication look like? Is it a valid
law in the algebra of numbers?
6. The distributive laws given in Figure 2.2 are sometimes called the left distributive laws. The
right distributive laws say that ( p ∨ q) ∧ r ≡ ( p ∧ r ) ∨ (q ∧ r ) and that ( p ∧ q) ∨ r ≡ ( p ∨ r ) ∧
(q ∨ r ). Show that the right distributive laws are also valid laws of Boolean algebra. (Note: In
practice, both the left and the right distributive laws are referred to simply as the distributive
laws, and both can be used freely in proofs.)
7. Show that p ∧ (q ∨ r ∨ s) ≡ ( p ∧ q) ∨ ( p ∧ r ) ∨ ( p ∧ s) for any propositions p, q, r, and s. In
words, we can say that conjunction distributes over a disjunction of three terms. (Recall that
the ∧ operator is called conjunction and ∨ is called disjunction.) Translate into logic and verify

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2

Chapter 2. Logic

the fact that conjunction distributes over a disjunction of four terms. Argue that, in fact, conjunction distributes over a disjunction of any number of terms.
8. There are two additional basic laws of logic, involving the two expression p ∧ F and p ∨ T.
What are the missing laws? Show that your answers are, in fact, laws.
9. For each of the following pairs of propositions, show that the two propositions are logically
equivalent by finding a chain of equivalences from one to the other. State which definition or
law of logic justifies each equivalence in the chain.
a) p ∧ (q ∧ p), p ∧ q
b) (¬ p) → q, p ∨ q
c) ( p ∨ q) ∧ ¬q, p ∧ ¬q
d) p → (q → r ), ( p ∧ q) → r
e) ( p → r ) ∧ (q → r ), ( p ∨ q) → r
f) p → ( p ∧ q), p → q
10. For each of the following compound propositions, find a simpler proposition that is logically
equivalent. Try to find a proposition that is as simple as possible.
a) ( p ∧ q) ∨ ¬q
b) ¬( p ∨ q) ∧ p
c) p → ¬ p
d) ¬ p ∧ ( p ∨ q)
e) (q ∧ p) → q
f) ( p → q) ∧ (¬ p → q)
11. Express the negation of each of the following sentences in natural English:
a) It is sunny and cold.
b) I will have stroopwafel or I will have appeltaart.
c) If today is Tuesday, this is Belgium.
d) If you pass the final exam, you pass the course.
12. Apply one of the laws of logic to each of the following sentences, and rewrite it as an equivalent
sentence. State which law you are applying.
a) I will have coffee and stroopwafel or appeltaart.
b) He has neither talent nor ambition.
c) You can have oliebollen, or you can have oliebollen.
13. Suppose it is simultaneously true that “All lemons are yellow” and “Not all lemons are yellow”.
Derive the conclusion “Unicorns exist”. (If you get stuck, check out en.wikipedia.org/wiki/
Principle_of_explosion.)

2.3

Application: Logic Circuits

As we saw in Chapter 1, computers have a reputation—not always deserved—for being
‘logical’. But fundamentally, deep down, they are made of logic in a very real sense.
The building blocks of computers are logic gates, which are electronic components that
compute the values of simple propositions such as p ∧ q and ¬ p. (Each gate is in turn
built of even smaller electronic components called transistors, but this needn’t concern
us here: see the course Computer Organisation.)



Don’t worry, logic circuits will be examined in Computer Organisation, not in
Reasoning & Logic. They are a good example and application of propositional
logic, and that’s why we’re talking about them in this section. Normal forms
(Section 2.3.4) are definitely on the syllabus, however, so pay attention!

2.3. Application: Logic Circuits

25

2.3.1 Logic gates
A wire in a computer can be in one of two states, which we can think of as being on and
off. These two states can be naturally associated with the Boolean values T and F. When
a computer computes, the multitude of wires inside it are turned on and off in patterns
that are determined by certain rules. The rules involved can be most naturally expressed
in terms of logic. A simple rule might be: “turn wire C on whenever wire A is on and
wire B is on”. This rule can be implemented in hardware as an AND gate. An and gate
is an electronic component with two input wires and one output wire, whose job is to
turn its output on when both of its inputs are on and to turn its output off in any other
case. If we associate ‘on’ with T and ‘off’ with F, and if we give the names A and B to the
inputs of the gate, then the gate computes the value of the logical expression A ∧ B. In
effect, A is a proposition with the meaning “the first input is on”, and B is a proposition
with the meaning “the second input is on”. The and gate functions to ensure that the
output is described by the proposition A ∧ B. That is, the output is on if and only if the
first input is on and the second input is on.
As you hopefully know from Computer Organisation, an OR gate is an electronic component with two inputs and one output which turns its output on if either (or both) of
its inputs is on. If the inputs are given names A and B, then the or gate computes the
logical value of A ∨ B. A NOT gate has one input and one output, and it turns its output
off when the input is on and on when the input is off. If the input is named A, then the
not gate computes the value of ¬ A.



As I mentioned earlier, other textbooks might use different notations to represent a negation. For instance a bar over the variable x̄ or a ∼ symbol. In
digital logic (and thus in your Computer Organisation course) you will also
often find the + symbol to represent an ‘or’ and a · (dot) symbol to represent
an ‘and’.

Other types of logic gates are, of course, possible. Gates could be made to compute
A → B or A ⊕ B, for example. However, any computation that can be performed by
logic gates can be done using only and, or, and not gates, as we will see below. (In
practice, however, nand gates and nor gates, which compute the values of ¬( A ∧ B) and
¬( A ∨ B) respectively, are often used because they are easier to build from transistors
than and and or gates.)

2.3.2 Combining gates to create circuits
The three types of logic gates are represented by standard symbols, as shown in Figure 2.3. Since the inputs and outputs of logic gates are just wires carrying on/off signals,
logic gates can be wired together by connecting outputs from some gates to inputs of
other gates. The result is a logic circuit. An example is also shown in Figure 2.3.

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Chapter 2. Logic

AND gate

2

A

OR gate

NOT gate

output

B
C
Figure 2.3: The standard symbols for the three basic logic gates, and a logic circuit that
computes the value of the logical expression (¬ A) ∧ ( B ∨ ¬( A ∧ C )). The input wires
to each logic gate are on the left, with the output wire on the right. Note that when wires
cross each other in a diagram such as this, the wires don’t actually intersect unless there
is a black circle at the point where they cross.

The logic circuit in the figure has three inputs, labeled A, B, and C. The circuit computes the value of the compound proposition (¬ A) ∧ ( B ∨ ¬( A ∧ C )). That is, when A
represents the proposition “the input wire labeled A is on,” and similarly for B and C,
then the output of the circuit is on if and only if the value of the compound proposition
(¬ A) ∧ ( B ∨ ¬( A ∧ C )) is true.
Given any compound proposition made from the operators ∧, ∨, and ¬, it is possible
to build a logic circuit that computes the value of that proposition. The proposition
itself is a blueprint for the circuit. As noted in Section 2.1, every logical operator that we
have encountered can be expressed in terms of ∧, ∨, and ¬, so in fact every compound
proposition that we know how to write can be computed by a logic circuit.
Given a proposition constructed from ∧, ∨, and ¬ operators, it is easy to build a circuit
to compute it. First, identify the main operator in the proposition—the one whose value
will be computed last. Consider ( A ∨ B) ∧ ¬( A ∧ B). This circuit has two input values,
A and B, which are represented by wires coming into the circuit. The circuit has an
output wire that represents the computed value of the proposition. The main operator
in ( A ∨ B) ∧ ¬( A ∧ B), is the first ∧, which computes the value of the expression as
a whole by combining the values of the subexpressions A ∨ B and ¬( A ∧ B). This ∧
operator corresponds to an and gate in the circuit that computes the final output of the
circuit.
Once the main operator has been identified and represented as a logic gate, you just
have to build circuits to compute the input or inputs to that operator. In the example,
the inputs to the main and gate come from two subcircuits. One subcircuit computes the
value of A ∨ B and the other computes the value of ¬( A ∧ B). Building each subcircuit

2.3. Application: Logic Circuits

27

1. We know that the final
output of the circuit is
computed by an AND gate,
whose inputs are as shown.

(A
(A

B)
B)

2
A
B
(A

B)

(A

B)

(A

2. These inputs, in
turn come from an
OR gate and a NOT
gate, with inputs as
shown.

B)

3. The circuit is completed by adding an AND gate
to compute the input for the NOT gate, and and connecting
the circuit inputs, A and B, to the apropriate gate inputs.
A
B

(A

(A

B)

B)

(A

B)

Figure 2.4: Stages in the construction of a circuit that computes the compound proposition ( A ∨ B) ∧ ¬( A ∧ B).

is a separate problem, but smaller than the problem you started with. Eventually, you’ll
come to a gate whose input comes directly from one of the input wires—A or B in this
case—instead of from a subcircuit.

2.3.3 From circuits to propositions
So, every compound proposition is computed by a logic circuit with one output wire. Is
the reverse true? That is, given a logic circuit with one output, is there a proposition
that expresses the value of the output in terms of the values of the inputs? Not quite.
When you wire together some logic gates to make a circuit, there is nothing to stop you
from introducing feedback loops. A feedback loop occurs when the output from a gate
is connected—possibly through one or more intermediate gates—back to an input of the
same gate. Figure 2.5 shows an example of a circuit with a feedback loop. Feedback loops

28

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Chapter 2. Logic

Figure 2.5: This circuit contains a feedback loop, so it is not a combinatorial logic circuit.
The feedback loop includes the and gate and the or gate on the right. This circuit does
not compute the value of a compound proposition. This circuit does, however, play an
important role in computer memories, since it can be used to store a logical value.

cannot be described by compound propositions, basically because there is no place to
start, no input to associate with a propositional variable. But feedback loops are the only
thing that can go wrong. A logic circuit that does not contain any feedback loops is called
a combinatorial logic circuit. Every combinatorial logic circuit with just one output
computes the value of some compound proposition. The propositional variables in the
compound proposition are just names associated with the input wires of the circuit. (Of
course, if the circuit has more than one output, you can simply use a different proposition
for each output.)
The key to understanding why this is true is to note that each wire in the circuit—not
just the final output wire—represents the value of some proposition. Furthermore, once
you know which proposition is represented by each input wire to a gate, it’s obvious
what proposition is represented by the output: You just combine the input propositions
with the appropriate ∧, ∨, or ¬ operator, depending on what type of gate it is. To find
the proposition associated with the final output, you just have to start from the inputs
and move through the circuit, labeling the output wire of each gate with the proposition
that it represents. Figure 2.6 illustrates this process.

2.3.4

Disjunctive Normal Form

Compound propositions, then, correspond naturally with combinatorial logic circuits.
But we have still not quite settled the question of just how powerful these circuits and
propositions are. We’ve looked at a number of logical operators and noted that they can
all be expressed in terms of ∧, ∨, and ¬. But might there be other operators that cannot
be so expressed? Equivalently, might there be other types of logic gates—possibly with
some large number of inputs—whose computations cannot be duplicated with and, or,
and not gates? Any logical operator or logic gate computes a value for each possible
combination of logical values of its inputs. We could always make a truth table showing
the output for each possible combination of inputs. As it turns out, given any such truth
table, it is possible to find a proposition, containing only the ∧, ∨, and ¬ operators, whose

2.3. Application: Logic Circuits

29

1
A

2
A

B

(A

B)
5
(A

B
C

B)
(B

B

C
3

C)

C

4

Figure 2.6: Finding the proposition whose value is computed by a combinatorial logic
circuit. Each wire in the circuit is labeled with the proposition that it represents. The
numbering of the labels shows one of the orders in which they can be associated with the
wires. The circuit as a whole computes the value of ¬( A ∧ B) ∧ ( B ∨ ¬C ).

value for each combination of inputs is given precisely by that table.
To see why this is true, it is useful to introduce a particular type of compound proposition. Define a simple term to be either a propositional variable or the negation of
a propositional variable. A conjunction of simple terms would then consist of one or
more simple terms put together with ∧ operators. (A ‘conjunction of one simple term’
is just a single simple term by itself. This might not make grammatical sense, but it’s the
way mathematicians think.) Some examples of conjunctions of simple terms would be
p ∧ q, p, ¬q, and p ∧ ¬r ∧ ¬w ∧ s ∧ t. Finally, we can take one or more such conjunctions
and join them into a ‘disjunction of conjunctions of simple terms’. This is the type of
compound proposition we need. We can avoid some redundancy by assuming that no
propositional variable occurs more than once in a single conjunction (since p ∧ p can be
replaced by p, and if p and ¬ p both occur in a conjunction, then the value of the conjuction is false, and it can be eliminated.) We can also assume that the same conjunction
does not occur twice in the disjunction.



Normal forms are part of the syllabus for Reasoning & Logic. These normal
forms, such as Disjunctive Normal Form (this subsection) and Conjunctive
Normal Form (see the exercises), are important in propositional logic. There
are normal forms for other logics, too, such as for predicate logic which we’ll
look at in the next Section 2.4.

Definition 2.6. A compound proposition is said to be in disjunctive normal form, or
DNF, if it is a disjunction of conjunctions of simple terms, and if, furthermore, each propositional variable occurs at most once in each conjunction and each conjunction occurs
at most once in the disjunction.

2

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Chapter 2. Logic

Using p, q, r, s, A, and B as propositional variables, here are a few examples of propositions that are in disjunctive normal form:

( p ∧ q ∧ r ) ∨ ( p ∧ ¬q ∧ r ∧ s) ∨ (¬ p ∧ ¬q)
( p ∧ ¬q)
( A ∧ ¬ B) ∨ (¬ A ∧ B)
p ∨ (¬ p ∧ q) ∨ (¬ p ∧ ¬q ∧ r ) ∨ (¬ p ∧ ¬q ∧ ¬r ∧ w)

2

Propositions in DNF are just what we need to deal with input/output tables of the type
that we have been discussing. Any such table can be computed by a proposition in disjunctive normal form. It follows that it is possible to build a circuit to compute that table
using only and, or, and not gates.
Theorem 2.3. Consider a table that lists a logical output value for every combination of values of
several propositional variables. Assume that at least one of the output values is true. Then there
is a proposition containing those variables such that the value of the proposition for each possible
combination of the values of the variables is precisely the value specified in the table. It is possible
to choose the proposition to be in disjunctive normal form.
Proof. Consider any row in the table for which the output value is T. Form a conjunction
of simple terms as follows: For each variable, p, whose value is T in that row, include p
itself in the conjunction; for each variable, q, whose value is F in the row, include ¬q in
the conjunction. The value of this conjunction is T for the combination of variable values
given in that row of the table, since each of the terms in the conjuction is true for that
combination of variables. Furthermore, for any other possible combination of variable
values, the value of the conjunction will be F, since at least one of the simple terms in
the conjunction will be false.
Take the disjunction of all such conjunctions constructed in this way, for each row in
the table where the output value is true. This disjunction has the value T if and only if
one of the conjunctions that make it up has the value T—and that is precisely when the
output value specified by the table is T. So, this disjunction of conjunctions satisfies the
requirements of the theorem.



This is the first proof of a non-trivial claim that we’ve seen. You will learn
about theorems and proofs, and proof techniques, at the end of this chapter
and in Chapter 3.

As an example, consider the table in Figure 2.7. This table specifies a desired output
value for each possible combination of values for the propositional variables p, q, and
r. Look at the second row of the table, where the output value is true. According to
the proof of the theorem, this row corresponds to the conjunction (¬ p ∧ ¬q ∧ r ). This
conjunction is true when p is false, q is false, and r is true; in all other cases it is false,

2.3. Application: Logic Circuits

31

p

q

r

output

F

F

F

F

F

F

T

T

F

T

F

F

F

T

T

T

T

F

F

F

T

F

T

F

T

T

F

F

(¬ p ∧ ¬q ∧ r )
(¬ p ∧ q ∧ r )

T
p∧q∧r
T T T
Figure 2.7: An input/output table specifying a desired output for each combination
of values of the propositional variables p, q, and r. Each row where the output is T
corresponds to a conjunction, shown next to that row in the table. The disjunction of
these conjunctions is a proposition whose output values are precisely those specified by
the table.

since in any other case at least one of the terms ¬ p, ¬q, or r is false. The other two
rows where the output is true give two more conjunctions. The three conjunctions are
combined to produce the DNF proposition (¬ p ∧ ¬q ∧ r ) ∨ (¬ p ∧ q ∧ r ) ∨ ( p ∧ q ∧ r ). This
proposition computes all the output values specified in the table. Using this proposition
as a blueprint, we get a logic circuit whose outputs match those given in the table.
Now, given any combinatorial logic circuit, there are many other circuits that have the
same input/output behaviour. When two circuits have the same input/output table, the
compound propositions associated with the two circuits are logically equivalent. To put
this another way, propositions that are logically equivalent produce circuits that have the
same input/output behaviour. As a practical matter, we will usually prefer the circuit
that is simpler. The correspondence between circuits and propositions allows us to apply
Boolean algebra to the simplification of circuits.



Our preference for simpler applies to compound propositions, whether or
not they correspond to circuits. We usually prefer the equivalent form of the
proposition that is simpler. Any proposition has an equivalent proposition
in DNF. So when proving a theorem about compound propositions, it is sufficient to consider only DNF propositions. This can make the proof easier to
write.

2

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2

Chapter 2. Logic

For example, consider the DNF proposition corresponding to the table in Figure 2.7.
In (¬ p ∧ ¬q ∧ r ) ∨ (¬ p ∧ q ∧ r ) ∨ ( p ∧ q ∧ r ), we can factor (q ∧ r ) from the last two terms,
giving (¬ p ∧ ¬q ∧ r ) ∨ ((¬ p ∨ p) ∧ (q ∧ r )). Since ¬ p ∨ p ≡ T, and T ∧ Q ≡ Q for any
proposition Q, this can be simplified to (¬ p ∧ ¬q ∧ r ) ∨ (q ∧ r ). Again, we can apply the
distributive law to this to factor out an r, giving ((¬ p ∧ ¬q) ∨ q) ∧ r ). This compound
proposition is logically equivalent to the one we started with, but implementing it in a
circuit requires only five logic gates, instead of the ten required by the original proposition.8
If you start with a circuit instead of a proposition, it is often possible to find the associated proposition, simplify it using Boolean algebra, and use the simplified proposition
to build an equivalent circuit that is simpler than the original. And simplifying a proposition to DNF is often a sensible approach.



2.3.5

One way to simplify propositions is using a Karnaugh-map (or K-map for
short) as you will learn in Computer Organisation. Using a K-map you can
find what they will call a ‘minimal sum of products’. Notice that a sum of
products is just a proposition written in DNF. For the course of Reasoning
& Logic we may ask you to translate propositions to a DNF form. You can
then choose to either do so using rewrite rules, but you are also free to use a
K-map if you prefer. In one of the pencasts of this course we show how both
methods lead to a result in DNF: youtu.be/GwVngCU9eYY.

Binary addition

All this explains nicely the relationship between logic and circuits, but it doesn’t explain
why logic circuits should be used in computers in the first place. Part of the explanation
is found in the fact that computers use binary numbers. A binary number is a string
of zeros and ones. Binary numbers are easy to represent in an electronic device like a
computer: Each position in the number corresponds to a wire. When the wire is on, it
represents one; when the wire is off, it represents zero. When we are thinking in terms
of logic, the same states of the wire represent true and false, but either representation
is just an interpretation of the reality, which is a wire that is on or off. The question is
whether the interpretation is fruitful.
Once wires are thought of as representing zeros and ones, we can build circuits to
do computations with binary numbers. Which computations? Any that we want! If we
know what the answer should be for each combination of inputs, then by Theorem 2.3 we
8 No, I didn’t count wrong. There are eleven logical operators in the original expression, but you can get
by with ten gates in the circuit: Use a single not gate to compute ¬ p, and connect the output of that gate to
two different and gates. Reusing the output of a logic gate is an obvious way to simplify circuits that does not
correspond to any operation on propositions.

2.3. Application: Logic Circuits

33

A

B

C

output

A

B

C

output

0
0
0
0
1
1
1
1

0
0
1
1
0
0
1
1

0
1
0
1
0
1
0
1

0
1
1
0
1
0
0
1

0
0
0
0
1
1
1
1

0
0
1
1
0
0
1
1

0
1
0
1
0
1
0
1

0
0
0
1
0
1
1
1

Figure 2.8: Input/output tables for the addition of three binary digits, A, B, and C.

can build a circuit to compute that answer. Of course, the procedure described in that
theorem is only practical for small circuits, but small circuits can be used as building
blocks to make all the calculating circuits in a computer.
For example, let’s look at binary addition. To add two ordinary, decimal numbers,
you line them up one on top of the other, and add the digits in each column. In each
column, there might also be a carry from the previous column. To add up a column, you
only need to remember a small number of rules, such as 7 + 6 + 1 = 14 and 3 + 5 + 0 = 8.
For binary addition, it’s even easier, since the only digits are 0 and 1. There are only eight
rules:
0 + 0 + 0 = 00

1 + 0 + 0 = 01

0 + 0 + 1 = 01

1 + 0 + 1 = 10

0 + 1 + 0 = 01

1 + 1 + 0 = 10

0 + 1 + 1 = 10

1 + 1 + 1 = 11

Here, I’ve written each sum using two digits. In a multi-column addition, one of these
digits is carried over to the next column. Here, we have a calculation that has three
inputs and two outputs. We can make an input/output table for each of the two outputs.
The tables are shown in Figure 2.8. We know that these tables can be implemented as
combinatorial circuits, so we know that circuits can add binary numbers. To add multidigit binary numbers, we just need one copy of the basic addition circuit for each column
in the sum.

Exercises

2

34

2

Chapter 2. Logic

1. Using only and, or, and not gates, draw circuits that compute the value of each of the propositions A → B, A ↔ B, and A ⊕ B.
2. For each of the following propositions, find a combinatorial logic circuit that computes that
proposition:
a) A ∧ ( B ∨ ¬C )
b) ( p ∧ q) ∧ ¬( p ∧ ¬q)
c) ( p ∨ q ∨ r ) ∧ (¬ p ∨ ¬q ∨ ¬r )
d) ¬( A ∧ ( B ∨ C )) ∨ ( B ∧ ¬ A)
3. Find the compound proposition computed by each of the following circuits:

A
B
C
A
B
C
4. This section describes a method for finding the compound proposition computed by any combinatorial logic circuit. This method fails if you try to apply it to a circuit that contains a feedback loop. What goes wrong? Give an example.
5. Show that every compound proposition which is not a contradiction is equivalent to a proposition in disjunctive normal form. (Note: We can eliminate the restriction that the compound
proposition is not a contradiction by agreeing that “F” counts as a proposition in disjunctive
normal form. F is logically equivalent to any contradiction.)
6. A proposition in conjunctive normal form (CNF) is a conjunction of disjunctions of simple
terms (with the proviso, as in the definition of DNF that a single item counts as a conjunction
or disjunction). Show that every compound proposition which is not a tautology is logically
equivalent to a compound proposition in conjunctive normal form. (Hint: What happens if
you take the negation of a DNF proposition and apply DeMorgan’s Laws?)
7. Use the laws of Boolean algebra to simplify each of the following circuits:

A
A
B

B
C
A
B
C

2.4. Predicate Logic

35

8. Design circuits to implement the input/output tables for addition, as given in Figure 2.8. Try to
make your circuits as simple as possible. (The circuits that are used in real computers for this
purpose are more simplified than the ones you will probably come up with, but the general
approach of using logic to design computer circuits is valid. If you are interested to learn
more about this, the second year variant course Digital Systems describes circuit design in more
detail.)

2.4 Predicate Logic
In propositional logic, we can let p stand for “Roses are red” and q stand for “Violets are
blue”. Then p ∧ q will stand for “Roses are red and violets are blue”. But we lose a lot in
the translation into logic. Since propositional logic only deals with truth values, there’s
nothing we can do with p and q in propositional logic that has anything to do with roses,
violets, or colour. To apply logic to such things, we need predicates. The type of logic
that uses predicates is called predicate logic or, when the emphasis is on manipulating
and reasoning with predicates, predicate calculus.

2.4.1 Predicates
A predicate is a kind of incomplete proposition, which becomes a proposition when it is
applied to some entity (or, as we’ll see later, to several entities). In the proposition “the
rose is red”, the predicate is is red. By itself, ‘is red’ is not a proposition. Think of it as
having an empty slot, that needs to be filled in to make a proposition: “— is red”. In the
proposition “the rose is red”, the slot is filled by the entity “the rose”, but it could just as
well be filled by other entities: “the barn is red”; “the wine is red”; “the banana is red”.
Each of these propositions uses the same predicate, but they are different propositions
and they can have different truth values.
If P is a predicate and a is an entity, then P( a) stands for the proposition that is formed
when P is applied to a. If P represents ‘is red’ and a stands for ‘the rose’, then P( a) is
‘the rose is red’. If M is the predicate ‘is mortal’ and s is ‘Socrates’, then M (s) is the
proposition “Socrates is mortal”.
Now, you might be asking, just what is an entity anyway? I am using the term here
to mean some specific, identifiable thing to which a predicate can be applied. Generally,
it doesn’t make sense to apply a given predicate to every possible entity, but only to
entities in a certain category. For example, it probably doesn’t make sense to apply the
predicate ‘is mortal’ to your living room sofa. This predicate only applies to entities in
the category of living things, since there is no way something can be mortal unless it is
alive. This category is called the domain of discourse for the predicate.9
9 In the language of set theory, which will be introduced in Chapter 4, we would say that a domain of
discourse is a set, U, and a predicate is a function from U to the set of truth values. The definition should be
clear enough without the formal language of set theory, and in fact you should think of this definition—and
many others—as motivation for that language.

2

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Chapter 2. Logic

We are now ready for a formal definition of one-place predicates. A one-place predicate, like all the examples we have seen so far, has a single slot which can be filled in
with one entity:

2
Definition 2.7. A one-place predicate associates a proposition with each entity in some
collection of entities. This collection is called the domain of discourse for the predicate.
If P is a predicate and a is an entity in the domain of discourse for P, then P( a) denotes
the proposition that is associated with a by P. We say that P( a) is the result of applying
P to a.

We can obviously extend this to predicates that can be applied to two or more entities.
In the proposition “John loves Mary”, loves is a two-place predicate. Besides John and
Mary, it could be applied to other pairs of entities: “John loves Jane”, “Bill loves Mary”,
“John loves Bill”, “John loves John”. If Q is a two-place predicate, then Q( a, b) denotes
the proposition that is obtained when Q is applied to the entities a and b. Note that each
of the ‘slots’ in a two-place predicate can have its own domain of discourse. For example,
if Q represents the predicate ‘owns’, then Q( a, b) will only make sense when a is a person
and b is an inanimate object. An example of a three-place predicate is “a gave b to c”, and
a four-place predicate would be “a bought b from c for d euros”. But keep in mind that
not every predicate has to correspond to an English sentence.
When predicates are applied to entities, the results are propositions, and all the operators of propositional logic can be applied to these propositions just as they can to any
propositions. Let R be the predicate ‘is red’, and let L be the two-place predicate ‘loves’.
If a, b, j, m, and b are entities belonging to the appropriate categories, then we can form
compound propositions such as:

R( a) ∧ R(b)
¬ R( a)
L( j, m) ∧ ¬ L(m, j)
L( j, m) → L(b, m)
R( a) ↔ L( j, j)

a is red and b is red
a is not red
j loves m, and m does not love j
if j loves m then b loves m
a is red if and only if j loves j

2.4. Predicate Logic



37

Predicate logic is founded on the ideas developed
by Charles Sanders Peirce (1839–1914), an American philosopher, logician, mathematician, and scientist.10 Many of his contributions to logic were
appreciated only years after he died. He has been
called “the most original and versatile of American
philosophers and America’s greatest logician.” and
“one of the greatest philosophers ever”. As early as
1886 he saw that logical operations could be carried
out by electrical switching circuits; the same idea
was used decades later to produce digital computers, as we saw in Section 2.3.
You can read about his colourful life at the link below.
Source: en.wikipedia.org/wiki/Charles_Sanders_Peirce.

2.4.2 Quantifiers
Let’s go back to the proposition with which we started this section: “Roses are red”. This
sentence is more difficult to handle than it might appear. We still can’t express it properly
in logic. The problem is that this proposition is not saying something about some particular entity. It really says that all roses are red (which happens to be a false statement,
but that’s what it means). Predicates can only be applied to individual entities.
Many other sentences raise similar difficulties: “All persons are mortal.” “Some roses
are red, but no roses are black.” “All maths courses are interesting.” “Every prime number greater than two is odd.” Words like all, no, some, and every are called quantifiers.
We need to be able to express similar concepts in logic.
Suppose that P is a predicate, and we want to express the proposition that P is true
when applied to any entity in the domain of discourse. That is, we want to say “for any
entity x in the domain of discourse, P( x ) is true”. In predicate logic, we write this in
symbols as ∀ x ( P( x )). The ∀ symbol, which looks like an upside-down A, is usually read
‘for all’, so that ∀ x ( P( x )) is read as “for all x, P( x )”. (It is understood that this means for
all x in the domain of discourse for P.) For example, if R is the predicate ‘is red’ and the
domain of discourse consists of all roses, then ∀ x ( R( x )) expresses the proposition “All
roses are red”. Note that the same proposition could be expressed in English as “Every
rose is red” or “Any rose is red”.
Now, suppose we want to say that a predicate, P, is true for some entity in its domain
of discourse. This is expressed in predicate logic as ∃ x ( P( x )). The ∃ symbol, which
looks like a backwards E, is usually read ‘there exists’ ,but a more exact reading would
10 And also on the ideas developed independently around the same time by German philosopher and logician Gottlob Frege, who we’ll see in Section 3.1.

2

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2

Chapter 2. Logic

be ‘there is at least one’. Thus, ∃ x ( P( x )) is read as “There exists an x such that P( x )”,
and it means “there is at least one x in the domain of discourse for P for which P( x ) is
true”. If, once again, R stands for ‘is red’ and the domain of discourse is ‘roses’, then
∃ x ( R( x )) could be expressed in English as “There is a red rose” or “At least one rose is
red” or “Some rose is red”. It might also be expressed as “Some roses are red”, but the
plural is a bit misleading since ∃ x ( R( x )) is true even if there is only one red rose. We
can now give the formal definitions:
Definition 2.8. Suppose that P is a one-place predicate. Then ∀ x ( P( x )) is a proposition,
which is true if and only if P( a) is true for every entity a in the domain of discourse for P.
And ∃ x ( P( x )) is a proposition which is true if and only if there is at least one entity, a, in
the domain of discourse for P for which P( a) is true. The ∀ symbol is called the universal
quantifier, and ∃ is called the existential quantifier.
The x in ∀ x ( P( x )) and ∃ x ( P( x )) is a variable. (More precisely, it is an entity variable,
since its value can only be an entity.) Note that a plain P( x )—without the ∀ x or ∃ x—is
not a proposition. P( x ) is neither true nor false because x is not some particular entity,
but just a placeholder in a slot that can be filled in with an entity. P( x ) would stand
for something like the statement ‘x is red’, which is not really a statement in English at
all. But it becomes a statement when the x is replaced by some particular entity, such as
‘the rose’. Similarly, P( x ) becomes a proposition if some entity a is substituted for the x,
giving P( a).11
An open statement is an expression that contains one or more entity variables, which
becomes a proposition when entities are substituted for the variables. (An open statement has open ‘slots’ that need to be filled in.) P( x ) and “x is red” are examples of
open statements that contain one variable. If L is a two-place predicate and x and y are
variables, then L( x, y) is an open statement containing two variables. An example in
English would be “x loves y”. The variables in an open statement are called free variables. An open statement that contains x as a free variable can be quantified with ∀ x or
∃ x. The variable x is then said to be bound. For example, x is free in P( x ) and is bound
in ∀ x ( P( x )) and ∃ x ( P( x )). The free variable y in L( x, y) becomes bound in ∀y( L( x, y))
and in ∃y( L( x, y)).
Note that ∀y( L( x, y)) is still an open statement, since it contains x as a free variable.
to apply
the quantifier ∀ x or ∃ x to ∀y( L( x, y)), giving
( Therefore,
) it is possible
(
)
∀ x ∀y( L( x, y)) and ∃ x ∀y( L( x, y)) . Since all the variables are bound in these expressions, they are propositions. If L( x, (y) represents) ‘x loves y’, then ∀y( L( x, y)) is something like “x loves everyone”, and ∃ x ∀y( L( x, y)) is the proposition, “There is someone
who loves everyone”. Of course, we could also have started
( with ∃ x ()L( x, y)): “There
is someone who loves y”. Applying ∀y to this gives ∀y ∃ x ( L( x, y)) , which means
“For every person, there is someone who loves that person”. Note in particular that
11 There is certainly room for confusion about names here. In this discussion, x is a variable and a is an entity.
But that’s only because I said so. Any letter could be used in either role, and you have to pay attention to the
context to figure out what is going on. Usually, x, y, and z will be variables.

2.4. Predicate Logic

39

(
)
(
)
∃ x ∀y( L( x, y)) and ∀y ∃ x ( L( x, y)) do not mean the same thing. Altogether, there are
eight different propositions that can be obtained from L( x, y) by applying quantifiers,
with six distinct meanings among them.



From now on, I will leave out parentheses when there is no ambiguity. For
example,
I will write
∀ x P( x ) instead of ∀ x ( P( x )) and ∃ x ∃y L( x, y) instead of
(
)
∃ x ∃y( L( x, y)) . Make sure though that when you leave out the parentheses
you do so only when no ambiguity exists. In one of the problems of this
chapter, you will see an example of two very similar statements where the
parentheses do change the meaning significantly!
Further, I will sometimes give predicates and entities names that are complete words instead of just letters, as in Red( x ) and Loves( john, mary). This
might help to make examples more readable.

2.4.3 Operators
In predicate logic, the operators and laws of Boolean algebra still apply. For example, if P
and Q are one-place predicates and a is an entity in the domain of discourse, then P( a) →
Q( a) is a proposition, and it is logically equivalent to ¬ P( a) ∨ Q( a). Furthermore, if
x is a variable, then P( x ) → Q( x ) is an open statement, and ∀ x ( P( x ) → Q( x )) is a
proposition. So are P( a) ∧ (∃ x Q( x )) and (∀ x P( x )) → (∃ xP( x )). Obviously, predicate
logic can be very expressive. Unfortunately, the translation between predicate logic and
English sentences is not always obvious.



One of the commonly-made mistakes in predicate logic is the difference in
translation between statements like: “All humans are mortal” and “There is
a human that is mortal”. We discuss the difference in translation of these
statements in one of our pencasts: youtu.be/FxR07brGvkw.

Let’s look one more time at the proposition “Roses are red”. If the domain of discourse
consists of roses, this translates into predicate logic as ∀ x Red( x ). However, the sentence
makes more sense if the domain of discourse is larger—for example if it consists of all
flowers. Then “Roses are red” has to be read as “All flowers which are roses are red”, or
“For any flower,( if that flower is a rose,
then it is red”. The last form translates directly
)
into logic as ∀ x Rose( x ) → Red( x ) . Suppose we want to say that all red roses are pretty.
The phrase ‘red rose’ is saying both that the flower is a rose and that it is red, and it must
be translated as( a conjunction, Rose( x ) ∧ Red( x )). So, “All red roses are pretty” can be
rendered as ∀ x ( Rose( x ) ∧ Red( x )) → Pretty( x ) .

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
2

Chapter 2. Logic
Here are a few more examples of translations from predicate logic to English.
Let H ( x ) represent ‘x is happy’, let C (y) represent ‘y is a computer’, and let
O( x, y) represent ‘x owns y’. Then we have the following translations:
(
)
• Jack owns a computer: ∃ x O( jack, x ) ∧ C ( x ) . (That is, there is at least
one thing such that Jack owns that thing and that thing is a computer.)
(
)
• Everything Jack owns is a computer: ∀ x O( jack, x ) → C ( x ) .
• If Jack owns
a computer, then he’s
(
) happy:
∃y(O( jack, y) ∧ C (y)) → H ( jack).
• Everyone who
( ( owns a computer)is happy:) )
∀ x ∃y(O( x, y) ∧ C (y) → H ( x ) .
(
)
• Everyone owns a computer: ∀ x ∃y C (y) ∧ O( x, y) . (Note that this
allows( each person to
) own a different computer. The proposition
∃y ∀ x C (y) ∧ O( x, y) would mean that there is a single computer
which is owned by everyone.)
• Everyone is happy: ∀ xH ( x ).
• Everyone is unhappy: ∀ x (¬ H ( x )).
• Someone is unhappy: ∃ x (¬ H ( x )).

(
)
• At least two people are happy: ∃ x ∃y H ( x ) ∧ H (y) ∧ ( x ̸= y) . (The
stipulation that x ̸= y is necessary because two different variables can
refer to the same entity. The proposition ∃ x ∃y( H ( x ) ∧ H (y)) is true
even if there is only one happy person.)
• There is (exactly one
person:
) happy
(
)
∃ xH ( x ) ) ∧ ∀y∀z(( H (y) ∧ H (z)) → (y = z)) . (The first
part of this conjunction says that there is at least one happy person. The
second part says that if y and z are both happy people, then they are
actually the same person. That is, it’s not possible to find two different
people who are happy. The statement can be simplified a little however,
to get: ∃ x ( H ( x ) ∧ ∀y( H (y) → ( x = y))) Do you see why this works as
well?)

2.4.4

Tarski’s world and formal structures

To help you reason about sets of predicate logic statements, or even arguments expressed
in predicate logic, we often use a ‘mathematical structure’. For some of these structures
a visualisation in the form of Tarski’s world can sometimes be useful.

2.4. Predicate Logic

41

e

d
c

a

b

Figure 2.9: An instance of a Tarski World.



What is truth? In 1933, Polish mathematician Alfred
Tarski (1901–1983) published a very long paper in Polish (titled Pojȩcie prawdy w jȩzykach nauk dedukcyjnych),
setting out a mathematical definition of truth for formal
languages. “Along with his contemporary, Kurt Gödel
[who we’ll see in Chapter 4], he changed the face of
logic in the twentieth century, especially through his
work on the concept of truth and the theory of models.”
Source: en.wikipedia.org/wiki/Alfred_Tarski.

In Tarski’s world, it is possible to describe situations using formulas whose truth can
be evaluated, which are expressed in a first-order language that uses predicates such as
Rightof( x, y), which means that x is situated—somewhere, not necessarily directly—to
the right of y, or Blue( x ), which means that x is blue. In the world in Figure 2.9, for
instance, the formula ∀ x (Triangle( x ) → Blue( x )) holds, since all triangles are blue, but
the converse of this formula, ∀ x (Blue( x ) → Triangle( x )), does not hold, since object c is
blue but not a triangle.
Such an instance of Tarski world can be more formally described as a ‘mathematical
structure’ (which we refer to as a formal structure occasionally). These structures allow
us to evaluate statements in predicate logic as being true or false. To formalise a structure,
we need to describe two things: the domain of discourse D of the structure and for all of
the predicates, for which objects of the domain they are true. We do so using set-notation
which we discuss in more depth in Chapter 4. The formal description of the structure S
depicted in Figure 2.9 is:
• D = { a, b, c, d, e}

• SquareS = { a}

• BlueS = {b, c}

• TriangleS = {b}

• GrayS = { a, d}

• CircleS = {c, d, e}

• RedS = {e}

• RightOf S = {(b, a), (c, a), (d, a), (e, a),

2

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2

Chapter 2. Logic

(b, c), (d, c), (b, e), (c, e), (d, e)}

• BelowOf S = {( a, c), ( a, d), ( a, e), (b, c),
(b, d), (b, e), (c, d), (c, e)}

• LeftOf S = {( a, b), (c, b), (e, b), ( a, c),
(e, c), ( a, d), (c, d), (e, d), ( a, e)}

• AboveOf S = {(c, a), (c, b), (d, a), (d, b),
(d, c), (e, a), (e, b), (e, c)}

Notice that for the one-place predicates we have a set of objects for which this predicate is true (e.g., only b and c are blue) and such a set is denoted using ‘{’ and ‘}’ symbols,
called ‘curly braces’ or just ‘braces’.12 For the two-place predicates we have a set of tuples
that are denoted using ‘(’ and ‘)’ symbols, called ‘parentheses’ or ‘round brackets’. In
this case, for instance, the fact that ( a, b) is in the set LeftOf S means that LeftOf ( a, b) is
true for this structure, i.e., a is left of b.
Such formal structures can also be defined to disprove arguments written in predicate
logic, as we will see in Section 2.5.3.

2.4.5

Logical equivalence

To calculate in predicate logic, we need a notion of logical equivalence. Clearly, there
are pairs of propositions in predicate logic that mean the same thing. Consider the propositions ¬(∀ xH ( x )) and ∃ x (¬ H ( x )), where H ( x ) represents ‘x is happy’. The first of
these propositions means “Not everyone is happy”, and the second means “Someone
is not happy”. These statements have the same truth value: If not everyone is happy,
then someone is unhappy and vice versa. But logical equivalence is much stronger than
just having the same truth value. In propositional logic, logical equivalence is defined
in terms of propositional variables: two compound propositions are logically equivalent
if they have the same truth values for all possible truth values of the propositional variables they contain. In predicate logic, two formulas are logically equivalent if they have
the same truth value for all possible predicates.
Consider ¬(∀ xP( x )) and ∃ x (¬ P( x )). These formulas make sense for any predicate
P, and for any predicate P they have the same truth value. Unfortunately, we can’t—
as we did in propositional logic—just check this fact with a truth table: there are no
subpropositions, connected by ∧, ∨, etc, out of which to build a table. So, let’s reason it
out: To say ¬(∀ xP( x )) is true is just to say that it is not the case that P( x ) is true for all
possible entities x. So, there must be some entity a for which P( a) is false. Since P( a) is
false, ¬ P( a) is true. But saying that there is an a for which ¬ P( a) is true is just saying that
∃ x (¬ P( x )) is true. So, the truth of ¬(∀ xP( x )) implies the truth of ∃ x (¬ P( x )). On the
other hand, if ¬(∀ xP( x )) is false, then ∀ xP( x ) is true. Since P( x ) is true for every x, ¬ P( x )
is false for every x; that is, there is no entity a for which the statement ¬ P( a) is true. But
this just means that the statement ∃ x (¬ P( x )) is false. In any case, then, the truth values
of ¬(∀ xP( x )) and ∃ x (¬ P( x )) are the same. Since this is true for any predicate P, we will
say that these two formulas are logically equivalent and write ¬(∀ xP( x )) ≡ ∃ x (¬ P( x )).
12 See

Chapter 4.

2.4. Predicate Logic

43

¬ (∀ xP( x )) ≡ ∃ x (¬ P( x ))
¬ (∃ xP( x )) ≡ ∀ x (¬ P( x ))
∀ x ∀yQ( x, y) ≡ ∀y∀ xQ( x, y)
∃ x ∃yQ( x, y) ≡ ∃y∃ xQ( x, y)
Figure 2.10: Four important rules of predicate logic. P can be any one-place predicate,
and Q can be any two-place predicate. The first two rules are called DeMorgan’s Laws
for predicate logic.

A similar argument would show that ¬(∃ xP( x )) ≡ ∀ x (¬ P( x )). These two equivalences, which explicate the relation between negation and quantification, are known as
DeMorgan’s Laws for predicate logic. (They are closely related to DeMorgan’s Laws for
propositional logic; see the exercises.) These laws can be used to help simplify expressions. For example,

¬ ∀y( R(y) ∨ Q(y)) ≡ ∃y(¬( R(y) ∨ Q(y)))
≡ ∃y((¬ R(y)) ∧ (¬ Q(y))

It might not be clear exactly why this qualifies as a ‘simplification’, but it’s generally considered simpler to have the negation operator applied to basic propositions such as R(y),
rather than to quantified expressions such as ∀y( R(y) ∨ Q(y)). For a more complicated
example:
(
)
¬ ∃ x P( x ) ∧ (∀y( Q(y) → Q( x )))
( (
)
≡ ∀ x ¬ P( x ) ∧ (∀y( Q(y) → Q( x )))
(
)
≡ ∀ x (¬ P( x )) ∨ (¬∀y( Q(y) → Q( x )))
(
)
≡ ∀ x (¬ P( x )) ∨ (∃y(¬( Q(y) → Q( x ))))
(
)
≡ ∀ x (¬ P( x )) ∨ (∃y(¬(¬ Q(y) ∨ Q( x ))))
(
)
≡ ∀ x (¬ P( x )) ∨ (∃y(¬¬ Q(y) ∧ ¬ Q( x )))
(
)
≡ ∀ x (¬ P( x )) ∨ (∃y( Q(y) ∧ ¬ Q( x )))

DeMorgan’s Laws are listed in Figure 2.10 along with two other laws of predicate logic.
The other laws allow you to interchange the order of the variables when two quantifiers
of the same type (both ∃ or ∀) occur together.

2

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
2

Chapter 2. Logic
Notice however that we may not change the order of quantifiers that are not
the same! For instance: ∀ x ∃y( R( x, y)) ̸≡ ∃y∀ x ( R( x, y)). If you are not convinced about this, try to draw up a Tarski’s world that shows this unequivalence.

To define logical equivalence in predicate logic more formally, we need to talk about
formulas that contain predicate variables, that is, variables that act as place-holders for
arbitrary predicates in the same way that propositional variables are place-holders for
propositions and entity variables are place-holders for entities. With this in mind, we
can define logical equivalence and the closely related concept of tautology for predicate
logic. We’ll see that these are crucial pieces of writing proofs.
Definition 2.9. Let P be a formula of predicate logic which contains one or more predicate variables. P is said to be a tautology if it is true whenever all the predicate variables
that it contains are replaced by actual predicates. Two formulas P and Q are said to be
logically equivalent if P ↔ Q is a tautology, that is if P and Q always have the same
truth value when the predicate variables they contain are replaced by actual predicates.
The notation P ≡ Q asserts that P is logically equivalent to Q.

Exercises
1. Simplify each of the following propositions. In your answer, the ¬ operator should be applied
only to individual predicates.
a) ¬ ∀ x (¬ P( x ))
b) ¬(∃ x ( P( x ) ∧ Q( x ))
)
c) ¬ ∀z( P(z) → Q(z))
d) ¬ (∀ xP( x )) ∧ (∀yQ(y))
e) ¬ ∀ x ∃yP( x, y)
f) ¬(∃ x ( R( x ) ∧ ∀yS( x, y))
)
g) ¬ ∃y( P(y) ↔ Q(y))
h) ¬ ∀ x ( P( x ) → (∃yQ( x, y)))
2. Give a careful argument to show that the second of DeMorgan’s laws for predicate calculus,
¬(∀ xP( x )) ≡ ∃ x (¬ P( x )), is valid.
3. Find the negation of each of the following propositions. Simplify the result; in your answer,
the ¬ operator should be applied only to individual predicates.
a) ∃n(∀sC (s, n))
b) ∃n(∀s( L(s, n) → P(s)))
c) ∃n(∀s( L(s, n) → (∃ x ∃y∃zQ( x, y, z)))).
d) ∃n(∀s( L(s, n) → (∃ x ∃y∃z(s = xyz ∧ R( x, y) ∧ T (y) ∧ U ( x, y, z)))).
4. Suppose that the domain of discourse for a predicate P contains only two entities. Show that
∀ xP( x ) is equivalent to a conjunction of two simple propositions, and ∃ xP( x ) is equivalent to a
disjunction. Show that in this case, DeMorgan’s Laws for propositional logic and DeMorgan’s
Laws for predicate logic actually say exactly the same thing. Extend the results to a domain of
discourse that contains exactly three entities.
5. Let H ( x ) stand for “x is happy”, where the domain of discourse consists of people. Express
the proposition “There are exactly three happy people” in predicate logic.

2.4. Predicate Logic

45

6. What is the difference between the following two statements?
∃ xRed( x ) ∧ ∃ xSquare( x ) and ∃ x (Red( x ) ∧ Square( x ))
7. Draw a Tarski world for the last exercise.
8. Express Johan Cruyff’s statement “There is only one ball, so you need to have it” in predicate
logic.
9. Let T ( x, y) stand for ‘x has taken y’, where the domain of discourse for x consists of students
and the domain of discourse for y consists of CS courses (at TUDelft). Translate each of the
following propositions into an unambiguous English sentence:
a) ∀ x ∀y T ( x, y)
b) ∀ x ∃y T ( x, y)
c) ∀y ∃ x T ( x, y)
d) ∃ x ∃y T ( x, y)
e) ∃ x ∀y T ( x, y)
f) ∃y ∀ x T ( x, y)
10. Let F ( x, t) stand for “You can fool person x at time t.” Translate the following sentence into
predicate logic: “You can fool some of the people all of the time, and you can fool all of the
people some of the time, but you can’t fool all of the people all of the time.”
11. Translate each of the following sentences into a proposition using predicate logic. Make up
any predicates you need. State what each predicate means and what its domain of discourse
is.
a) All crows are black.
b) Any white bird is not a crow.
c) Not all politicians are honest.
d) All purple elephants have green feet.
e) There is no one who does not like pizza.
f) Anyone who passes the final exam will pass the course.13
g) If x is any positive number, then there is a number y such that y2 = x.
12. Consider the following description of a Tarski World. Does an instance of a Tarski World exist
with these properties? If so, give one with a domain of at most 5 elements. If no such instance
exists, explain why not.
• ∀ x (Circle( x ) → ¬Blue( x ))
• ∃ x (Circle( x )) ∧ ∃ x (Blue( x ))
• RightOf ( a, b)
• LeftOf ( a, b) ∨ Square(c)
13. The sentence “Someone has the answer to every question” is ambiguous. Give two translations
of this sentence into predicate logic, and explain the difference in meaning.
14. The sentence “Jane is looking for a dog” is ambiguous. One meaning is that there is some
particular dog—maybe the one she lost—that Jane is looking for. The other meaning is that
Jane is looking for any old dog—maybe because she wants to buy one. Express the first meaning in predicate logic. Explain why the second meaning is not expressed by ∀ x ( Dog( x ) →
LooksFor ( jane, x )). In fact, the second meaning cannot be expressed in predicate logic. Philosophers of language spend a lot of time thinking about things like this. They are especially
fond of the sentence “Jane is looking for a unicorn”, which is not ambiguous when applied to
the real world. Why is that?
13 This

is not true for Reasoning & Logic: see the syllabus.

2

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2.5

2

Chapter 2. Logic

Deduction

Logic can be applied to draw conclusions from a set of premises. A premise is just a
proposition that is known to be true or that has been accepted to be true for the sake
of argument, and a conclusion is a proposition that can be deduced logically from the
premises. The idea is that if you believe that the premises are true, then logic forces you
to accept that the conclusion is true. An argument is a claim that a certain conclusion
follows from a given set of premises. Here is an argument laid out in a traditional format:
If today is Tuesday, then this is Belgium
Today is Tuesday
∴ This is Belgium
The premises of the argument are shown above the line, and the conclusion below. The
symbol ∴ is read ‘therefore’. The claim is that the conclusion, “This is Belgium”, can be
deduced logically from the two premises, “If today is Tuesday, then this is Belgium” and
“Today is Tuesday”. In fact, this claim is true. Logic forces you to accept this argument.
Why is that?

2.5.1

Arguments

Let p stand for the proposition “Today is Tuesday”, and let q stand for the proposition
“This is Belgium”. Then the above argument has the form
p→q
p
∴ q
Now, for any propositions p and q—not just the ones in this particular argument—if
p → q is true and p is true, then q must also be true. This is easy to check in a truth table:
p

q

p→q

0
0
1
1

0
1
0
1

1
1
0
1

The only case where both p → q and p are true is on the last line of the table, and in this
case, q is also true. If you believe p → q and p, you have no logical choice but to believe
q. This applies no matter what p and q represent. For example, if you believe “If Jill is
breathing, then Jill pays taxes”, and you believe that “Jill is breathing”, logic forces you
to believe that “Jill pays taxes”. Note that we can’t say for sure that the conclusion is true,
only that if the premises are true, then the conclusion must be true.

2.5. Deduction

47

(
)
This fact can be rephrased by saying that ( p → q) ∧ p → q is a tautology. More
generally, for any compound propositions P and Q, saying “P → Q is a tautology” is
the same as saying that “in all cases where P is true, Q is also true”.14 We will use the
notation P =⇒ Q to mean that P → Q is a tautology. Think of P as being the premise
of an argument or the conjunction of several premises. To say P =⇒ Q is to say that Q
follows logically from P. We will use the same notation in both propositional logic and
predicate logic. (Note that the relation of =⇒ to → is the same as the relation of ≡ to
↔.)
Definition 2.10. Let P and Q be any formulas in either propositional logic or predicate
logic. The notation P =⇒ Q is used to mean that P → Q is a tautology. That is, in all
cases where P is true, Q is also true. We then say that Q can be logically deduced from P
or that P logically implies Q.
An argument in which the conclusion follows logically from the premises is said to
be a valid argument. To test whether an argument is valid, you have to replace the particular propositions or predicates that it contains with variables, and then test whether
the conjunction of the premises logically implies the conclusion. We have seen that any
argument of the form
p→q
p
∴ q

(
)
is valid, since ( p → q) ∧ p → q is a tautology. This rule of deduction is called modus
ponens. It plays a central role in logic. Another, closely related rule is modus tollens,
which applies to arguments of the form
p→q
¬q
∴ ¬p

(
)
To verify that this is a valid argument, just check that ( p → q) ∧ ¬q =⇒ ¬ p, that is,
(
)
that ( p → q) ∧ ¬q → ¬ p is a tautology. As an example, the following argument has
the form of modus tollens and is therefore a valid argument:
If Feyenoord is a great team, then I’m the king of the Netherlands
I am not the king of the Netherlands
∴ Feyenoord is not a great team
You might remember this argument from page 10. You should note carefully that the
validity of this argument has nothing to do with whether or not Feyenoord can play
14 Here, “in all cases” means for all combinations of truth values of the propositional variables in P and Q,
i.e., in every situation. Saying P → Q is a tautology means it is true in all cases. But by definition of →, it is
automatically true in cases where P is false. In cases where P is true, P → Q will be true if and only if Q is true.

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Chapter 2. Logic

football well. The argument forces you to accept the conclusion only if you accept the
premises. You can logically believe that the conclusion is false, as long as you believe
that at least one of the premises is false.15
Another named rule of deduction is the Law of Syllogism, which has the form

2

p→q
q→r
∴ p→r
For example:
If you study hard, you do well in school
If you do well in school, you get a good job
∴ If you study hard, you get a good job
There are many other rules. Here are a few that might prove useful. Some of them
might look trivial, but don’t underestimate the power of a simple rule when it is combined with other rules.
p∨q
¬p
∴ q

p
q
∴ p∧q

p∧q
∴ p

p
∴ p∨q

Logical deduction is related to logical equivalence. We defined P and Q to be logically
equivalent if P ↔ Q is a tautology. Since P ↔ Q is equivalent to (P → Q) ∧ (Q → P),
we see that P ≡ Q if and only if both Q =⇒ P and P =⇒ Q. Thus, we can show that
two statements are logically equivalent if we can show that each of them can be logically
deduced from the other. Also, we get a lot of rules about logical deduction for free—two
rules of deduction for each logical equivalence we know. For example, since ¬( p ∧ q) ≡
(¬ p ∨ ¬q), we get that ¬( p ∧ q) =⇒ (¬ p ∨ ¬q). For example, if we know “It is not both
sunny and warm”, then we can logically deduce “Either it’s not sunny or it’s not warm.”
(And vice versa.)

2.5.2

Valid arguments and proofs

In general, arguments are more complicated that those we’ve considered so far. Here,
for example, is an argument that has five premises:

( p ∧ r) → s
q→p
t→r
q
t
∴ s
15 Unless the conclusion is a tautology. If that’s the case, then even when a premise is false the conclusion
will still be true. You do always know that if the conclusion is false then at least one of the premises is false.

2.5. Deduction

49

Is this argument valid? Of course, you could use a truth table to check whether the
conjunction of the premises logically implies the conclusion. But with five propositional
variables, the table would have 32 lines, and the size of the table grows quickly when
more propositional variables are used. So, in general, truth tables are not practical when
we have a large number of variables.



For a relatively small number of variables (say three or fewer) a truth table
can be a rather efficient method to test validity of an argument. In one of
the pencast of this course we show how you can use truth tables to test for
validity as well as how you can use them to find counterexamples for invalid
arguments: youtu.be/lSZS3qbA88o

Fortunately, there is another way to proceed, based on the fact that it is possible to
chain several logical deductions together. That is, if P =⇒ Q and Q =⇒ R, it follows that
P =⇒ R. This means we can demonstrate the validity of an argument by deducing the
conclusion from the premises in a sequence of steps. These steps can be presented in the
form of a proof:
Definition 2.11. A formal proof that an argument is valid consists of a sequence of propositions such that the last proposition in the sequence is the conclusion of the argument,
and every proposition in the sequence is either a premise of the argument or follows by
logical deduction from propositions that precede it in the list.
The existence of such a proof shows that the conclusion follows logically from the
premises, and therefore that the argument is valid. Here is a formal proof that the argument given above is valid. The propositions in the proof are numbered, and each
proposition has a justification.
Proof.
1.
2.
3.
4.
5.
6.
7.
8.
9.

q→p
q
p
t→r
t
r
p∧r
( p ∧ r) → s
s

premise
premise
from 1 and 2 (modus ponens)
premise
premise
from 4 and 5 (modus ponens)
from 3 and 6
premise
from 7 and 8 (modus ponens)

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Chapter 2. Logic


2

Once a formal proof has been constructed, it is convincing. Unfortunately,
it’s not necessarily easy to come up with the proof. Usually, the best method
is a combination of working forward (“Here’s what I know, what can I deduce from that?”) and working backwards (“Here’s what I need to prove,
what other things would imply that?”). For this proof, I might have thought:
I want to prove s. I know that p ∧ r implies s, so if I can prove p ∧ r, I’m okay.
But to prove p ∧ r, it’ll be enough to prove p and r separately….

Of course, not every argument is valid, so the question also arises, how can we show
that an argument is invalid? Let’s assume that the argument has been put into general
form, with all the specific propositions replaced by propositional variables. The argument is valid if in all cases where all the premises are true, the conclusion is also true.
The argument is invalid if there is even one case where all the premises are true and
the conclusion is false. We can prove that an argument is invalid by finding an assignment of truth values to the propositional variables which makes all the premises true but
makes the conclusion false. We call such an assignment a counterexample . To disprove
the validity of an argument you should always provide a counterexample. This holds in
propositional logic, predicate logic, and any other type of argument you may be asked
to disprove.
For example, consider an argument of the form:
p→q
q → ( p ∧ r)
r
∴ p
In the case where p is false, q is false, and r is true, the three premises of this argument
are all true, but the conclusion is false. This counterexample shows that the argument is
invalid.
To apply all this to arguments stated in English, we have to introduce propositional
variables to represent all the propositions in the argument. For example, consider:
John will be at the party if Mary is there and Bill is not there. Mary will be at
the party if it’s on Friday or Saturday. If Bill is at the party, Tom will be there.
Tom won’t be at the party if it’s on Friday. The party is on Friday. Therefore,
John will be at the party.
Let j stand for “John will be at the party”, m for “Mary will be there”, b for “Bill will be
there”, t for “Tom will be there”, f for “The party is on Friday”, and s for “The party is
on Saturday”. Then this argument has the form

2.5. Deduction

51

(m ∧ ¬b) → j
( f ∨ s) → m
b→t
f → ¬t
f
∴ j
This is a valid argument, as the following proof shows:
Proof.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.



f → ¬t
f
¬t
b→t
¬b
f ∨s
( f ∨ s) → m
m
m ∧ ¬b
(m ∧ ¬b) → j
j

premise
premise
from 1 and 2 (modus ponens)
premise
from 4 and 3 (modus tollens)
from 2
premise
from 6 and 7 (modus ponens)
from 8 and 5
premise
from 10 and 9 (modus ponens)

You may have noticed that we start our proofs with the word ‘proof’ and end
it with a little square. This is done to illustrate clearly where our proof starts
and ends. Historically different symbols and expressions have been used to
indicate that a proof is done. You may have heard of the abbreviation Q.E.D.
for instance for ‘Quod Erat Demonstrandum’, which translates to: ‘what was
to be shown’. Even in ancient Greece a Greek version of Q.E.D. was used by
Greek mathematicians like Euclid. You are free to choose between Q.E.D.
and the open square, so long as you remember that no proof is complete if
it does not have either one of them.

2.5.3 Proofs in predicate logic
So far in this section, we have been working mostly with propositional logic. But the
definitions of valid argument and logical deduction apply to predicate logic as well.
One of the most basic rules of deduction in predicate logic says that (∀ xP( x )) =⇒
P( a) for any entity a in the domain of discourse of the predicate P. That is, if a predicate is

2

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Chapter 2. Logic

true of all entities, then it is true of any given particular entity. This rule can be combined
with rules of deduction for propositional logic to give the following valid arguments:

∀ x ( P( x ) → Q( x ))
P( a)
∴ Q( a)

2

∀ x ( P( x ) → Q( x ))
¬ Q( a)
∴ ¬ P( a)

These valid arguments go by the names of modus ponens and modus tollens for predicate
logic. Note that from the premise ∀ x ( P( x ) → Q( x )) we can deduce P( a) → Q( a). From
this and from the premise that P( a), we can deduce Q( a) by modus ponens. So the first
argument above is valid. The second argument is similar, using modus tollens.
The most famous logical deduction of them all is an application of modus ponens for
predicate logic:
All humans are mortal
Socrates is human
∴ Socrates is mortal
This has the form of modus ponens with P( x ) standing for “x is human”, Q( x ) standing
for “x is mortal”, and a standing for the noted entity, Socrates.
To disprove validity of arguments in predicate logic, you again need to provide a
counterexample. These are most easily given in the form of a mathematical structure.
Consider for instance the following argument:

∃ xP( x )
∀ x ( P( x ) → Q( x ))
∴ ∀ xQ( x )
This argument is not valid and we can prove that using the following structure A.
• D = { a, b}
• PA = { a}
• QA = { a}
As you can see, the first premise is true. There is an x such that P( x ) holds, namely
x = a. The second premis is also true, as for all x for which P( x ) holds (so only x = a),
Q( x ) also holds (and indeed Q( a)) holds. However the conclusion is false, as Q(b) does
not hold, so the Q( x ) does not hold for all x.
There is a lot more to say about logical deduction and proof in predicate logic, and
we’ll spend the whole of the next chapter on the subject.

2.5. Deduction

53

Exercises
1. Verify the validity of modus tollens and the Law of Syllogism.
2. Each of the following is a valid rule of deduction. For each one, give an example of a valid
argument in English that uses that rule.
p∨q
¬p
∴ q

p
q
∴ p∧q

p∧q
∴ p

p
∴ p∨q

3. There are two notorious invalid arguments that look deceptively like modus ponens and modus
tollens:
p→q
q
∴ p

p→q
¬p
∴ ¬q

Show that each of these arguments is invalid. Give an English example that uses each of these
arguments.
4. Decide whether each of the following arguments is valid. If it is valid, give a formal proof.
If it is invalid, show that it is invalid by finding an appropriate assignment of truth values to
propositional variables.
a) p → q
q→s
s
∴ p

b) p ∧ q
q → (r ∨ s )
¬r
∴ s

c) p ∨ q
q → (r ∧ s )
¬p
∴ s

d) (¬ p) → t
q→s
r→q
¬(q ∨ t)
∴ p

e) p
s→r
q∨r
q → ¬p
∴ ¬s

f) q → t
p → (t → s)
p
∴ q→s

5. For each of the following English arguments, express the argument in terms of propositional
logic and determine whether the argument is valid or invalid.
a) If it is Sunday, it rains or snows. Today, it is Sunday and it’s not raining. Therefore, it
must be snowing.
b) If there is herring on the pizza, Jack won’t eat it. If Jack doesn’t eat pizza, he gets angry.
Jack is angry. Therefore, there was herring on the pizza.
c) At 8:00, Jane studies in the library or works at home. It’s 8:00 and Jane is not studying
in the library. So she must be working at home.

2

Chapter 3

Proof
athematics is unique in that it claims a certainty that is beyond all possible doubt

M or argument. A mathematical proof shows how some result follows by logic alone

from a given set of assumptions, and once the result has been proven, it is as solid as
the foundations of logic themselves. Of course, mathematics achieves this certainty by
restricting itself to an artificial, mathematical world, and its application to the real world
does not carry the same degree of certainty.
Within the world of mathematics, consequences follow from assumptions with the
force of logic, and a proof is just a way of pointing out logical consequences. Of course,
the fact that mathematical results follow logically does not mean that they are obvious
in any normal sense. Proofs are convincing once they are discovered, but finding them
is often very difficult. They are written in a language and style that can seem obscure to
the uninitiated. Often, a proof builds on a long series of definitions and previous results,
and while each step along the way might be ‘obvious’ the end result can be surprising
and powerful. This is what makes the search for proofs worthwhile.
In this chapter, we’ll look at some approaches and techniques that can be used for
proving mathematical results, including two important proof techniques known as proof
by contradiction and mathematical induction. Along the way, we’ll encounter a few new
definitions and notations. Hopefully, you will be left with a higher level of confidence
for exploring the mathematical world on your own.

3.1 A Little Historical Background
The mathematical world and the real world weren’t always quite so separate. Until some
time near the middle of the nineteenth century, the statements of mathematics were
regarded as statements about the world. A proof was simply a convincing argument,
rather than a chain forged of absolute logic. It was something closer to the original mean55

56

Chapter 3. Proof

ing of the word ‘proof’, as a test or trial: to prove something was to test its truth by putting
it to the trial of logical argument.


3

A commonly made mistake centres around the difference between ‘proof‘ (a
noun) and ‘to prove‘ (a verb). You can prove a certain claim using a proof.
But grammatically speaking you cannot proof a certain claim using a prove.
Historically, in the course Reasoning & Logic this is one of the most common
spelling/grammar mistakes on exams. By including it in this book, we hope
that you will now know better.

The first rumble of trouble came in the form of non-Euclidean geometry. For two
thousand years, the geometry of the Greek mathematician Euclid had been accepted,
simply, as the geometry of the world. In the middle of the nineteenth century, it was
discovered that there are other systems of geometry, which are at least as valid and selfconsistent as Euclid’s system. Mathematicians can work in any of these systems, but they
cannot all claim to be working in the real world.
Near the end of the nineteenth century came another shock, in the form of cracks
in the very foundation of mathematics. At that time, mathematician Gottlieb Frege was
finishing a book on set theory that represented his life’s work. In Frege’s set theory, a
set could be defined by any property. You could have, for example, the set consisting of
all sets that contain three objects. As he was finishing his book, Frege received a letter
from a young mathematician named Bertrand Russell which described what became
known as Russell’s Paradox. Russell pointed out that the set of all sets—that is, the
set that contains every entity that satisfies the property of being a set—cannot logically
exist. We’ll see Russell’s reasoning in the following chapter. Frege could only include a
postscript in his book stating that the basis of the work had been swept away.



A contemporary of Peirce (see page 37), Friedrich Ludwig
Gottlob Frege (1848–1925) was a German philosopher, logician,
and mathematician. Peirce and Frege were apparently mostly
unaware of each other’s work. Frege is understood by many to
be the father of analytic philosophy, concentrating on the philosophy of language and mathematics. Like Peirce, his work was
largely ignored during his lifetime, and came to be recognised
by later mathematicans—including Russell.
Source: en.wikipedia.org/wiki/Gottlob_Frege.

Mathematicians responded to these problems by banishing appeals to facts about the

3.2. Mathematical Proof

57

real world from mathematical proof. Mathematics was to be its own world, built on its
own secure foundation. The foundation would be a basic set of assumptions or ‘axioms’
from which everything else would follow by logic. It would only be necessary to show
that the axioms themselves were logically consistent and complete, and the world of
mathematics would be secure. Unfortunately, even this was not to be. In the 1930s, Kurt
Gödel showed that there is no consistent, finite set of axioms that completely describes
even the corner of the mathematical world known as arithmetic. Gödel showed that
given any finite, consistent set of axioms, there are true statements about arithmetic that
do not follow logically from those axioms.
We are left with a mathematical world in which iron chains of logic still bind conclusions to assumptions. But the assumptions are no longer rooted in the real world. Nor
is there any finite core of axioms to which the rest of the mathematical world can be
chained. In this world, axioms are set up as signposts in a void, and then structures of
logic are built around them. For example, in the next chapter, instead of talking about
the set theory that describes the real world, we have a set theory, based on a given set
of axioms. That set theory is necessarily incomplete, and it might differ from other set
theories which are based on other sets of axioms.

3.2 Mathematical Proof
Understandably, mathematicians are very picky about getting their proofs right. It’s how
they construct their world. Students sometimes object that mathematicians are too picky
about proving things that are ‘obvious’. But the fact that something is obvious in the real
world counts for very little in the constructed world of mathematics. Too many obvious
things have turned out to be dead wrong. (For that matter, even things in the real world
that seem ‘obviously’ true are not necessarily true at all.)



For example, consider the quantity f (n) = n2 + n + 41. When n = 0,
f (n) = 41 which is prime; when n = 1, f (n) = 43 which is prime;
when n = 2, f (n) = 47, which is prime. By the time you had calculated
f (3), f (4), . . . , f (10) and found that they were all prime, you might conclude
that it is ‘obvious’ that f (n) is prime for all n ≥ 0. But this is not in fact the
case! (See exercises.)

As we saw in Section 2.5, a formal proof consists of a sequence of statements where
each statement is either an assumption or follows by a rule of logic from previous statements. The examples in that section all worked with unspecified generic propositions
(p, q, etc). Let us now look at how one might use the same techniques to prove a specific
proposition about the mathematical world. We will prove that for all integers n, if n is
even then n2 is even. (Definition: an integer n is even iff n = 2k for some integer k. For

3

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Chapter 3. Proof

example, 2 is even since 2 = 2 · 1; 66 is even since 66 = 2 · 33; 0 is even since 0 = 2 · 0.)
Proof. This is a proposition of the form ∀n( P(n) → Q(n)) where P(n) is “n is even” and
Q(n) is “n2 is even.” We need to show that P(n) → Q(n) is true for all values of n. Or
alternatively we can phrase it as: ∀n( E(n) → E(n2 )) where E( x ) is ‘x is even’.
In the language of Section 2.5, we need to show that for any n, E(n) logically implies
E(n2 ); or, equivalently, that E(n2 ) can be logically deduced from E(n); or, equivalently,
that

3

n is even
∴ n2 is even
is a valid argument. Here is a formal proof that
n is even
∴ n2 is even
is in fact a valid argument for any value of n:
Let n be an arbitrary integer.
1. n is even
2. if n is even, then n = 2k
for some integer k
3. n = 2k for some integer k
4. if n = 2k for some integer k,
then n2 = 4k2 for that integer k
2
5. n = 4k2 for some integer k
6. if n2 = 4k2 for some integer k,
then n2 = 2(2k2 ) for that k
2
7. n = 2(2k2 ) for some integer k
8. if n2 = 2(2k2 ) for some integer k,
then n2 = 2k′ for some integer k′
2
9. n = 2k′ for some integer k′
10. if n2 = 2k′ for some integer k′ ,
then n2 is even
2
11. n is even

premise
definition of even
from 1, 2 (modus ponens)
basic algebra
from 3, 4 (modus ponens)
basic algebra
from 5, 6 (modus ponens)
basic fact about integers
from 7, 8 (modus ponens)
definition of even
from 9, 10 (modus ponens)

(The “basic fact about integers” referred to above is that the product of integers is again
an integer.) Since n could be replaced by any integer throughout this argument, we have
proved the statement “if n is even then n2 is even” is true for all integers n. (You might
worry that the argument is only valid for even n; see the disclaimer about Feyenoord’s
football ability on page 47, or remind yourself that P(n) → Q(n) is automatically true if
P(n) is false.)

3.2. Mathematical Proof

59

Mathematical proofs are rarely presented with this degree of detail and formality. A
slightly less formal proof of our proposition might leave out the explicit implications and
instances of modus ponens and appear as follows:
Proof. Let n be an arbitrary integer.
1.
2.
3.
4.
5.
6.

n is even
n = 2k for some integer k
n2 = 4k2 for that integer k
n2 = 2(2k2 ) for that k
n2 = 2k′ for some integer k′
n2 is even

premise
definition of even
basic algebra
basic algebra
substituting k′ = 2k2
definition of even

Since n was an arbitrary integer, the statement is true for all integers.
A more typical proof would take the argument above and present it in prose rather
than list form:
Proof. Let n be an arbitrary integer and assume n is even. Then n = 2k for some integer
k by the definition of even, and n2 = 4k2 = 2(2k2 ). Since the product of integers is
an integer, we have n2 = 2k′ for some integer k′ . Therefore n2 is even. Since n was an
arbitrary integer, the statement is true for all integers.
Typically, in a ‘formal’ proof, it is this kind of (relatively) informal discussion that is
given, with enough details to convince the reader that a complete, formal proof could be
constructed. Of course, how many details the reader can be expected to fill in depends
on the reader, and reading proofs is a skill that must be developed and practiced.



In the course Reasoning & Logic you are learning to write proper formal
proofs, and as a part of that we also need to evaluate your performance. To
this end we ask you to write proofs similar to the second example given above
(the list form rather than the prose form) as this shows more clearly that you
are aware of the formalisms required in a proof.

Writing a proof is even more difficult. Every proof involves a creative act of discovery,
in which a chain of logic that leads from assumptions to conclusion is discovered. It
also involves a creative act of expression, in which that logic is presented in a clear and
convincing way. There is no algorithm for producing correct, coherent proofs. There are,
however, some general guidelines for discovering and writing proofs. Let’s look at some
of these next.

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3.2.1

3

Chapter 3. Proof

How to write a proof

One of the most important pieces of advice to keep in mind is: “Use the definition”.
In the world of mathematics, terms mean exactly what they are defined to mean and
nothing more. Definitions allow very complex ideas to be summarized as single terms.
When you are trying to prove things about those terms, you generally need to ‘unwind’
the definitions. In our example above, we used the definition of even to write n = 2k, and
then we worked with that equation. When you are trying to prove something about equivalence relations in Chapter 4, you can be pretty sure that you will need to use the fact
that equivalence relations, by definition, are symmetric, reflexive, and transitive. (And,
of course, you’ll need to know how the term ‘relation’ is defined in the first place! We
mean something quite different than the idea that ‘relations’ are something like your
aunt and uncle.)
More advice along the same line is to check whether you are using the assumptions
of the theorem. An assumption that is made in a theorem is called an hypothesis. The
hypotheses of the theorem state conditions whose truth will guarantee the conclusion of
the theorem. To prove the theorem means to assume that the hypotheses are true, and
to show, under that assumption, that the conclusion must be true. It’s likely (though
not guaranteed) that you will need to use the hypotheses explicitly at some point in the
proof, as we did in our example above.1 Also, you should keep in mind that any result
that has already been proved is available to be used in your proof.
A proof is a logical argument, based on the rules of logic. Since there are really not all
that many basic rules of logic, the same patterns keep showing up over and over. Let’s
look at some of the patterns.
The most common pattern arises in the attempt to prove that something is true ‘for all’
or ‘for every’ or ‘for any’ entity in a given category. In terms of logic, the statement you
are trying to prove is of the form ∀ x P( x ). In this case, the most likely way to begin the
proof is by saying something like, “Let x be an arbitrary entity in the domain of discourse.
We want to show that P( x ).” We call this a proof by generalisation In the rest of the
proof, x refers to some unspecified but definite entity in the domain of discourse. Since
x is arbitrary, proving P( x ) amounts to proving ∀ x P( x ). You only have to be careful that
you don’t use any facts about x beyond what you have assumed. For example, in our
proof above, we cannot make any assumptions about the integer n except that it is even;
if we for instance also assume x = 6 or that x is divisible by 3, then the proof would have
been incorrect, or at least incomplete.
Sometimes, you have to prove that an entity exists that satisfies certain stated properties. Such a proof is called an existence proof . In this case, you are attempting to prove
a statement of the form ∃ x P( x ). The way to do this is to find an example, that is, to find
a specific entity a for which P( a) is true. One way to prove the statement “There is an
even prime number” is to find a specific number that satisfies this description. The same
1 Of course, if you set out to discover new theorems on your own, you aren’t given the hypotheses and
conclusion in advance, which makes things quite a bit harder—and more interesting.

3.2. Mathematical Proof

61

statement could also be expressed “Not every prime number is odd.” This statement has
the form ¬(∀ x P( x )), which is equivalent to the statement ∃ x (¬ P( x )). An example that
proves the statement ∃ x (¬ P( x )) also proves ¬(∀ x P( x )). Such an example is called a
counterexample to the statement ∀ x P( x ): A counterexample proves that the statement
∀ x P( x ) is false. The number 2 is a counterexample to the statement “All prime numbers
are odd.” In fact, 2 is the only counterexample to this statement; many statements have
multiple counterexamples.
Note that we have now discussed how to prove and disprove universally quantified statements, and how to prove existentially quantified statements. How do you disprove ∃ x P( x )? Recall that ¬∃ x P( x ) is logically equivalent to ∀ x (¬ P( x )), so to disprove
∃ x P( x ) you need to prove ∀ x (¬ P( x )).
Many statements, like that in our example above, have the logical form of an implication, p → q. (More accurately, they are of the form “∀ x ( P( x ) → Q( x ))”, but as
discussed above the strategy for proving such a statement is to prove P( x ) → Q( x ) for
an arbitrary element x of the domain of discourse.) The statement might be “For all natural numbers n, if n is even then n2 is even,” or “For all strings x, if x is in the language
L then x is generated by the grammar G,”2 or “For all elements s, if s ∈ A then s ∈ B.”
Sometimes the implication is implicit rather than explicit: for example, “The sum of two
rationals is rational” is really short for “For any numbers x and y, if x and y are rational
then x + y is rational.” A proof of such a statement often begins something like this:
“Assume that p. We want to show that q.” In the rest of the proof, p is treated as an
assumption that is known to be true. As discussed above, the logical reasoning behind
this is that you are essentially proving that
p
∴ q
is a valid argument. Another way of thinking about it is to remember that p → q is automatically true in the case where p is false, so there is no need to handle that case explicitly.
In the remaining case, when p is true, we can show that p → q is true by showing that
the truth of q follows from the truth of p. So remember than proving an implication you
should assume the antecedent and prove the consequent (you can refresh your memory
of what those words mean on page 10).
A statement of the form p ∧ q can be proven by proving p and q separately. A statement of the form p ∨ q can be proved by proving the logically equivalent statement
(¬ p) → q: to prove p ∨ q, you can assume that p is false and prove, under that assumption, that q is true. For example, the statement “Every integer is even or odd” is
equivalent to the statement “Every integer that is not even is odd”.
Since p ↔ q is equivalent to ( p → q) ∧ (q → p), a statement of the form p ↔ q is
often proved by giving two proofs, one of p → q and one of q → p. In English, p ↔ q
can be stated in several forms such as “p if and only if q”, “if p then q and conversely,”
2 You

will learn about this in the course Automata, Languages, and Computability in your second year.

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Chapter 3. Proof

and “p is necessary and sufficient for q”. The phrase ‘if and only if’ is so common in
mathematics that it is often abbreviated iff .
You should also keep in mind that you can prove p → q by displaying a chain of valid
implications p → r → s → · · · → q. Similarly, p ↔ q can be proved with a chain of valid
biconditionals p ↔ r ↔ s ↔ · · · ↔ q.

3.2.2

3

Some terminology

Before we look at some sample proofs, here is some terminology that we will use throughout our sample proofs and the rest of the course of Reasoning & Logic.
• The natural numbers (denoted N) are the numbers 0, 1, 2, . . .. Note that the sum
and product of natural numbers are natural numbers.
• The integers (denoted Z) are the numbers 0, −1, 1, −2, 2, −3, 3, . . .. Note that the
sum, product, and difference of integers are integers.
• The rational numbers (denoted Q) are all numbers that can be written in the form
1
−65
m
n where m and n are integers and n ̸ = 0. So 3 and 7 are rationals; so, less
√

√

√27
obviously, are 6 and √27 since 6 = 61 (or, for that matter, 6 = −−12
2 ), and 12 =
12
√
√
27
9
3
12 =
4 = 2 . Note the restriction that the number in the denominator cannot

be 0: 30 is not a number at all, rational or otherwise; it is an undefined quantity.
Note also that the sum, product, difference, and quotient of rational numbers are
rational numbers (provided you don’t attempt to divide by 0).
• The real numbers (denoted R) are numbers that can be written in decimal form,
possibly with an infinite number of digits after the decimal point. Note that the
sum, product, difference, and quotient of real numbers are real numbers (provided
you don’t attempt to divide by 0).
• The irrational numbers are real numbers that are not√
rational, i.e., that cannot be
written as a ratio of integers. Such numbers include 3 (which we will prove is
22
not rational) and π (if anyone ever told you that π = 22
7 , remember that 7 is only
an approximation of the value of π). Later you will learn that we can describe this
set of irrational numbers as R − Q, that is: it is all the numbers that are in R but
are not in Q.
• An integer n is divisible by m iff n = mk for some integer k. This can also be
expressed by saying that m evenly divides n, which has the mathematical notation
m | n. So for example, 2 | 8, but 8 ∤ 2. 2 | n iff n = 2k for some integer k; n is
divisible by 3 iff n = 3k for some integer k, and so on. Note that if 2 ∤ n (i.e., n is
not divisible by 2), then n must be 1 more than a multiple of 2 so n = 2k + 1 for

3.2. Mathematical Proof

63

some integer k. Similarly, if n is not divisible by 3 then n must be 1 or 2 more than
a multiple of 3, so n = 3k + 1 or n = 3k + 2 for some integer k.
• An integer is even iff it is divisible by 2 and odd iff it is not.
• An integer n > 1 is prime if it is divisible by exactly two positive integers, namely
1 and itself. Note that a number must be greater than 1 to even have a chance of
being termed ‘prime’. In particular, neither 0 nor 1 is prime.

3

3.2.3 Examples
Let’s look now at another example of a proof. We set out to prove that the sum of any
two rational numbers is rational.
Proof. We start by assuming that x and y are arbitrary rational numbers. Here’s a formal
proof that the inference rule
x is rational
y is rational
∴ x + y is rational
is a valid rule of inference:
1. x is rational
2. if x is rational, then x = ba
for some integers a and b ̸= 0
3. x = ba for some integers a and b ̸= 0
4. y is rational
5. if y is rational, then y = dc for
some integers c and d ̸= 0
6. y = dc for some c and d ̸= 0
7. x = ba for some a and b ̸= 0 and
y = dc for some c and d ̸= 0
8. if x = ba for some a and b ̸= 0 and
y = dc for c and d ̸= 0 then
+bc where a, b, c, d
x + y = adbd
are integers and b, d ̸= 0
+bc for some a, b, c, d
9. x + y = adbd
where b, d ̸= 0
+bc for some a, b, c, d
10. if x + y = adbd
where b, d ̸= 0 then x + y = m
n
where m, n are integers and n ̸= 0
11. x + y = m
n where m and n

premise
definition of rationals
from 1,2 (modus ponens)
premise
definition of rational
from 4,5 (modus ponens)
from 3,6

basic algebra
from 7,8 (modus ponens)

properties of integers

64

Chapter 3. Proof
are integers and n ̸= 0
12. if x + y = m
n where m and n are
integers and n ̸= 0
then x + y is rational
13. x + y is rational

3

from 9,10 (modus ponens)

definition of rational
from 11,12 (modus ponens)

So the rule of inference given above is valid. Since x and y are arbitrary rationals, we have
proved that the rule is valid for all rationals, and hence the sum of any two rationals is
rational.
Again, a more informal presentation that we expect from you during the course
would look like:
Proof. Proof by generalisation:
• Let x and y be arbitrary rational numbers.
• By the definition of rational, there are integers a, b ̸= 0, c, d ̸= 0 such that x =
and y = dc .
• Then x + y =

a
b

ad+bc
bd ;

• We know ad + bc and bd are integers since the sum and product of integers are
integers, and we also know bd ̸= 0 since neither b nor d is 0.
• So we have written x + y as the ratio of two integers, the denominator being nonzero.
• Therefore, by the definition of rational numbers, x + y is rational.
• Since x and y were arbitrary rationals, the sum of any two rationals is rational.

And one more example: we will prove that any 4-digit number d1 d2 d3 d4 is divisible
by 3 iff the sum of the four digits is divisible by 3.
Proof. This statement is of the form p ↔ q; recall that p ↔ q is logically equivalent to
( p → q) ∧ (q → p). So we need to prove for any 4-digit number d1 d2 d3 d4 that (1) if
d1 d2 d3 d4 is divisible by 3 then d1 + d2 + d3 + d4 is divisible by 3, and (2) if d1 + d2 + d3 +
d4 is divisible by 3 then d1 d2 d3 d4 is divisible by 3. So let d1 d2 d3 d4 be an arbitrary 4-digit
number.

3.2. Mathematical Proof

65

(1) Assume d1 d2 d3 d4 is divisible by 3, i.e., d1 d2 d3 d4 = 3k for some integer k. The
number d1 d2 d3 d4 is actually d1 × 1000 + d2 × 100 + d3 × 10 + d4 , so we have the equation
d1 × 1000 + d2 × 100 + d3 × 10 + d4 = 3k.
Since 1000 = 999 + 1, 100 = 99 + 1, and 10 = 9 + 1, this equation can be rewritten
999d1 + d1 + 99d2 + d2 + 9d3 + d3 + d4 = 3k.
Rearranging gives
d1 + d2 + d3 + d4 = 3k − 999d1 − 99d2 − 9d3

= 3k − 3(333d1 ) − 3(33d2 ) − 3(3d3 ).
We can now factor a 3 from the right side to get
d1 + d2 + d3 + d4 = 3(k − 333d1 − 33d2 − d3 ).
Since (k − 333d1 − 33d2 − d3 ) is an integer, we have shown that d1 + d2 + d3 + d4 is
divisible by 3.
(2) Assume d1 + d2 + d3 + d4 is divisible by 3. Consider the number d1 d2 d3 d4 . As
remarked above,
d1 d2 d3 d4 = d1 × 1000 + d2 × 100 + d3 × 10 + d4
so
d1 d2 d3 d4 = 999d1 + d1 + 99d2 + d2 + 9d3 + d3 + d4

= 999d1 + 99d2 + 9d3 + (d1 + d2 + d3 + d4 ).
We assumed that d1 + d2 + d3 + d4 = 3k for some integer k, so we can substitute into the
last equation to get
d1 d2 d3 d4 = 999d1 + 99d2 + 9d3 + 3k = 3(333d1 + 33d2 + 3d3 + k).
Since the quantity in parentheses is an integer, we have proved that d1 d2 d3 d4 is divisible
by 3.
In (1) and (2) above, the number d1 d2 d3 d4 was an arbitrary 4-digit integer, so we
have proved that for all 4-digit integers, d1 d2 d3 d4 is divisible by 3 iff the sum of the four
digits is divisible by 3.
Now suppose we wanted to prove the statement “For all integers n, n2 is even if and
only if n is even.” We have already proved half of this statement (“For all integers n, if n
is even then n2 is even”), so all we need to do is prove the statement “For all integers n, if
n2 is even then n is even” and we’ll be done. Unfortunately, this is not as straightforward

3

66

Chapter 3. Proof

as it seems: suppose we started in our standard manner and let n be an arbitrary integer
and assumed that n2 = 2k for some integer k. Then we’d be stuck!
√ Taking the square
root of both sides would give us n on the left but would leave a 2k√ on the right. This
√
quantity is not of the form 2k′ for any integer k′ ; multiplying it by
√
√k
2

3

√2
2

would give 2 √k

2

but there is no way for us to prove that
is an integer. So we’ve hit a dead end. What
do we do now?
The answer is that we need a different proof technique. The proofs we have written
so far are what are called direct proofs: to prove p → q you assume p is true and prove
that the truth of q follows. Sometimes, when a direct proof of p → q fails, an indirect
proof will work. Recall that the contrapositive of the implication p → q is the implication
¬q → ¬ p, and that this proposition is logically equivalent to p → q. An indirect proof
of p → q, then, is a direct proof of the contrapositive ¬q → ¬ p. In our current example,
instead of proving “if n2 is even then n is even” directly, we can prove its contrapositive
“if n is not even (i.e., n is odd) then n2 is not even (i.e., n2 is odd.)” We call this method
a proof by contrapositive. The proof of this contrapositive is a routine direct argument
which we leave to the exercises.
Alternatively we sometimes need a proof by division into cases. Consider for instance that we want to prove that 3 | (n3 + 3n2 + 2n) for all integers n. What we can do
is split our proof into three different case based on the divisibility by 3. Recall from the
definition of divisibility that every number can be written as either n = 3k, n = 3k + 1,
or n = 3k + 2. In a proof by division into cases, we prove that the claim holds for all of
these cases and thereby prove the claim holds for all numbers.



We have also created a pencast about several of the different proof techniques outlined in this chapter. This includes one in which we prove
that 3 | (n3 + 3n2 + 2n) using a proof by division into cases, found here:
youtu.be/4OHyyGY_WpI

Exercises
1. Find a natural number n for which n2 + n + 41 is not prime.
2. Show that the propositions p ∨ q and (¬ p) → q are logically equivalent.
3. Show that the proposition ( p ∨ q) → r is equivalent to ( p → r ) ∧ (q → r ).
4. Determine whether each of the following statements is true. If it true, prove it. If it is false,
give a counterexample.
a) Every prime number is odd.
b) Every prime number greater than 2 is odd.
c) If x and y are integers with x < y, then there is an integer z such that x < z < y.
d) If x and y are real numbers with x < y, then there is a real number z such that x < z < y.

3.3. Proof by Contradiction

67

5. Suppose that r, s, and t are integers, such that r evenly divides s and s evenly divides t. Prove
that r evenly divides t.
6. Prove that for all integers n, if n is odd then n2 is odd.
7. Prove that an integer n is divisible by 3 iff n2 is divisible by 3. (Hint: give an indirect proof of
“if n2 is divisible by 3 then n is divisible by 3.”)
8. Prove or disprove each of the following statements. Remember that to disprove a statement
we always expect a counterexample!
a) The product of two even integers is even.
b) The product of two integers is even only if both integers are even.
c) The product of two rational numbers is rational.
d) The product of two irrational numbers is irrational.
e) For all integers n, if n is divisible by 4 then n2 is divisible by 4.
f) For all integers n, if n2 is divisible by 4 then n is divisible by 4.

3.3 Proof by Contradiction
Suppose that we start with some set of assumptions and apply rules of logic to derive
a sequence of statements that can be proved from those assumptions, and suppose that
we derive a statement that we know to be false. When the laws of logic are applied to
true statements, the statements that are derived will also be true. If we derive a false
statement by applying rules of logic to a set of assumptions, then at least one of the
assumptions must be false. This observation leads to a powerful proof technique, which
is known as proof by contradiction.
Suppose that you want to prove some proposition, p. To apply proof by contradiction,
assume that ¬ p is true, and apply the rules of logic to derive conclusions based on this
assumption. If it is possible to derive a statement that is known to be false, it follows
that the assumption, ¬ p, must be false. (Of course, if the derivation is based on several
assumptions, then you only know that at least one of the assumptions must be false.) The
fact that ¬ p is false proves that p is true. Essentially, you are arguing that p must be true,
because if it weren’t, then some statement that is known to be false could be proved to
be true. Generally, the false statement that is derived in a proof by contradiction is of the
form q ∧ ¬q. This statement is a contradiction in the sense that it is false no matter what
the value of q. Note that deriving the contradiction q ∧ ¬q is the same as showing that
the two statements, q and ¬q, both follow from the assumption that ¬ p.
As a first example of proof by contradiction, consider the following theorem:
√
Theorem 3.1. The number 3 is irrational.
Proof. Proof by contradiction:
• Assume for the sake of contradiction that

√

3 is rational.

3

68

Chapter 3. Proof

√
√
• Then 3 can be written as the ratio of two integers, 3 =
and n′ .

m′
n′

for some integers m′

′

• Furthermore, the fraction m
be reduced to lowest terms by canceling all comn′ can
√
mon factors of m′ and n′ . So 3 = m
n for some integers m and n which have no
common factors.
• Squaring both sides of this equation gives 3 =

3

m2
n2

and re-arranging gives 3n2 = m2 .

• From this equation we see that m2 is divisible by 3; you proved in the previous
section (Exercise 6) that m2 is divisible by 3 iff m is divisible by 3. Therefore m is
divisible by 3 and we can write m = 3k for some integer k.
• Substituting m = 3k into the last equation above gives 3n2 = (3k )2 or 3n2 = 9k2 ,
which in turn becomes n2 = 3k2 . From this we see that n2 is divisible by 3, and
again we know that this implies that n is divisible by 3.
• But now we have (i) m and n have no common factors, and (ii) m and n have a
common factor, namely 3. It is impossible for both these things to be true, yet our
argument has been logically correct.
√
• Therefore our original assumption, namely that 3 is rational, must be incorrect.
√
• Therefore 3 must be irrational.

One of the oldest mathematical proofs, which goes all the way back to Euclid, is a
proof by contradiction. Recall that a prime number is an integer n, greater than 1, such
that the only positive integers that evenly divide n are 1 and n. We will show that there
are infinitely many primes. Before we get to the theorem, we need a lemma. (A lemma
is a theorem that is introduced only because it is needed in the proof of another theorem.
Lemmas help to organize the proof of a major theorem into manageable chunks.)
Lemma 3.2. If N is an integer and N > 1, then there is a prime number which evenly divides
N.
Proof. Let D be the smallest integer which is greater than 1 and which evenly divides N.
(D exists since there is at least one number, namely N itself, which is greater than 1 and
which evenly divides N. We use the fact that any non-empty subset of N has a smallest
member.) I claim that D is prime, so that D is a prime number that evenly divides N.
Suppose that D is not prime. We show that this assumption leads to a contradiction.
Since D is not prime, then, by definition, there is a number k between 2 and D − 1, inclusive, such that k evenly divides D. But since D evenly divides N, we also have that k
evenly divides N (by exercise 5 in the previous section). That is, k is an integer greater
than one which evenly divides N. But since k is less than D, this contradicts the fact that
D is the smallest such number. This contradiction proves that D is a prime number.

3.3. Proof by Contradiction

69

Theorem 3.3. There are infinitely many prime numbers.
Proof. Suppose that there are only finitely many prime numbers. We will show that this
assumption leads to a contradiction.
Let p1 , p2 , …, pn be a complete list of all prime numbers (which exists under the
assumption that there are only finitely many prime numbers). Consider the number N
obtained by multiplying all the prime numbers together and adding one. That is,
N = ( p1 · p2 · p3 · · · pn ) + 1.
Now, since N is larger than any of the prime numbers pi , and since p1 , p2 , …, pn is a
complete list of prime numbers, we know that N cannot be prime. By the lemma, there
is a prime number p which evenly divides N. Now, p must be one of the numbers p1 ,
p2 , …, pn . But in fact, none of these numbers evenly divides N, since dividing N by any
pi leaves a remainder of 1. This contradiction proves that the assumption that there are
only finitely many primes is false.
This proof demonstrates the power of proof by contradiction. The fact that is proved
here is not at all obvious, and yet it can be proved in just a few paragraphs.



It is easy to get a proof by contradiction wrong however. In one of the pencasts of this course we treat a commonly-made mistake when using proofs
by contradiction: youtu.be/OqKvBWxanok.

Exercises
1. Suppose that a1 , a2 , …, a10 are real numbers, and suppose that a1 + a2 + · · · + a10 > 100. Use a
proof by contradiction to conclude that at least one of the numbers ai must be greater than 10.
2. Prove that each of the following statements is true. In each case, use a proof by contradiction.
Remember that the negation of p → q is p ∧ ¬q.
a) Let n be an integer. If n2 is an even integer, then n is an even integer.
√
b) 2 is irrational.
c) If r is a rational number and x is an irrational number, then r + x is an irrational number.
(That is, the sum of a rational number and an irrational number is irrational.)
d) If r is a non-zero rational number and x is an irrational number, then rx is an irrational
number.
e) If r and r + x are both rational, then x is rational.
3. The pigeonhole principle is the following obvious observation: If you have n pigeons in k pigeonholes and if n > k, then there is at least one pigeonhole that contains more than one
pigeon. Even though this observation seems obvious, it’s a good idea to prove it. Prove the
pigeonhole principle using a proof by contradiction.

3

70

Chapter 3. Proof

3.4

3

Mathematical Induction

The structure of the natural numbers—0, 1, 2, 3, and on to infinity—makes possible a
powerful proof technique known as induction or mathematical induction. Although
the idea behind induction is simple, students often struggle with it. Take your time and
study this material thoroughly! You will have opportunity to practice in the labs after
we discuss induction in the lectures.
Let P be a one-place predicate whose domain of discourse includes the natural numbers. Suppose that we can
prove that P(0) is true. Suppose that we can also prove the
statements P(0) → P(1), P(1) → P(2), P(2) → P(3), and
so on. The principle of mathematical induction is the observation that we can then conclude that P(n) is true for
all natural numbers n. This should be clear. It’s like a line
of dominos, lined up and ready to fall, one after the next.3
Since P(0) and P(0) → P(1) are true, we can apply the rule
of modus ponens to conclude that P(1) is true. Then, since P(1) and P(1) → P(2) are
true, we can conclude by modus ponens that P(2) is true. From P(2) and P(2) → P(3),
we conclude that P(3) is true. For any given n in the set N, we can continue this chain
of deduction for n steps to prove that P(n) is true.
When applying induction, we don’t actually prove each of the implications P(0) →
P(1), P(1) → P(2), and so on, individually. That would require an infinite amount of
work. The whole point of induction is to avoid any infinitely long process. Instead, we
prove ∀k ( P(k) → P(k + 1)) (where the domain of discourse for the predicate P is N.
The statement ∀k ( P(k) → P(k + 1)) summarizes all the infinitely many implications in
a single statement. Stated formally, the
induction says that
( principle of mathematical
)
if we can prove the statement P(0) ∧ ∀k ( P(k ) → P(k + 1) , then we can deduce that
∀n P(n) (again, with N as the domain of discourse).
Khan Academy offers a video with a clear explanation of mathematical
induction:
www.khanacademy.org/math/algebra-home/
alg-series-and-induction/alg-induction/v/proof-by-induction.
I highly recommend that you watch it to also hear this information from
another perspective.



3.4.1

How to write a proof by induction

It should be intuitively clear that the principle of induction is valid. It follows from
the fact that the list 0, 1, 2, 3, …, if extended long enough, will eventually include any
3 Image:

commons.wikimedia.org/wiki/File:Dominoeffect.png.

3.4. Mathematical Induction

71

given natural number. If we start from P(0) and take enough steps of the form P(k) →
P(k + 1), we can get P(n) for any given natural number n. However, whenever we deal
with infinity, we are courting the possibility of paradox. We will prove the principle of
induction rigorously in the next chapter (see Theorem 4.3), but for now we just state it
as a theorem:
Theorem 3.4. Let P be a one-place
includes the natural
( predicate whose domain of discourse
)
numbers. Suppose that P(0) ∧ ∀k ∈ N ( P(k) → P(k + 1)) .4 Then P(n) is true for all
natural numbers n. (That is, the statement ∀n P(n) is true, where the domain of discourse for P
is the set of natural numbers.)
Mathematical induction can be applied in many situations: you can prove things
about strings of characters by doing induction on the length of the string, things about
graphs by doing induction on the number of nodes in the graph, things about grammars
by doing induction on the number of productions in the grammar, and so on. We’ll be
looking at applications of induction for the rest of this chapter, and treat a form called
structural induction in the next chapter.
Although proofs by induction can be very different from one another, they all follow
just a few basic structures. A proof based on the preceding theorem always has two parts.
First, P(0) is proved. This is called the base case of the induction. Then the statement
∀k ( P(k) → P(k + 1)) is proved. This statement can be proved by letting k be an arbitrary
element of N and proving P(k ) → P(k + 1). This in turn can be proved by assuming
that P(k ) is true and proving that the truth of P(k + 1) follows from that assumption.
This case is called the inductive case, and P(k ) is called the inductive hypothesis or the
induction hypothesis. Note that the base case is just as important as the inductive case.
By itself, the truth of the statement ∀k ( P(k ) → P(k + 1)) says nothing at all about the
truth of any of the individual statements P(n). The chain of implications P(0) → P(1),
P(1) → P(2), …, P(n − 1) → P(n) says nothing about P(n) unless the chain is anchored
at the other end by the truth of P(0).

3.4.2 Examples
Let’s look at a few examples.
Theorem 3.5. The number 22n − 1 is divisible by 3 for all natural numbers n.
Proof. Here, P(n) is the statement that 22n − 1 is divisible by 3.
Base case: When n = 0, 22n − 1 = 20 − 1 = 1 − 1 = 0 and 0 is divisible by 3 (since
0 = 3 · 0.) Therefore the statement holds when n = 0.
Inductive case: We want to show that if the statement is true for n = k (where k is an
arbitrary natural number), then it is true for n = k + 1 also. That is, we must prove the
4 We will encounter this notation of k ∈ N again in Chapter 4. For now you should just remember that
k ∈ N means: k is an integer from the set N.

3

72

Chapter 3. Proof

implication P(k) → P(k + 1). So we assume P(k), that is, we assume that 22k is divisible
by 3. This means that 22k − 1 = 3m for some integer m. We want to prove P(k + 1), that
is, that 22(k+1) − 1 is also divisible by 3:
22(k+1) − 1 = 22k+2 − 1

= 22k · 22 − 1

properties of exponents

= 4 · 22k − 1

3

= 4 · 22k − 4 + 4 − 1
= 4(22k − 1) + 3
= 4(3m) + 3
= 3(4m + 1)

algebra
the inductive hypothesis
algebra

and from the last line we see that 22k+1 is in fact divisible by 3. (The third step—subtracting
and adding 4—was done to enable us to use our inductive hypothesis.)
Altogether, we have proved that P(0) holds and that, for all k, P(k) → P(k + 1) is
true. Therefore, by the principle of induction, P(n) is true for all n in N, i.e. 22n − 1 is
divisible by 3 for all n in N.
The principle of mathematical induction gives a method for proving P(n) for all n in
the set N. It should be clear that if M is any natural number, a similar method can be
used to show that P(n) is true for all natural numbers n that satisfy n ≥ M. Just start
the induction with a base case of n = M instead of with a base case of n = 0. I leave
the proof of this extension of the principle of induction as an exercise. We can use the
extended principle of induction to prove a result that was first mentioned in Section 2.1.
Theorem 3.6. Suppose that a compound proposition contains exactly n propositional variables,
where n ≥ 1. Then there are exactly 2n different ways of assigning truth values to the n variables.
Proof. Let P(n) be the statement “There are exactly 2n different ways of assigning truth
values to n propositional variables.” We will use induction to prove the P(n) is true for
all n ≥ 1.
Base case: First, we prove the statement P(1). If there is exactly one variable, then
there are exactly two ways of assigning a truth value to that variable. Namely, the variable can be either true or false. Since 2 = 21 , P(1) is true.
Inductive case: Suppose that P(k) is already known to be true. We want to prove
that, under this assumption, P(k + 1) is also true. Suppose that p1 , p2 , …, pk+1 are k + 1
propositional variables. Since we are assuming that P(k) is true, we know that there are
2k ways of assigning truth values to p1 , p2 , …, pk . But each assignment of truth values to
p1 , p2 , …, pk can be extended to the complete list p1 , p2 , …, pk , pk+1 in two ways. Namely,
pk+1 can be assigned the value true or the value false. It follows that there are 2 · 2k ways
of assigning truth values to p1 , p2 , …, pk+1 . Since 2 · 2k = 2k+1 , this finishes the proof.

3.4. Mathematical Induction

73

3.4.3 More examples
The sum of an arbitrary number of terms is written using the symbol ∑. (This symbol
is the Greek letter sigma, which is equivalent to the Latin letter S and stands for ‘sum’.)
Thus, we have

5

∑ i2 = 12 + 22 + 32 + 42 + 52

i =1
7

∑ a k = a3 + a4 + a5 + a6 + a7

k =3
N

1
1
1
1
1
=
+
+
+···+
n
+
1
0
+
1
1
+
1
2
+
1
N
+1
n =0

∑

This notation for a sum, using the ∑ operator, is called summation notation. A similar
notation for products uses the symbol ∏. (This is the Greek letter pi, which is equivalent
to the Latin letter P and stands for ‘product’.) For example,

5

∏ (3k + 2) = (3 · 2 + 2)(3 · 3 + 2)(3 · 4 + 2)(3 · 5 + 2)

k =2
n

1

∏i

i =1

=

1 1
1
· ··· ·
1 2
n

Induction can be used to prove many formulas that use these notations. Here are two
examples:

n

Theorem 3.7.

∑

i =1

i=

n ( n + 1)
for any integer n greater than zero.
2

We’ll prove this theorem in class. Can you do it yourself?

3

74

Chapter 3. Proof



3

The summation of Theorem 3.7 is often attributed
to mathematician and physicist Carl Friedrich Gauss
(1777–1855). Gauss has been exceptionally influential
in a variety of fields; he is called “the greatest mathematician since antiquity”. For example, you might
have heard of the Gaussian probability distribution, or
perhaps of the Gauss unit for magnetic flux (although
we now commonly use the Tesla for that, with 1 Tesla
equalling 105 Gauss). His brilliance was already apparent in primary school when he allegedly used the ‘Gauss
sum’ from Theorem 3.7 to solve the maths homework the teacher had set the
class—to great astonishment of the teacher. You will see this summation
again in the course Algorithms & Data Structures when analysing the runtime
of algorithms containing loops.
Source: en.wikipedia.org/wiki/Carl_Friedrich_Gauss.

n

Theorem 3.8.

∑

i2i−1 = (n − 1) · 2n + 1 for any natural number n > 0.

i =1
n

Proof. Let P(n) be the statement

∑

i2i−1 = (n − 1) · 2n + 1. We use induction to show

i =1

that P(n) is true for all n > 0
1

Base case: Consider the case n = 1. P(1) is the statement that

∑

i2i−1 = (1 − 1) ·

i =1

21 + 1. Since each side of this equation is equal to one, this is true.
Inductive case: Let k > 1 be arbitrary, and assume that P(k) is true. We want to show
k +1

that P(k + 1) is true. P(k + 1) is the statement
we can compute that
(
k +1

∑

i =1

i2

i −1

(

i2i−1 = ((k + 1) − 1) · 2k+1 + 1. But,

i =1

)

k

∑

=

∑

i2

i −1

+ (k + 1)2(k+1)−1

i =1

)
= ( k − 1 ) · 2k + 1 + ( k + 1 )2k
(
)
= ( k − 1 ) + ( k + 1 ) 2k + 1

= ( k · 2 ) · 2k + 1
= k2k+1 + 1

(inductive hypothesis)

3.5. Strong Mathematical Induction

75

which is what we wanted to show. This completes the induction.
100

For example, these theorems show that ∑ i = 1 + 2 + 3 + 4 + · · · + 100 =
i =1

100(100 + 1)
=
2

5050 and that 1 · 20 + 2 · 21 + 3 · 22 + 4 · 23 + 5 · 24 = (5 − 1)25 + 1 = 129, as well as infinitely many other such sums.

3.5 Strong Mathematical Induction
There is a second form of the principle of mathematical induction which is useful in
some cases. To apply the first form of induction, we assume P(k) for an arbitrary natural
number k and show that P(k + 1) follows from that assumption. In the second form
of induction, the assumption is that P( x ) holds for all x between 0 and k inclusive, and
we show that P(k + 1) follows from this. This gives us a lot more to work with when
deducing P(k + 1). We will need this second, stronger form of induction in the next two
sections. A proof will be given in the next chapter.
Theorem 3.9. Let P be a one-place predicate whose domain of discourse includes the natural
numbers. Suppose that P(0) is true and that

( P(0) ∧ P(1) ∧ · · · ∧ P(k)) → P(k + 1)
is true for each natural number k ≥ 0. Then P(n) is true for every natural number n.
For example, we can use this theorem to prove that every integer greater than one
can be written as a product of prime numbers (where a number that is itself prime is
considered to be a product of one prime number). The proof illustrates an important
point about applications of this theorem: When proving P(k + 1), you don’t necessarily
have to use the assumptions that P(0), P(1), …, and P(k) are true. If P(k + 1) is proved
by any means—possibly including the assumptions—then the statement ( P(0) ∧ P(1) ∧
· · · ∧ P(k)) → P(k + 1) has been shown to be true. It follows from this observation
that several numbers, not just zero, can be ‘base cases’ in the sense that P( x + 1) can be
proved independently of P(0) through P( x ). In this sense, 0, 1, and every prime number
are base cases in the following theorem.
Theorem 3.10. Every natural number greater than one can be written as a product of prime
numbers.
Proof. Let P(n) be the statement “if n > 1, then n can be written as a product of prime
numbers”. We will prove that P(n) is true for all n by applying the second form of the
principle of induction.
Note that P(0) and P(1) are both automatically true, since n = 0 and n = 1 do not
satisfy the condition that n > 1, and P(2) is true since 2 is the product of the single prime

3

76

3

Chapter 3. Proof

number 2. Suppose that k is an arbitrary natural number with k > 1, and suppose that
P(0), P(1), …, P(k) are already known to be true; we want to show that P(k + 1) is true.
In the case where k + 1 is a prime number, then k + 1 is a product of one prime number,
so P(k + 1) is true.
Consider the case where k + 1 is not prime. Then, according to the definition of prime
number, it is possible to write k + 1 = ab where a and b are numbers in the range from 2
to k inclusive. Since P(0) through P(k) are known to be true, a and b can each be written
as a product of prime numbers. Since k + 1 = ab, k + 1 can also be written as a product
of prime numbers. We have shown that P(k + 1) follows from P(0) ∧ P(1) ∧ · · · ∧ P(k),
and this completes the induction.



In two of the pencasts of this course, we treat two flawed induction proofs
and examine what mistakes have been made. You can find them here: youtu.
be/m0EIgQyukdQ and youtu.be/2c-zw-ENNss.

Exercises
1. Use induction to prove that n3 + 3n2 + 2n is divisible by 3 for all natural numbers n.
2. Use induction to prove that
n

∑ ri =

i =0

1 − r n +1
1−r

for any natural number n and for any real number r such that r ̸= 1.
3. Use induction to prove that for any natural number n,
n

1

∑ 2i

i =0

= 2−

1
2n

In addition to proving this by induction, show that it follows as a corollary of Exercise 2.
4. Use induction to prove that for any natural number n,
n

∑ 2i = 2 n + 1 − 1

i =0

In addition to proving this by induction, show that it follows as a corollary of Exercise 2.
5. Use induction to prove that for any positive integer n,
n

∑ i2 =

i =1

n(n + 1)(2n + 1)
6

3.6. Application: Recursion and Induction

77

6. Use induction to prove that for any positive integer n,
n

∑ (2i − 1) = n2

i =1

7. Evaluate the following sums, using results proved in this section and in the previous exercises:
a) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
1
1
1
1
1
1
b) 1 + + 2 + 3 + 4 + 5 + 6
3 3
3
3
3
3
c) 50 + 51 + 52 + 53 + · · · + 99 + 100
d) 1 + 4 + 9 + 16 + 25 + 36 + 49 + 81 + 100
1
1
1
e) 2 + 3 + · · · + 99
2
2
2
8. Write each of the sums in the preceding problem using summation notation.
9. Rewrite the proof of Theorem 3.8 without using summation notation.
10. Use induction to prove the following generalized distributive laws for propositional logic: For
any natural number n > 1 and any propositions q, p1 , p2 , …, pn ,
a) q ∧ ( p1 ∨ p2 ∨ · · · ∨ pn ) = (q ∧ p1 ) ∨ (q ∧ p2 ) ∨ · · · ∨ (q ∧ pn )
b) q ∨ ( p1 ∧ p2 ∧ · · · ∧ pn ) = (q ∨ p1 ) ∧ (q ∨ p2 ) ∧ · · · ∧ (q ∨ pn )

3.6 Application: Recursion and Induction
In computer programming, there is a technique called recursion that is closely related
to induction. In a computer program, a subroutine is a named sequence of instructions
for performing a certain task. When that task needs to be performed in a program, the
subroutine can be called by name. A typical way to organize a program is to break down
a large task into smaller, simpler subtasks by calling subroutines to perform each of the
subtasks. A subroutine can perform its task by calling other subroutines to perform
subtasks of the overall task. A subroutine can also call itself. That is, in the process of
performing some large task, a subroutine can call itself to perform a subtask. This is
known as recursion, and a subroutine that does this is said to be a recursive subroutine.
Recursion is appropriate when a large task can be broken into subtasks where some or
all of the subtasks are smaller, simpler versions of the main task.



Prolog is an example of a programming language that uses recursion to
powerful effect. Classical Prolog has no loop construct: loops are defined
using recursive subroutines.

Like induction, recursion is often considered to be a ‘hard’ topic by students. Experienced computer scientists, on the other hand, often say that they can’t see what all the
fuss is about, since induction and recursion are elegant methods which ‘obviously’ work.

3

78

Chapter 3. Proof

In fairness, students have a point, since induction and recursion both manage to pull infinite rabbits out of very finite hats. But the magic is indeed elegant, and learning the
trick is very worthwhile.


3
3.6.1

In your course Computer Organisation you will be tasked to write a small recursive program to compute the factorial in Assembly. You can have a sneakpreview below of of what the pseudocode for such a program might be. In
future courses like Algorithms & Data Structures and Algorithm Design you
will also be tasked to write and analyse recursive algorithms.

Recursive factorials

A simple example of a recursive subroutine is a function that computes n! for a nonnegative integer n. n!, which is read “n factorial”, is defined as follows:
0! = 1
n

n! =

∏i

for n > 0

i =1

For example, 5! = 1 · 2 · 3 · 4 · 5 = 120. Note that for n > 1,
n

n! =

(

∏i= ∏

i =1

)

n −1

i

(
)
· n = ( n − 1) ! · n

i =1

(
)
It is also true that n! = (n − 1)! · n when n = 1. This observation makes it possible to
write a recursive function to compute n!.



All the programming examples in this section are written in the Java programming language. I won’t put these blue boxes around them.

To compute factorial(n) for n > 0, we can write a function (in Java). This function
computes factorial(n − 1) first by calling itself recursively. The answer from that computation is then multiplied by n to give the value of n!. The recursion has a base case,
namely the case when n = 0. For the base case, the answer is computed directly rather
than by using recursion. The base case prevents the recursion from continuing forever,
in an infinite chain of recursive calls.
Now, as it happens, recursion is not the best way to compute n!. It can be computed
more efficiently using a loop. Furthermore, except for small values of n, the value of n! is

3.6. Application: Recursion and Induction

79

outside the range of numbers that can be represented as 32-bit ints. However, ignoring
these problems, the factorial function provides a first example of the interplay between
recursion and induction. We can use induction to prove that factorial(n) does indeed
compute n! for n ≥ 0.5
Theorem 3.11. Assume that the data type int can represent arbitrarily large integers. Under this
assumption, the factorial function defined above correctly computes n! for any natural number n.
Proof. Let P(n) be the statement “factorial(n) correctly computes n!”. We use induction
to prove that P(n) is true for all natural numbers n.
Base case: In the case n = 0, the if statement in the function assigns the value 1 to
the answer. Since 1 is the correct value of 0!, factorial(0) correctly computes 0!.
Inductive case: Let k be an arbitrary natural number, and assume that P(k) is true.
From this assumption, we must show that P(k + 1) is true. The assumption is that
factorial(k) correctly computes k!, and we want to show that factorial(k + 1) correctly
computes (k + 1)!.
When the function computes factorial(k + 1), the value of the parameter n is k + 1.
Since k + 1 > 0, the if statement in the function computes the value of factorial(k + 1)
by applying the computation factorial(k) ∗ (k + 1). We know, by the induction hypothesis, that the value computed by factorial(k) is k!. It follows that the value computed
by factorial(k + 1) is (k!) · (k + 1). As we observed above, for any k + 1 > 0, (k!) · (k + 1) =
(k + 1)!. We see that factorial(k + 1) correctly computes (k + 1)!. This completes the induction.
In this proof, we see that the base case of the induction corresponds to the base case
of the recursion, while the inductive case corresponds to a recursive subroutine call. A
recursive subroutine call, like the inductive case of an induction, reduces a problem to a
‘simpler’ or ‘smaller’ problem, which is closer to the base case.

3.6.2 Towers of Hanoi
Another standard example of recursion is the Towers of Hanoi problem. Let n be a positive integer. Imagine a set of n discs of decreasing size, piled up in order of size, with
the largest disc on the bottom and the smallest disc on top. The problem is to move this
tower of discs to a second pile, following certain rules: Only one disc can be moved at
a time, and a disc can only be placed on top of another disc if the disc on top is smaller.
While the discs are being moved from the first pile to the second pile, discs can be kept
in a third, spare pile. All the discs must at all times be in one of the three piles.
5 In the proof, we pretend that the data type int is not limited to 32 bits. In reality, the function only gives
the correct answer when the answer can be represented as a 32-bit binary number. This is the kind of implementation issue that matters in practice, especially in lower-level languages like Assembly.

3

80

Chapter 3. Proof



3

The Towers of Hanoi puzzle was first
published by Édouard Lucas in 1883.
The puzzle is based on a legend of
temple wherein there initially was one
pile of discs neatly sorted from largest
to smallest. In Lucas’s story, monks
have since been continuously moving
discs from this pile of 64 discs according to the rules of the puzzle to again created a sorted stack at the other end
of the temple. It is said that when the last disc is placed, the world will end.
But on the positive side, even if the monks move one disc every second, it
will take approximately 42 times the age of the universe until they are done.
And that is assuming they are using the optimal strategy…
Source: en.wikipedia.org/wiki/Tower_of_Hanoi

For example, if there are two discs, the problem can be solved by the following sequence of moves:
Move disc 1 from pile 1 to pile 3
Move disc 2 from pile 1 to pile 2
Move disc 1 from pile 3 to pile 2
A simple recursive subroutine can be used to write out the list of moves to solve the
problem for any value of n. The recursion is based on the observation that for n > 1, the
problem can be solved as follows: Move n − 1 discs from pile number 1 to pile number
3 (using pile number 2 as a spare). Then move the largest disc, disc number n, from
pile number 1 to pile number 2. Finally, move the n − 1 discs from pile number 3 to pile
number 2, putting them on top of the nth disc (using pile number 1 as a spare). In both
cases, the problem of moving n − 1 discs is a smaller version of the original problem and
so can be done by recursion. Here is the subroutine, written in Java:
1

3

5

void
//
//
//
if

Hanoi(int n, int A, int B, int C) {
List the moves for moving n discs from
pile number A to pile number B, using
pile number C as a spare. Assume n > 0.
(n == 1) {
System.out.println("Move disc 1 from pile " + A + " to pile " + B);

}
else {
Hanoi(n-1, A, C, B);
System.out.println("Move disc " + n + " from pile " +
A + " to pile " + B);
Hanoi(n-1, C, B, A);
}

7

9

11

13

}

3.6. Application: Recursion and Induction

81

This problem and its fame have led to implementations in a variety of languages, including a language called Brainf*ck.6 In the Computer Organisation
course, you can implement an interpreter for this language and test it on the
implementation of the Hanoi algorithm.



We can use induction to prove that this subroutine does in fact solve the Towers of
Hanoi problem.
Theorem 3.12. The sequence of moves printed by the Hanoi subroutine as given above correctly
solves the Towers of Hanoi problem for any integer n ≥ 1.
Proof. We prove by induction that whenever n is a positive integer and A, B, and C are the
numbers 1, 2, and 3 in some order, the subroutine call Hanoi(n, A, B, C) prints a sequence
of moves that will move n discs from pile A to pile B, following all the rules of the Towers
of Hanoi problem.
In the base case, n = 1, the subroutine call Hanoi(1, A, B, C) prints out the single step
“Move disc 1 from pile A to pile B”, and this move does solve the problem for 1 disc.
Let k be an arbitrary positive integer, and suppose that Hanoi(k, A, B, C) correctly
solves the problem of moving the k discs from pile A to pile B using pile C as the spare,
whenever A, B, and C are the numbers 1, 2, and 3 in some order. We need to show
that Hanoi(k + 1, A, B, C) correctly solves the problem for k + 1 discs. Since k + 1 > 1,
Hanoi(k + 1, A, B, C) begins by calling Hanoi(k, A, C, B). By the induction hypothesis,
this correctly moves k discs from pile A to pile C. disc number k + 1 is not moved during
this process. At that point, pile C contains the k smallest discs and pile A still contains the
(k + 1)st disc, which has not yet been moved. So the next move printed by the subroutine,
“Move disc (k + 1) from pile A to pile B”, is legal because pile B is empty. Finally, the
subroutine calls Hanoi(k, C, B, A), which, by the induction hypothesis, correctly moves
the k smallest discs from pile C to pile B, putting them on top of the (k + 1)st disc, which
does not move during this process. At that point, all (k + 1) discs are on pile B, so the
problem for k + 1 discs has been correctly solved.

3.6.3 Binary trees
Recursion is often used with linked data structures, which are data structures that are
constructed by linking several objects of the same type together with pointers. For an
example, we’ll look at the data structure known as a binary tree.


6 See

If you don’t already know about objects and pointers, you will not be able to
follow the rest of this section. Time to read about it on the internet?

en.wikipedia.org/wiki/Brainfuck

3

82

Chapter 3. Proof

A binary tree consists of nodes linked together in a tree-like structure. The nodes can
contain any type of data, but we will consider binary trees in which each node contains
an integer. A binary tree can be empty, or it can consist of a node (called the root of the
tree) and two smaller binary trees (called the left subtree and the right subtree of the
tree). You can already see the recursive structure: a tree can contain smaller trees. In
Java, the nodes of a tree can be represented by objects belonging to the class
2

3

4

class BinaryTreeNode {
int item;
// An integer value stored in the node.
BinaryTreeNode left;
// Pointer to left subtree.
BinaryTreeNode right; // Pointer to right subtree.
}

An empty tree is represented by a pointer that has the special value null. If root is a
pointer to the root node of a tree, then root.left is a pointer to the left subtree and root.right
is a pointer to the right subtree. Of course, both root.left and root.right can be null if the
corresponding subtree is empty. Similarly, root.item is a name for the integer in the root
node.



Binary trees are not part of the material for Reasoning & Logic and only serve
as examples of induction on algorithms here. You will study this material in
more detail in the course Algorithms & Data Structures.

Let’s say that we want a function that will find the sum of all the integers in all the
nodes of a binary tree. We can do this with a simple recursive function. The base case
of the recursion is an empty tree. Since there are no integers in an empty tree, the sum
of the integers in an empty tree is zero. For a non-empty tree, we can use recursion to
find the sums of the integers in the left and right subtrees, and then add those sums to
the integer in the root node of the tree. In Java, this can be expressed as follows:
1

3

5

7

9

11

13

int TreeSum( BinaryTreeNode root ) {
// Find the sum of all the integers in the
// tree that has the given root.
int answer;
if (root == null) { // The tree is empty.
answer = 0;
} else {
answer = TreeSum(root.left);
answer += TreeSum(root.right);
answer += root.item;
}
return answer;
}

We can use the second form of the principle of mathematical induction to prove that this
function is correct.

3.6. Application: Recursion and Induction

83

Theorem 3.13. The function TreeSum, defined above, correctly computes the sum of all the integers in a binary tree.
Proof. We use induction on the number of nodes in the tree. Let P(n) be the statement
“TreeSum correctly computes the sum of the nodes in any binary tree that contains exactly
n nodes”. We show that P(n) is true for every natural number n.
Consider the case n = 0. A tree with zero nodes is empty, and an empty tree is
represented by a null pointer. In this case, the if statement in the definition of TreeSum
assigns the value 0 to the answer, and this is the correct sum for an empty tree. So, P(0)
is true.
Let k be an arbitrary natural number, with k > 0. Suppose we already know P( x )
for each natural number x with 0 ≤ x < k. That is, TreeSum correctly computes the sum
of all the integers in any tree that has fewer than k nodes. We must show that it follows
that P(k) is true, that is, that TreeSum works for a tree with k nodes. Suppose that root
is a pointer to the root node of a tree that has a total of k nodes. Since the root node
counts as a node, that leaves a total of k − 1 nodes for the left and right subtrees, so each
subtree must contain fewer than k nodes. By the induction hypothesis, we know that
TreeSum(root.left) correctly computes the sum of all the integers in the left subtree, and
TreeSum(root.right) correctly computes the sum of all the integers in the right subtree.
The sum of all the integers in the tree is root.item plus the sums of the integers in the
subtrees, and this is the value computed by TreeSum. So, TreeSum does work for a tree
with k nodes. This completes the induction.
Note how closely the structure of the inductive proof follows the structure of the
recursive function. In particular, the second principle of mathematical induction is very
natural here, since the size of subtree could be anything up to one less than the size of
the complete tree. It would be very difficult to use the first principle of induction in a
proof about binary trees.

Exercises
1. The Hanoi subroutine given in this section does not just solve the Towers of Hanoi problem.
It solves the problem using the minimum possible number of moves. Use induction to prove
this fact.
2. Use induction to prove that the Hanoi subroutine uses 2n − 1 moves to solve the Towers of
Hanoi problem for n discs.
3. Consider the following recursive function:
1

3

5

int power(int x, int n) {
// Compute x raised to the power n.
// Assume that n >= 0.
int answer;
if (n == 0) {
answer = 1;

3

84
} else if (n % 2 == 0) {
answer = power(x * x, n / 2);
} else {
answer = x * power(x * x, (n-1) / 2);
}
return answer;

7

9

11

13

Chapter 3. Proof

}

Show that for any integer x and any non-negative integer n, the function power(x,n) correctly
computes the value of x n . (Assume that the int data type can represent arbitrarily large integers.) Note that the test “if (n % 2 == 0)” tests whether n is evenly divisible by 2. That is,
the test is true if n is an even number. (This function is actually a very efficient way to compute
x n .)

3

4. A leaf node in a binary tree is a node in which both the left and the right subtrees are empty.
Prove that the following recursive function correctly counts the number of leaves in a binary
tree:
1

3

5

7

9

11

13

int LeafCount(BinaryTreeNode root) {
// Counts the number of leaf nodes in
// the tree with the specified root.
int count;
if (root == null) {
count = 0;
} else if (root.left == null && root.right == null) {
count = 1;
} else {
count = LeafCount(root.left);
count += LeafCount(root.right);
}
return count;
}

5. A binary sort tree satisfies the following property: If node is a pointer to any node in the tree,
then all the integers in the left subtree of node are less than node.item and all the integers in the
right subtree of node are greater than or equal to node.item. Prove that the following recursive
subroutine prints all the integers in a binary sort tree in non-decreasing order:
2

4

6

8

10

12

void SortPrint(BinaryTreeNode root) {
// Assume that root is a pointer to the
// root node of a binary sort tree. This // subroutine prints the integers in
the
// tree in non -decreasing order.
if (root == null) {
// There is nothing to print.
}
else {
SortPrint(root.left);
System.out.println(root.item);
SortPrint(root.right);
}
}

3.7. Recursive Definitions

85

3

Figure 3.1: Fibonacci numbers occur in nature, as in this model of the florets in the
head of a sunflower.

3.7 Recursive Definitions
Recursion occurs in programming when a subroutine is defined—or at least partially
defined—in terms of itself. But recursion also occurs outside of programming. A recursive definition is a definition that includes a reference to the term that is being defined. A
recursive definition defines something at least partially in terms of itself. As in the case
of recursive subroutines, mathematical induction can often be used to prove facts about
things that are defined recursively.
As I already noted, there is a recursive definition for n!, for n in N, and we can use this
definition to prove facts about the factorials. We can define 0! = 1 and n! = n · (n − 1)!
for n > 0. Do you see how the base case and the inductive case in an inductive proof can

86

Chapter 3. Proof

correspond to the two parts of the recursive definition?
Other sequences of numbers can also be defined recursively. For example, the famous
Fibonacci sequence is the sequence of numbers f 0 , f 1 , f 2 , …, defined recursively by
f0 = 0
f1 = 1
f n = f n −1 + f n −2

3

for n > 1

Using this definition, we compute that
f2 = f1 + f0 = 0 + 1 = 1
f3 = f2 + f1 = 1 + 1 = 2
f4 = f3 + f2 = 2 + 1 = 3
f5 = f4 + f3 = 3 + 2 = 5
f6 = f5 + f4 = 5 + 3 = 8
f 7 = f 6 + f 5 = 8 + 5 = 13
and so on. Based on this definition, we can use induction to prove facts about the Fibonacci sequence. We can prove, for example, that f n grows exponentially with n, even
without finding an exact formula for f n :
( ) n −1
Theorem 3.14. The Fibonacci sequence, f 0 , f 1 , f 2 , …, satisfies f n > 32
, for n ≥ 6.
Proof. We prove this by induction on n. For n = 6, we have that f n = 8 while 1.5n−1 =
1.55 , which is about 7.6. So f n > 1.5n−1 for n = 6. Similarly, for n = 7, we have f n = 13
and 1.5n−1 = 1.56 , which is about 11.4. So f n > 1.5n−1 for n = 7.
Now suppose that k is an arbitrary integer with k > 7. Suppose that we already know
that f n > 1.5n−1 for n = k − 1 and for n = k − 2. We want to show that the inequality
then holds for n = k as well. But
f k = f k −1 + f k −2

> 1.5(k−1)−1 + 1.5(k−2)−1
= 1.5

k −2

+ 1.5

(by the induction hypothesis)

k −3

= (1.5) · (1.5k−3 ) + (1.5k−3 )
= (2.5) · (1.5k−3 )
> (1.52 ) · (1.5k−3 )

(since 1.52 = 2.25)

= 1.5k−1
This string of equalities and inequalities shows that f k > 1.5k−1 . This completes the
induction and proves the theorem.

3.8. Invariants

87

Exercises
1. Prove that the Fibonacci sequence, f 0 , f 1 , f 2 , …, satisfies f n < 2n for all natural numbers n ≥ 1.
2. Suppose that a1 , a2 , a3 , …, is a sequence of numbers which is defined recursively by a1 = 1
and an = 2an−1 + 2n−1 for n > 1. Prove that an = n2n−1 for every positive integer n.

3.8 Invariants
Recursion is closely linked to iteration. In fact, a while loop can be written as a recursive
subroutine (and this how the language Prolog achieves ‘iteration’: see page 77). In computer science we would like to prove correctness and other properties about algorithms.
Proofs about algorithms can be more difficult than the proofs about simple properties
of the integers that I often use as examples this book.
A tool that helps us to prove properties about algorithms is an invariant. For example, a loop invariant is a property P of a loop such that:
1. P is true before the loop is executed, and
2. P remains true after each execution of the body of the loop (but not necessarily
in-between the steps within the body).
So to prove that an algorithm A has the property Q (the post-condition), we can find an
invariant P of A such that Q follows from P, together with the fact that A has terminated.
This last fact that A has terminated means that the loop condition (the guard of the loop)
has become false.
In more detail, we need to find an invariant and show four things about it:
1. Initialization or basis property. The invariant holds before the first iteration of the
loop.
2. Maintenance or inductive property. If the invariant holds before an iteration, then it
also holds before the next iteration.
3. Termination and falsity of guard. After a finite number of iterations the guard becomes false and the loop terminates.
4. Correctness of the post-condition. The invariant together with the negation of the
guard imply that the post-condition holds, in which case the program is correct.



In one of the pencasts of this course, we prove the correctness of an algorithm
using an invariant. You can find that pencast here: youtu.be/GSvqF48TVM4.

As an example, consider the simple loop:

3

88

1

Chapter 3. Proof

while (x < 10)
x = x+1;

What does this loop achieve? What is an invariant that helps us to prove the loop
correctly achieves this? The invariant is x ≤ 10—check that it does satisfy the above four
properties!
While invariants can be useful, a suitable invariant can be difficult to find.

3



There is a form of logic, Floyd–Hoare logic, in which we can express invariants
and can formally prove the (partial) correctness of a program. Read about
it on wikipedia: en.wikipedia.org/wiki/Loop_invariant.

Chapter 4

Sets, Functions, and Relations
e deal with the complexity of the world by putting things into categories. There

W are not just hordes of individual creatures. There are dogs, cats, elephants, and

mice. There are mammals, insects, and fish. Animals, vegetables, and minerals. Solids,
liquids, and gases. Things that are red. Big cities. Pleasant memories…. Categories
build on categories. They are the subject and the substance of thought.
In mathematics, which operates in its own abstract and rigorous world, categories
are modelled by sets. A set is just a collection of elements. Along with logic, sets form
the ‘foundation’ of mathematics, just as categories are part of the foundation of day-today thought. In this chapter, we study sets and relationships among sets. And, yes, that
means we’ll prove theorems about sets!

4.1 Basic Concepts
A set is a collection of elements. A set is defined entirely by the elements that it contains.
An element can be anything, including another set. You will notice that this is not a
precise mathematical definition. Instead, it is an intuitive description of what the word
‘set’ is supposed to mean: any time you have a bunch of entities and you consider them as
a unit, you have a set. Mathematically, sets are really defined by the operations that can
be performed on them. These operations model things that can be done with collections
of objects in the real world. These operations are the subject of the branch of mathematics
known as set theory.
The most basic operation in set theory is forming a set from a given list of specific
entities. The set that is formed in this way is denoted by enclosing the list of entities
between a left brace, ‘{’, and a right brace, ‘}’. The entities in the list are separated by
commas. For example, the set denoted by

{ 17, π, New York City, King Willem-Alexander, Euromast }
89

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Chapter 4. Sets, Functions, and Relations
Figure 4.1: Venn diagram of an example set.

4

is the set that contains the entities 17, π, New York City, King Willem-Alexander, and
Euromast. These entities are the elements of the set. Since we assume that a set is completely defined by the elements that it contains, the set is well-defined. Of course, we
still haven’t said what it means to be an ‘entity’. Something as definite as ‘New York
City’ should qualify, except that it doesn’t seem like New York City really belongs in the
world of mathematics. The problem is that mathematics is supposed to be its own selfcontained world, but it is supposed to model the real world. When we use mathematics
to model the real world, we admit entities such as New York City and even Euromast.
But when we are doing mathematics per se, we’ll generally stick to obviously mathematical entities such as the integer 17 or the real number π. We will also use letters such
as a and b to refer to entities. For example, when I say something like “Let A be the set
{ a, b, c}”, I mean a, b, and c to be particular, but unspecified, entities.



It’s important to understand that a set is defined by the elements that it contains, and not by the order in which those elements might be listed. For
example, the notations { a, b, c, d} and {b, c, a, d} define the same set. Furthermore, a set can only contain one copy of a given element, even if the notation
that specifies the set lists the element twice. This means that { a, b, a, a, b, c, a}
and { a, b, c} specify exactly the same set. Note in particular that it’s incorrect
to say that the set { a, b, a, a, b, c, a} contains seven elements, since some of
the elements in the list are identical. The notation { a, b, c} can lead to some
confusion, since it might not be clear whether the letters a, b, and c are assumed to refer to three different entities. A mathematician would generally
not make this assumption without stating it explicitly, so that the set denoted
by { a, b, c} could actually contain either one, two, or three elements. When
it is important that different letters refer to different entities, I will say so
explicitly, as in “Consider the set { a, b, c}, where a, b, and c are distinct.”

4.1. Basic Concepts

91

4.1.1 Elements of sets
The symbol ∈ is used to express the relation ‘is an element of’. That is, if a is an entity
and A is a set, then a ∈ A is a statement that is true if and only if a is one of the elements
of A. In that case, we also say that a is a member of the set A. The assertion that a is not
an element of A is expressed by the notation a ̸∈ A. Note that both a ∈ A and a ̸∈ A are
statements in the sense of propositional logic. That is, they are assertions which can be
either true or false. The statement a ̸∈ A is equivalent to ¬( a ∈ A).



As you may have noticed by now, it is convention for sets to be denoted using
capital letters (e.g. ‘A’) and elements within sets to be denoted using lowercase letters (e.g. ‘a’). You should adhere to the same convention to prevent
misunderstandings!

It is possible for a set to be empty, that is, to contain no elements whatsoever. Since
a set is completely determined by the elements that it contains, there is only one set that
contains no elements. This set is called the empty set, and it is denoted by the symbol ∅.
Note that for any element a, the statement a ∈ ∅ is false. The empty set, ∅, can also be
denoted by an empty pair of braces, i.e., { }.
If A and B are sets, then, by definition, A is equal to B if and only if they contain
exactly the same elements. In this case, we write A = B. Using the notation of predicate
logic, we can say that A = B if and only if ∀ x ( x ∈ A ↔ x ∈ B).



Later, when proving theorems in set theory, we will find it can often help to
use this predicate logic notation to simplify our proofs. To avoid having to
look them up later, make sure that you understand why the predicate logic
notation is equivalent to the set notation.

Suppose now that A and B are sets such that every element of A is an element of
B. In that case, we say that A is a subset of B, i.e. A is a subset of B if and only if
∀ x ( x ∈ A → x ∈ B). The fact that A is a subset of B is denoted by A ⊆ B. Note that ∅
is a subset of every set B: x ∈ ∅ is false for any x, and so given any B, ( x ∈ ∅ → x ∈ B)
is true for all x.
If A = B, then it is automatically true that A ⊆ B and that B ⊆ A. The converse is
also true: If A ⊆ B and B ⊆ A, then A = B. This follows from the fact that for any x,
the statement ( x ∈ A ↔ x ∈ B) is logically equivalent to the statement ( x ∈ A → x ∈
B) ∧ ( x ∈ B → x ∈ A). This fact is important enough to state as a theorem.
Theorem 4.1. Let A and B be sets. Then A = B if and only if both A ⊆ B and B ⊆ A.
This theorem expresses the following advice: If you want to check that two sets, A
and B, are equal, you can do so in two steps. First check that every element of A is also

4

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Chapter 4. Sets, Functions, and Relations

an element of B, and then check that every element of B is also an element of A.
If A ⊆ B but A ̸= B, we say that A is a proper subset of B. We use the notation A ⊊ B
to mean that A is a proper subset of B. That is, A ⊊ B if and only if A ⊆ B ∧ A ̸= B.
We will sometimes use A ⊇ B as an equivalent notation for B ⊆ A, and A ⊋ B as an
equivalent for B ⊊ A. Other text books also sometimes use the ⊂ symbol to represent
proper subsets, e.g., A ⊂ B ≡ A ⊊ B. Additionally, you may come across A ̸⊆ B which
means that A is not a subset of B. Notice that (especially in written text) the difference
between A ⊊ B and A ̸⊆ B can be small, so make sure to read properly and to write
clearly!

4.1.2

4

Set-builder notation

A set can contain an infinite number of elements. In such a case, it is not possible to list all
the elements in the set: we cannot give an extensional definition of the set. Sometimes
the ellipsis ‘…’ is used to indicate a list that continues on infinitely. For example, N, the
set of natural numbers, can be specified as
N = {0, 1, 2, 3, . . . }
However, this is an informal notation, which is not really well-defined, and it should
only be used in cases where it is clear what it means. It’s not very useful to say that “the
set of prime numbers is {2, 3, 5, 7, 11, 13, . . . }”, and it is completely meaningless to talk
about “the set {17, 42, 105, . . . }”. Clearly, we need another way to specify sets besides
listing their elements. The need is fulfilled by predicates.
If P( x ) is a predicate, then we can form the set that contains all entities a such that a
is in the domain of discourse for P and P( a) is true. The notation { x | P( x )} is used to
denote this set. This is the intensional definition of the set. The name of the variable, x,
is arbitrary, so the same set could equally well be denoted as {z | P(z)} or {r | P(r )}. The
notation { x | P( x )} can be read “the set of x such that P( x )”. We call this the set-builder
notation, as you can think of the predicate as a building material for the elements of the
set. For example, if E( x ) is the predicate ‘x is an even number’, and if the domain of
discourse for E is the set N, then the notation { x | E( x )} specifies the set of even natural
numbers. That is,
{ x | E( x )} = {0, 2, 4, 6, 8, . . . }



It turns out, for deep and surprising reasons that we will discuss later, that
we have to be a little careful about what counts as a predicate. In order for
the notation { x | P( x )} to be valid, we have to assume that the domain of
discourse of P is in fact a set. (You might wonder how it could be anything
else. That’s the surprise!)

4.1. Basic Concepts

93

Often, it is useful to specify the domain of discourse explicitly in the notation that
defines a set. In the above example, to make it clear that x must be a natural number, we
could write the set as { x ∈ N | E( x )}. This notation can be read as “the set of all x in
N such that E( x )”. More generally, if X is a set and P is a predicate whose domain of
discourse includes all the elements of X, then the notation

{ x ∈ X | P( x )}
is the set that consists of all entities a that are members of the set X and for which P( a) is
true. In this notation, we don’t have to assume that the domain of discourse for P is a set,
since we are effectively limiting the domain of discourse to the set X. The set denoted
by { x ∈ X | P( x )} could also be written as { x | x ∈ X ∧ P( x )}.



We can use this notation to define the set of prime numbers in a rigorous
way. A prime number is a natural number n which is greater than 1 and
which satisfies the property that for any factorization n = xy, where x and y
are natural numbers, either x or y must be n. We can express this definition
as a predicate and define the set of prime numbers as

{ n ∈ N | ( n > 1) ∧
(
)
∀ x ∀y ( x ∈ N ∧ y ∈ N ∧ n = xy) → ( x = n ∨ y = n) }
Admittedly, this definition is hard to take in in one gulp. But this example
shows that it is possible to define complex sets using predicates.

4.1.3 Operations on sets
Now that we have a way to express a wide variety of sets, we turn to operations that can
be performed on sets. The most basic operations on sets are union and intersection. If
A and B are sets, then we define the union of A and B to be the set that contains all the
elements of A together with all the elements of B. The union of A and B is denoted by
A ∪ B. The union can be defined formally as
A ∪ B = { x | x ∈ A ∨ x ∈ B }.
The intersection of A and B is defined to be the set that contains every entity that is both
a member of A and a member of B. The intersection of A and B is denoted by A ∩ B.
Formally,
A ∩ B = { x | x ∈ A ∧ x ∈ B }.

4

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Chapter 4. Sets, Functions, and Relations

An entity gets into A ∪ B if it is in either A or B. It gets into A ∩ B if it is in both A and B.
Note that the symbol for the logical ‘or’ operator, ∨, is similar to the symbol for the union
operator, ∪, while the logical ‘and’ operator, ∧, is similar to the intersection operator, ∩.
The set difference of two sets, A and B, is defined to be the set of all entities that are
members of A but are not members of B. The set difference of A and B is denoted A ∖ B
or alternatively as A − B. The idea is that A ∖ B is formed by starting with A and then
removing any element that is also found in B. Formally,
A ∖ B = { x | x ∈ A ∧ x ̸ ∈ B }.
Union and intersection are clearly commutative operations. That is, A ∪ B = B ∪ A and
A ∩ B = B ∩ A for any sets A and B. However, set difference is not commutative. In
general, A ∖ B ̸= B ∖ A.

4



Suppose that A = { a, b, c}, that B = {b, d}, and that C = {d, e, f }. Then we
can apply the definitions of union, intersection, and set difference to compute, for example, that:
A ∪ B = { a, b, c, d}

A ∩ B = {b}

A ∖ B = { a, c}

A ∪ C = { a, b, c, d, e, f }

A∩C = ∅

A ∖ C = { a, b, c}

In this example, the sets A and C have no elements in common, so that A ∩ C = ∅.
There is a term for this: Two sets are said to be disjoint if they have no elements in
common. That is, for any sets A and B, A and B are said to be disjoint if and only if
A ∩ B = ∅.
Of course, the set operations can also be applied to sets that are defined by predicates.
The next example illustrates this.



let L( x ) be the predicate ‘x is lucky’, and let W ( x ) be the predicate ‘x is wise’,
where the domain of discourse for each predicate is the set of people. Let
X = { x | L( x )}, and let Y = { x | W ( x )}. Then
X ∪ Y = { x | L( x ) ∨ W ( x )} = {people who are lucky or wise}
X ∩ Y = { x | L( x ) ∧ W ( x )} = {people who are lucky and wise}
X ∖ Y = { x | L( x ) ∧ ¬W ( x )} = {people who are lucky but not wise}
Y ∖ X = { x | W ( x ) ∧ ¬ L( x )} = {people who are wise but not lucky}

4.1. Basic Concepts

Notation

95

Definition

a∈A

a is a member (or element) of A

a ̸∈ A

¬( a ∈ A), a is not a member of A

∅

the empty set, which contains no elements

A⊆B

A is a subset of B, ∀ x ( x ∈ A → x ∈ B)

A⊊B

A is a proper subset of B, A ⊆ B ∧ A ̸= B

A⊇B

A is a superset of B, same as B ⊆ A

A⊋B

A is a proper superset of B, same as B ⊊ A

A=B

A and B have the same members, A ⊆ B ∧ B ⊆ A

A∪B

union of A and B, { x | x ∈ A ∨ x ∈ B}

A∩B

intersection of A and B, { x | x ∈ A ∧ x ∈ B}

A∖B

set difference of A and B, { x | x ∈ A ∧ x ̸∈ B}

P( A )

power set of A, { X | X ⊆ A}

Figure 4.2: Some of the notations that are defined in this section. A and B are sets, and
a is an entity.



You have to be a little careful with the English word ‘and’. We might say that
the set X ∪ Y contains people who are lucky and people who are wise. But
what this means is that a person gets into the set X ∪ Y either by being lucky
or by being wise, so X ∪ Y is defined using the logical ‘or’ operator, ∨.

4.1.4 Visualising sets1
A Venn diagram shows all possible logical relations between a finite collection of different
sets. These diagrams depict elements as points in the plane, and sets as regions inside
closed curves. So a Venn diagram consists of multiple overlapping closed curves, usually
circles, each representing a set. The points inside a curve (circle) labelled S represent
elements of the set S, while points outside the boundary represent elements not in the
set S. Figure 4.1 shows our example set which opened the section.
Venn diagrams help us to visualise sets and set operations. For example, the set of
all elements that are members of both sets S and T, S ∩ T, is represented visually by the
1 This

subsection is derived from en.wikipedia.org/wiki/Venn_diagram.

4

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Chapter 4. Sets, Functions, and Relations

4
Figure 4.3: Venn diagram of the intersection of two sets.

area of overlap of the regions S and T: see Figure 4.3. In Venn diagrams the curves are
overlapped in every possible way, showing all possible relations between the sets. You
can find it useful to draw a Venn diagram to gain intuition of what’s happening. On their
own, Venn diagrams do not offer a proof for theorems in set theory however.

4.1.5

Sets of sets

Sets can contain other sets as elements. For example, the notation { a, {b}} defines a set
that contains two elements, the entity a and the set {b}. Since the set {b} is a member
of the set { a, {b}}, we have that {b} ∈ { a, {b}}. On the other hand, provided that a ̸= b,
the statement {b} ⊆ { a, {b}} is false, since saying {b} ⊆ { a, {b}} is equivalent to saying
that b ∈ { a, {b}}, and the entity b is not one of the two members of { a, {b}}. For the
entity a, it is true that { a} ⊆ { a, {b}} and for the set {b}, it is true that {{b}} ⊆ { a, {b}}.
Study these examples carefully before you continue, as many students struggle with the
notion and notation of putting sets in sets.
Given a set A, we can construct the set that contains all the subsets of A. This set is
called the power set of A, and is denoted P( A). Formally, we define
P( A ) = { X | X ⊆ A }.

4.1. Basic Concepts



97

For example, if A = { a, b}, then the subsets of A are the empty set, { a}, {b},
and { a, b}, so the power set of A is set given by
P( A) = { ∅, { a}, {b}, { a, b} }.
So the power set of of A has four elements. Does this surprise you?

Note that since the empty set is a subset of any set, the empty set is an element of the
power set of any set. That is, for any set A, ∅ ⊆ A and ∅ ∈ P( A). Since the empty set
is a subset of itself, and is its only subset, we have that P(∅) = {∅}. The set {∅} is not
empty. It contains one element, namely ∅.



4
The Nobel Prize was won by Bertrand Russell (1872–
1970), a dominant figure in British thought during
the twentieth century. Russell was a philosopher and
mathematician, and also a historian, social critic and
political activist. With A. N. Whitehead, Russell wrote
Principia Mathematica, an epic attempt to create a logical basis for mathematics. His work has had a considerable influence on computer science, and not just for
his contributions to logic and set theory: he proposed
the beginnings of what are now type systems.
Source: en.wikipedia.org/wiki/Bertrand_Russell.

We remarked earlier in this section that the notation { x | P( x )} is only valid if the
domain of discourse of P is a set. This might seem a rather puzzling thing to say—after
all, why and how would the domain of discourse be anything else? The answer is related
to Russell’s Paradox, which we mentioned briefly in Chapter 3 and which shows that it is
logically impossible for the set of all sets to exist. This impossibility can be demonstrated
using a proof by contradiction. In the proof, we use the existence of the set of all sets
to define another set which cannot exist because its existence would lead to a logical
contradiction.
Theorem 4.2. There is no set of all sets.
Proof. Suppose that the set of all sets exists. We will show that this assumption leads to

98

Chapter 4. Sets, Functions, and Relations

a contradiction. Let V be the set of all sets. We can then define the set R to be the set
which contains every set that does not contain itself. That is,
R = { X ∈ V | X ̸∈ X }

4

Now, we must have either R ∈ R or R ̸∈ R. We will show that either case leads to a
contradiction.
Consider the case where R ∈ R. Since R ∈ R, R must satisfy the condition for membership in R. A set X is in R iff X ̸∈ X. To say that R satisfies this condition means that
R ̸∈ R. That is, from the fact that R ∈ R, we deduce the contradiction that R ̸∈ R.
Now consider the remaining case, where R ̸∈ R. Since R ̸∈ R, R does not satisfy
the condition for membership in R. Since the condition for membership is that R ̸∈ R,
and this condition is false, the statement R ̸∈ R must be false. But this means that the
statement R ∈ R is true. From the fact that R ̸∈ R, we deduce the contradiction that
R ∈ R.
Since both possible cases, R ∈ R and R ̸∈ R, lead to contradictions, we see that it is
not possible for R to exist. Since the existence of R follows from the existence of V, we
see that V also cannot exist.



This (in)famous contradiction has been adapted to natural language to make
it easier to convey the problem to laymen. Unfortunately many of these translations are flawed. Can you think of a solution for the following for instance?
“The barber of Seville shaves all men who do not shave themselves. Who
shaves the barber?”

To avoid Russell’s paradox, we must put limitations on the construction of new sets.
We can’t force the set of all sets into existence simply by thinking of it. We can’t form the
set { x | P( x )} unless the domain of discourse of P is a set. Any predicate Q can be used
to form a set { x ∈ X | Q( x )}, but this notation requires a pre-existing set X. Predicates
can be used to form subsets of existing sets, but they can’t be used to form new sets
completely from scratch.
The notation { x ∈ A | P( x )} is a convenient way to effectively limit the domain of
discourse of a predicate, P, to members of a set, A, that we are actually interested in. We
will use a similar notation with the quantifiers ∀ and ∃. The proposition (∀ x ∈ A)( P( x ))
is true if and only if P( a) is true for every element a of the set A. And the proposition
(∃ x ∈ A)( P( x )) is true if and only if there is some element a of the set A for which P( a)
is true. These notations are valid only when A is contained in the domain of discourse
for P. As usual, we can leave out parentheses when doing so introduces no ambiguity.
So, for example, we might write ∀ x ∈ A P( x ).

4.1. Basic Concepts

99

4.1.6 Mathematical induction revisited
We end this section by returning to the topic of mathematical induction. First, I will give
proofs of the two forms of the principle of mathematical induction. These proofs were
omitted from the previous chapter, but only for the lack of a bit of set notation. In fact,
the principle of mathematical induction is valid only because it follows from one of the
basic axioms that define the natural numbers, namely the fact that any non-empty set
of natural numbers has a smallest element. Given this axiom, we can use it to prove the
following two theorems:
Theorem 4.3. Let P be a one-place
predicate whose domain )of discourse includes the natural
(
numbers. Suppose that P(0) ∧ ∀k ∈ N ( P(k) → P(k + 1)) . Then ∀n ∈ N, P(n).
(
)
(
Proof. Suppose that both P(0) and ∀k ∈ N ( P(k) → P(k + 1)) are true, but that ∀n ∈
)
N, P(n) is false. We show that this assumption leads to a contradiction.
Since the statement ∀n ∈ N, P(n) is false, its negation, ¬(∀n ∈ N, P(n)), is true.
The negation is equivalent to ∃n ∈ N, ¬ P(n). Let X = {n ∈ N | ¬ P(n)}. Since ∃n ∈
N, ¬ P(n) is true, we know that X is not empty. Since X is a non-empty set of natural
numbers, it has a smallest element. Let x be the smallest element of X. That is, x is
the smallest natural number such that P( x ) is false. Since we know that P(0) is true, x
cannot be 0. Let y = x − 1. Since x ̸= 0, y is a natural number. Since y < x, we know,
by the definition of x, that P(y) is true. We also know that ∀k ∈ N ( P(k ) → P(k + 1))
is true. In particular, taking k = y, we know that P(y) → P(y + 1). Since P(y) and
P(y) → P(y + 1), we deduce by modus ponens that P(y + 1) is true. But y + 1 = x,
so we have deduced that P( x ) is true. This contradicts the fact that P( x ) is false. This
contradiction proves the theorem.
Theorem 4.4. Let P be a one-place predicate whose domain of discourse includes the natural
numbers. Suppose that P(0) is true and that

( P(0) ∧ P(1) ∧ · · · ∧ P(k)) → P(k + 1)
is true for each natural number k ≥ 0. Then it is true that ∀n ∈ N, P(n).
Proof. Suppose that P is a predicate that satisfies the hypotheses of the theorem, and
suppose that the statement ∀n ∈ N, P(n) is false. We show that this assumption leads
to a contradiction.
Let X = {n ∈ N | ¬ P(n)}. Because of the assumption that ∀n ∈ N, P(n) is false, X is
non-empty. It follows that X has a smallest element. Let x be the smallest element of X.
The assumption that P(0) is true means that 0 ̸∈ X, so we must have x > 0. Since x is the
smallest natural number for which P( x ) is false, we know that P(0), P(1), …, and P( x − 1)
are all true. From this and the fact that ( P(0) ∧ P(1) ∧ · · · ∧ P( x − 1)) → P( x ), we deduce
that P( x ) is true. But this contradicts the fact that P( x ) is false. This contradiction proves
the theorem.

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4.1.7

4

Chapter 4. Sets, Functions, and Relations

Structural Induction

Next, while we are on the topic of induction, let’s generalise the idea of induction to also
apply it to sets. This more general form of induction is often called structural induction.
Structural induction is used to prove that some proposition P( x ) holds for all x of some
sort of recursively defined structure, such as formulae, lists, or trees—or recursivelydefined sets. In a proof by structural induction we show that the proposition holds for
all the ‘minimal’ structures, and that if it holds for the immediate substructures of a
certain structure S, then it must hold for S also. Structural induction is useful for proving properties about algorithms; sometimes it is used together with invariants for this
purpose.
To get an idea of what a ‘recursively defined set’ might look like, consider the following definition of the set of natural numbers N.
Basis: 0 ∈ N.
Succession: x ∈ N → x + 1 ∈ N.
Exclusivity: No other elements other than those outlined by the rules above are in N.
This definition is similar to one we have seen before, first stating that 0 ∈ N and then
saying that we can add 1 to an element in N to get another element of N. The final clause
is needed to ensure that other items are not part of N. Without it, you and me, as well
as π, ‘New York City’, and ‘King Willem-Alexander’ might have been in the set. After
all there was no reason for those elements not to be in there.
Now compare that recursive definition, with the method for mathematical induction
we have seen before:
Base case: Prove that P(0) holds.
Inductive case: Prove that ∀k ∈ N( P(k) → P(k + 1)) holds.
Conclusion: ∀n ∈ NP(n)) holds.
As we can see mathematical induction and this recursive definition show large similarities. The base case of the induction proves the property for the basis of our recursive
definition and the inductive step proves the property for the succession rule. In fact,
this similarity is no coincidence and we can generalise this method to get to structural
induction.
Consider for instance the set PROP, which represents all valid formula in propositional logic:
Atoms: pi ∈ PROP for all i ∈ N.
Negation: x ∈ PROP → ¬ x ∈ PROP.
Binary connective: x, y ∈ PROP → ( x ∗ y) ∈ PROP, s.t. ∗ ∈ {∧, ∨, →, ↔}.

4.1. Basic Concepts

101

Exclusivity: Nothing else is in PROP.
Using this definition of the set PROP we can use structural induction to prove certain
claims about PROP. For instance we can prove that every formula in PROP has equally
many left parentheses ‘(’ and right parentheses ‘)’.
Proof. Let l (ϕ) denote the number of left parentheses in a formula ϕ. Similarly let r (ϕ)
denote the number of right parentheses. Let P(ϕ) be the statement that l (ϕ) = r (ϕ). We
need to prove that ∀ϕ ∈ PROP( P(ϕ)).
Base case: Consider the Atoms rule of the definition of PROP: l ( pi ) = 0 = r ( pi ).
Therefore P( pi ) holds.
Inductive case: We want to show that if the statement is true for x, y ∈ PROP (where
x and y are arbitrary formula), then it is true for ¬ x and ( x ∗ y) for all ∗ ∈ {∨, ∧, →, ↔}.
That is, we must prove the implication ( P( x ) ∧ P(y)) → ( P(¬ x ) ∧ P(( x ∗ y))). So we
assume P( x ) ∧ P(y), that is, we assume that for both formula x and y: l ( x ) = r ( x ) and
l (y) = r (y). We want to prove P(¬ x ), that is, that for ¬ x l (¬ x ) = r (¬ x )
l (¬ x ) = l ( x )

= r(x)
= r (¬ x )

by the Negation rule of PROP
by the inductive hypothesis
by the Negation rule of PROP

Secondly we prove that P(( x ∗ y)) holds for all ∗ ∈ {∨, ∧, →, ↔}:
l (( x ∗ y)) = 1 + l ( x ) + l (y)

by the Binary connective rule of PROP

= 1 + r ( x ) + r (y)
= r (( x ∗ y))

by the inductive hypothesis
by the Binary connective rule of PROP

Altogether, we have shown that P( pi ) holds and that, for all x, y ∈ PROP and ∗ ∈
{∨, ∧, →, ↔}, ( P( x ) ∧ P(y)) → ( P(¬ x ) ∧ P(( x ∗ y)) is true. Therefore, by the principle
of structural induction, P(ϕ) is true for all ϕ ∈ PROP, so for all propositional formula
the number of left parentheses equals the number of right parentheses. This completes
the proof by structural induction.
Such structural induction proofs can be applied on any recursively defined set of
numbers, formulae or even strings (pieces of text) or lists or trees, making this a very
powerful generalised proof method.

Exercises
1. If we don’t make the assumption that a, b, and c are distinct, then the set denoted by { a, b, c}
might actually contain either 1, 2, or 3 elements. How many different elements might the set
{ a, b, { a}, { a, c}, { a, b, c} } contain? Explain your answer.
2. Compute A ∪ B, A ∩ B, and A ∖ B for each of the following pairs of sets

4

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Chapter 4. Sets, Functions, and Relations
a)
b)
c)
d)

A
A
A
A

= { a, b, c}, B = ∅
= {1, 2, 3, 4, 5}, B = {2, 4, 6, 8, 10}
= { a, b}, B = { a, b, c, d}
= { a, b, { a, b}}, B = {{ a}, { a, b}}

3. Draw a Venn diagram for each of the four sets of the last exercise.
4. Recall that N represents the set of natural numbers. That is, N = {0, 1, 2, 3, . . . }. Let X =
{n ∈ N | n ≥ 5}, let Y = {n ∈ N | n ≤ 10}, and let Z = {n ∈ N | n is an even number}. Find
each of the following sets:
a) X ∩ Y
b) X ∪ Y
c) X ∖ Y
d) N ∖ Z
e) X ∩ Z
f) Y ∩ Z
g) Y ∪ Z
h) Z ∖ N
(
)
5. Find P {1, 2, 3} . (Hint: It has eight elements.)

4

6. Assume that a and b are entities and that a ̸= b. Let A and B be the sets defined by A =
{ a, {b}, { a, b} } and B = { a, b, { a, {b}} }. Determine whether each of the following statements is true or false. Explain your answers.
a) b ∈ A
b) { a, b} ⊆ A
c) { a, b} ⊆ B
d) { a, b} ∈ B
e) { a, {b}} ∈ A
f) { a, {b}} ∈ B
(
)
(
)
(
)
7. Since P( A) is a set, it is possible to form the set P P( A) . What is P P(∅) ? What is P P({ a, b}) ?
(Hint: It has sixteen elements.)
8. In the English sentence, “She likes men who are tall, dark, and handsome”, does she like an
intersection or a union of sets of men? How about in the sentence, “She likes men who are tall,
men who are dark, and men who are handsome”?
9. If A is any set, what can you say about A ∪ A ? About A ∩ A ? About A ∖ A ? Why?
10. Suppose that A and B are sets such that A ⊆ B. What can you say about A ∪ B ? About A ∩ B ?
About A ∖ B ? Why?
11. Suppose that A, B, and C are sets. Show that C ⊆ A ∩ B if and only if (C ⊆ A) ∧ (C ⊆ B).
12. Suppose that A, B, and C are sets, and that A ⊆ B and B ⊆ C. Show that A ⊆ C.
13. Suppose that A and B are sets such that A ⊆ B. Is it necessarily true that P( A) ⊆ P( B) ? Why
or why not?
14. Let M be any natural number, and let P(n) be a predicate whose domain of discourse includes
all natural numbers greater than or equal to M. Suppose that P( M) is true, and suppose that
P(k) → P(k + 1) for all k ≥ M. Show that P(n) is true for all n ≥ M.
15. Prove that the number of propositional variables is always at most one more than the number
of connectives for every formula ϕ ∈ PROP.

4.2

The Boolean Algebra of Sets

It is clear that set theory is closely related to logic. The intersection and union of sets can
be defined in terms of the logical ‘and’ and logical ‘or’ operators. The notation { x | P( x )}
makes it possible to use predicates to specify sets. And if A is any set, then the formula
x ∈ A defines a one place predicate that is true for an entity x if and only if x is a member

4.2. The Boolean Algebra of Sets

103

of A. So it should not be a surprise that many of the rules of logic have analogues in set
theory.
For example, we have already noted that ∪ and ∩ are commutative operations. This
fact can be verified using the rules of logic. Let A and B be sets. According to the (definition of equality of sets, we can show that A ∪ B = B ∪ A by showing that ∀ x ( x ∈
)
A ∪ B) ↔ ( x ∈ B ∪ A) . But for any x,
x ∈ A∪B ↔ x ∈ A∨x ∈ B

↔ x ∈ B∨x ∈ A
↔ x ∈ B∪A

(definition of ∪)
(commutativity of ∨)
(definition of ∪)

The commutativity of ∩ follows in the same way from the definition of ∩ in terms of ∧
and the commutativity of ∧, and a similar argument shows that union and intersection
are associative operations.
The distributive laws for propositional logic give rise to two similar rules in set theory.
Let A, B, and C be any sets. Then
A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C )
and
A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C )
These rules are called the distributive laws for set theory. To verify the first of these laws,
we just have to note that for any x,
x ∈ A ∪ (B ∩ C)

↔ ( x ∈ A) ∨ (( x ∈ B) ∧ ( x ∈ C ))
↔ (( x ∈ A) ∨ ( x ∈ B)) ∧ (( x ∈ A) ∨ ( x ∈ C ))
↔ ( x ∈ A ∪ B) ∧ ( x ∈ A ∪ C )
↔ x ∈ (( A ∪ B) ∩ ( A ∪ C ))

(definition of ∪, ∩)
(distributivity of ∨)
(definition of ∪)
(definition of ∩)

The second distributive law for sets follows in exactly the same way.

4.2.1 Set complement
While ∪ is analogous to ∨ and ∩ is analogous to ∧, we have not yet seen any operation in
set theory that is analogous to the logical ‘not’ operator, ¬. Given a set A, it is tempting
to try to define { x | ¬( x ∈ A)}, the set that contains everything that does not belong to A.
Unfortunately, the rules of set theory do not allow us to define such a set. The notation
{ x | P( x )} can only be used when the domain of discourse of P is a set, so there must be
an underlying set from which the elements that are/are not in A are chosen, i.e., some

4

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Chapter 4. Sets, Functions, and Relations

underlying set of which A is a subset. We can get around this problem by restricting
the discussion to subsets of some fixed set. This set will be known as the universal set.
Keep in mind that the universal set is only universal for some particular discussion. It is
simply some set that is large enough to contain all the sets under discussion as subsets.
Given a universal set U and any subset A of U, we can define the set { x ∈ U | ¬( x ∈ A)}.
Definition 4.1. Let U be a given universal set, and let A be any subset of U. We define
the complement of A in U to be the set A that is defined by A = { x ∈ U | x ̸∈ A}.

4

Usually, we will refer to the complement of A in U simply as the complement of A, but
you should remember that whenever complements of sets are used, there must be some
universal set in the background. Other textbooks may use Ac to denote the complement
of A instead.
Given the complement operation on sets, we can look for analogues to the rules of
logic that involve negation. For example, we know that p ∧ ¬ p = F for any proposition p.
It follows that for any subset A of U,
A ∩ A = { x ∈ U | ( x ∈ A) ∧ ( x ∈ A)}

= { x ∈ U | ( x ∈ A) ∧ ( x ̸∈ A)}
= { x ∈ U | ( x ∈ A) ∧ ¬( x ∈ A)}
=∅

(definition of ∩)
(definition of complement)
(definition of ̸∈)

the last equality following because the proposition ( x ∈ A) ∧ ¬( x ∈ A) is false for any
x. Similarly, we can show that A ∪ A = U and that A = A (where A is the complement
of the complement of A, that is, the set obtained by taking the complement of A.)
The most important laws for working with complements of sets are DeMorgan’s Laws
for sets. These laws, which follow directly from DeMorgan’s Laws for logic, state that
for any subsets A and B of a universal set U,
A∪B = A∩B
and
A∩B = A∪B
For example, we can verify the first of these laws with the calculation
A ∪ B = { x ∈ U | x ̸∈ ( A ∪ B)}

= {x
= {x
= {x
= {x

∈ U | ¬( x ∈ A ∪ B)}
∈ U | ¬( x ∈ A ∨ x ∈ B)}
∈ U | (¬( x ∈ A)) ∧ (¬( x ∈ B))}
∈ U | ( x ̸∈ A) ∧ ( x ̸∈ B)}

(definition of complement)
(definition of ̸∈)
(definition of ∪)
(DeMorgan’s Law for logic)
(definition of ̸∈)

= { x ∈ U | ( x ∈ A) ∧ ( x ∈ B)}

(definition of complement)

= A∩B

(definition of ∩)

4.2. The Boolean Algebra of Sets

105

Double complement

A=A

Miscellaneous laws

A∪A = U
A∩A = ∅
∅∪A = A
∅∩A = ∅

Idempotent laws

A∩A = A
A∪A = A

Commutative laws

A∩B = B∩A
A∪B = B∪A

Associative laws

A ∩ ( B ∩ C ) = ( A ∩ B) ∩ C
A ∪ ( B ∪ C ) = ( A ∪ B) ∪ C

Distributive laws

A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C )
A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C )

DeMorgan’s laws

A∩B = A∪B

A∪B = A∩B
Figure 4.4: Some Laws of Boolean Algebra for sets. A, B, and C are sets. For the laws
that involve the complement operator, they are assumed to be subsets of some universal
set, U. For the most part, these laws correspond directly to laws of Boolean Algebra for
propositional logic as given in Figure 2.2.

An easy inductive proof can be used to verify generalized versions of DeMorgan’s
Laws for set theory. (In this context, all sets are assumed to be subsets of some unnamed
universal set.) A simple calculation verifies DeMorgan’s Law for three sets:
A ∪ B ∪ C = ( A ∪ B) ∪ C

= ( A ∪ B) ∩ C

(by DeMorgan’s Law for two sets)

= ( A ∩ B) ∩ C

(by DeMorgan’s Law for two sets)

= A∩B∩C
From there, we can derive similar laws for four sets, five sets, and so on. However, just
saying ‘and so on’ is not a rigorous proof of this fact. Whereas we may have excused
ourselves about that in Chapter 2, we can now prove this fact. Here is a rigorous inductive proof of a generalized DeMorgan’s Law:

4

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Chapter 4. Sets, Functions, and Relations

Theorem 4.5. For any natural number n ≥ 2 and for any sets X1 , X2 , …, Xn ,
X1 ∪ X2 ∪ · · · ∪ X n = X1 ∩ X2 ∩ · · · ∩ X2
Proof. We give a proof by induction. In the base case, n = 2, the statement is that
X1 ∪ X2 = X1 ∩ Xn . This is true since it is just an application of DeMorgan’s law for
two sets.
For the inductive case, suppose that the statement is true for n = k. We want to show
that it is true for n = k + 1. Let X1 , X2 , …, Xk+1 be any k + 1 sets. Then we have:
X1 ∪ X2 ∪ · · · ∪ X k + 1 = ( X1 ∪ X2 ∪ · · · ∪ X k ) ∪ X k + 1

= ( X1 ∪ X2 ∪ · · · ∪ X k ) ∩ X k + 1

4

= ( X1 ∩ X2 ∩ · · · ∩ X k ) ∩ X k + 1
= X1 ∩ X2 ∩ · · · ∩ X k + 1
In this computation, the second step follows by DeMorgan’s Law for two sets, while the
third step follows from the induction hypothesis. Therefore by the principle of induction
we have proven the theorem.

4.2.2

Link between logic and set theory

Just as the laws of logic allow us to do algebra with logical formulas, the laws of set
theory allow us to do algebra with sets. Because of the close relationship between logic
and set theory, their algebras are very similar. The algebra of sets, like the algebra of
logic, is Boolean algebra. When George Boole wrote his 1854 book about logic, it was
really as much about set theory as logic. In fact, Boole did not make a clear distinction
between a predicate and the set of objects for which that predicate is true. His algebraic
laws and formulas apply equally to both cases. More exactly, if we consider only subsets
of some given universal set U, then there is a direct correspondence between the basic
symbols and operations of propositional logic and certain symbols and operations in set
theory, as shown in this table:
Logic

Set Theory

T

U

F

∅

p∧q

A∩B

p∨q

A∪B

¬p

A

4.2. The Boolean Algebra of Sets

107

4

Figure 4.5: The correspondence between set operators and propositional operators.
Source: commons.wikimedia.org/wiki/Template:Operations_and_
relations_in_set_theory_and_logic

108

Chapter 4. Sets, Functions, and Relations

Any valid logical formula or computation involving propositional variables and the symbols T, F, ∧, ∨, and ¬ can be transformed into a valid formula or computation in set
theory by replacing the propositions in the formula with subsets of U and replacing the
logical symbols with U, ∅, ∩, ∪, and the complement operator. Figure 4.5 illustrates.

4

Just as in logic, the operations of set theory can be combined to form complex expressions such as ( A ∪ C ) ∩ ( B ∪ C ∪ D ). Parentheses can always be used in such expressions
to specify the order in which the operations are to be performed. In the absence of parentheses, we need precedence rules to determine the order of operation. The precedence
rules for the Boolean algebra of sets are carried over directly from the Boolean algebra of
propositions. When union and intersection are used together without parentheses, intersection has precedence over union. Furthermore, when several operators of the same
type are used without parentheses, then they are evaluated in order from left to right.
(Of course, since ∪ and ∩ are both associative operations, it really doesn’t matter whether
the order of evaluation is left-to-right or right-to-left.) For example, A ∪ B ∩ C ∪ D is
evaluated as ( A ∪ ( B ∩ C )) ∪ D. The complement operation is a special case. Since it is
denoted by drawing a line over its operand, there is never any ambiguity about which
part of a formula it applies to.



Unfortunately in hand-written work this is not always true. Make sure to
write neatly and unambiguously when working with complements. Also
note that, similarly to parentheses in propositional logic, although parentheses may not be necessary we strongly encourage you to add them them
to improve readability of the order of operations.

The laws of set theory can be used to simplify complex expressions involving sets.
(As usual, of course, the meaning of ‘simplification’ is partly in the eye of the beholder.)
For example, for any sets X and Y,

( X ∪ Y ) ∩ (Y ∪ X ) = ( X ∪ Y ) ∩ ( X ∪ Y )
= (X ∪ Y)

(Commutative Law)
(Idempotent Law)

where in the second step, the Idempotent Law, which says that A ∩ A = A, is applied
with A = X ∪ Y. For expressions that use the complement operation, it is usually considered to be simpler to apply the operation to an individual set, as in A, rather than to
a formula, as in A ∩ B. DeMorgan’s Laws can always be used to simplify an expression

4.2. The Boolean Algebra of Sets

109

in which the complement operation is applied to a formula. For example,
A ∩ B ∪ A = A ∩ ( B ∩ A)

(DeMorgan’s Law)

= A ∩ ( B ∩ A)

(Double Complement)

= A ∩ ( A ∩ B)

(Commutative Law)

= ( A ∩ A) ∩ B)

(Associative Law)

= A∩B

(Idempotent Law)

As a final example of the relationship between set theory and logic, consider the
set-theoretical expression A ∩ ( A ∪ B) and the corresponding compound proposition
p ∧ ( p ∨ q). (These correspond since for any x, x ∈ A ∩ ( A ∪ B) ≡ ( x ∈ A) ∧ (( x ∈
A) ∨ ( x ∈ B)).) You might find it intuitively clear that A ∩ ( A ∪ B) = A. Formally,
this follows from the fact that p ∧ ( p ∨ q) ≡ p, which might be less intuitively clear and
is surprising difficult to prove algebraically from the laws of logic. However, there is
another way to check that a logical equivalence is valid: Make a truth table. Consider a
truth table for p ∧ ( p ∨ q):
p

q

p∨q

p ∧ ( p ∨ q)

0
0
1
1

0
1
0
1

0
1
1
1

0
0
1
1

The fact that the first column and the last column of this table are identical shows that
p ∧ ( p ∨ q) ≡ p. Taking p to be the proposition x ∈ A and q to be the proposition x ∈ B, it
follows that the sets A and A ∩ ( A ∪ B) have the same members and therefore are equal.



In one of the pencasts of the course, we describe how Venn diagrams can also
be used as alternatives to truth tables based on an old exam question. You
can find that pencast here: youtu.be/5xle5qfrh0k.

Exercises
1. Use the laws of logic to verify the associative laws for union and intersection. That is, show
that if A, B, and C are sets, then A ∪ ( B ∪ C ) = ( A ∪ B) ∪ C and A ∩ ( B ∩ C ) = ( A ∩ B) ∩ C.
2. Show that for any sets A and B, A ⊆ A ∪ B and A ∩ B ⊆ A.
3. Recall that the symbol ⊕ denotes the logical exclusive or operation. If A and B sets, define the
set A △ B by A △ B = { x | ( x ∈ A) ⊕ ( x ∈ B)}. Show that A △ B = ( A ∖ B) ∪ ( B ∖ A). (A △ B
is known as the symmetric difference of A and B.)

4

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Chapter 4. Sets, Functions, and Relations

4. Choose three non-empty sets A, B, C. Draw a Venn diagram of A △ B △ C.
5. Let A be a subset of some given universal set U. Verify that A = A and that A ∪ A = U.
6. Verify the second of DeMorgan’s Laws for sets, A ∩ B = A ∪ B. For each step in your verification, state why that step is valid.

4

7. The subset operator, ⊆, is defined in terms of the logical implication operator, →. However, ⊆
differs from the ∩ and ∪ operators in that A ∩ B and A ∪ B are sets, while A ⊆ B is a statement.
So the relationship between ⊆ and → isn’t quite the same as the relationship between ∪ and ∨
or between ∩ and ∧. Nevertheless, ⊆ and → do share some similar properties. This problem
shows one example.
a) Show that the following three compound propositions are logically equivalent: p → q,
( p ∧ q) ↔ p, and ( p ∨ q) ↔ q.
b) Show that for any sets A and B, the following three statements are equivalent: A ⊆ B,
A ∩ B = A, and A ∪ B = B.
8. DeMorgan’s Laws apply to subsets of some given universal set U. Show that for a subset X of U,
X = U ∖ X. It follows that DeMorgan’s Laws can be written as U ∖ ( A ∪ B) = (U ∖ A) ∩ (U ∖
B) and U ∖ ( A ∩ B) = (U ∖ A) ∪ (U ∖ B). Show that these laws hold whether or not A and B
are subsets of U. That is, show that for any sets A, B, and C, C ∖ ( A ∪ B) = (C ∖ A) ∩ (C ∖ B)
and C ∖ ( A ∩ B) = (C ∖ A) ∪ (C ∖ B).
9. Show that A ∪ ( A ∩ B) = A for any sets A and B.
10. Let X and Y be sets. Simplify each of the following expressions. Justify each step in the simplification with one of the rules of set theory.
a) X ∪ (Y ∪ X )
b) ( X ∩ Y ) ∩ X
c) ( X ∪ Y ) ∩ Y
d) ( X ∪ Y ) ∪ ( X ∩ Y )
11. Let A, B, and C be sets. Simplify each of the following expressions. In your answer, the complement operator should only be applied to the individual sets A, B, and C.
a) A ∪ B ∪ C

b) A ∪ B ∩ C

c) A ∪ B

d) B ∩ C

e) A ∩ B ∩ C

f) A ∩ A ∪ B

12. Use induction to prove the following generalized DeMorgan’s Law for set theory: For any
natural number n ≥ 2 and for any sets X1 , X2 , …, Xn ,
X1 ∩ X2 ∩ · · · ∩ X n = X1 ∪ X2 ∪ · · · ∪ X n
13. State and prove generalized distributive laws for set theory.

4.3

Application: Programming with Sets

On a computer, all pieces of data are represented, ultimately, as strings of zeros and ones.
At times, computers need to work with sets. How can sets be represented as strings of
zeros and ones? And once we have represented sets on a computer, how do we program
with them?

4.3. Application: Programming with Sets *



111

In this section we go into detail about representing and computing with sets.
You won’t be examined on this in Reasoning & Logic. You will find that there
is again quite some overlap with your study materials from Computer Organisation however, as we discuss different numeric systems.

4.3.1 Representing sets
A set is determined by its elements. Given a set A and an entity x, the fundamental
question is, does x belong to A or not? If we know the answer to this question for each
possible x, then we know the set. For a given x, the answer to the question, “Is x a
member of A”, is either yes or no. The answer can be encoded by letting 1 stand for yes
and 0 stand for no. The answer, then, is a single bit, that is, a value that can be either
zero or one. To represent the set A as a string of zeros and ones, we could use one bit for
each possible member of A. If a possible member x is in the set, then the corresponding
bit has the value one. If x is not in the set, then the corresponding bit has the value zero.
Now, in cases where the number of possible elements of the set is very large or infinite,
it is not practical to represent the set in this way. It would require too many bits, perhaps
an infinite number. In such cases, some other representation for the set can be used.
However, suppose we are only interested in subsets of some specified small set. Since
this set plays the role of a universal set, let’s call it U. To represent a subset of U, we just
need one bit for each member of U. If the number of members of U is n, then a subset
of U is represented by a string of n zeros and ones. Furthermore, every string of n zeros
and ones determines a subset of U, namely that subset that contains exactly the elements
of U that correspond to ones in the string. You know by now from Computer Organisation
that a string of n zeros and ones is called an n-bit binary number. So, we see that if U is
a set with n elements, then the subsets of U correspond to n-bit binary numbers.
To make things more definite, let U be the set {0, 1, 2, . . . , 31}. This set consists of
the 32 integers between 0 and 31, inclusive. Then each subset of U can be represented
by a 32-bit binary number. We use 32 bits because most computer languages can work
directly with 32-bit numbers. For example, the programming languages Java has a data
type named int. A value of type int is a 32-bit binary number.2 Before we get a definite
correspondence between subsets of U and 32-bit numbers, we have to decide which bit
in the number will correspond to each member of U. Following tradition, we assume
that the bits are numbered from right to left. That is, the rightmost bit corresponds to
the element 0 in U, the second bit from the right corresponds to 1, the third bit from the
right to 2, and so on. For example, the 32-bit number
1000000000000000000001001110110
2 If in a future version of Java, a value of type int is, for instance, a 64-bit number—which can be used to
represent a subset of the set {0, 1, 2, . . . , 63}—the principle remains the same.

4

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Chapter 4. Sets, Functions, and Relations

Hex.

Binary

Hex.

Binary

0

00002

8

10002

1

00012

9

10012

2

00102

A

10102

3

00112

B

10112

4

01002

C

11002

5

01012

D

11012

6

01102

E

11102

7

01112

F

11112

Figure 4.6: The 16 hexadecimal digits and the corresponding binary numbers. Each
hexadecimal digit corresponds to a 4-bit binary number. Longer binary numbers can be
written using two or more hexadecimal digits. For example, 1010000111112 = 0xA1F.

corresponds to the subset {1, 2, 4, 5, 6, 9, 31}. Since the leftmost bit of the number is 1, the
number 31 is in the set; since the next bit is 0, the number 30 is not in the set; and so on.
From now on, I will write binary numbers with a subscript of 2 to avoid confusion
with ordinary numbers. Furthermore, I will often leave out leading zeros. For example,
11012 is the binary number that would be written out in full as
00000000000000000000000000001101
and which corresponds to the set {0, 2, 3}. On the other hand 1101 represents the ordinary number one thousand one hundred and one.
Even with this notation, it can be very annoying to write out long binary numbers—
and almost impossible to read them. So binary numbers are never written out as sequences of zeros and ones in computer programs. An alternative is to use hexadecimal
numbers. Hexadecimal numbers are written using the sixteen symbols 0, 1, 2, 3, 4, 5, 6, 7,
8, 9, A, B, C, D, E, and F. These symbols are knows as the hexadecimal digits. Each hexadecimal digit corresponds to a 4-bit binary number, as shown in Figure 4.6. To represent
a longer binary number, several hexadecimal digits can be strung together. For example,
the hexadecimal number C7 represents the binary number 110001112 . In Java and many
related languages, a hexadecimal number is written with the prefix 0x. Thus, the hexadecimal number C7 would appear in the program as 0xC7. I will follow the same convention here. Any 32-bit binary number can be written using eight hexadecimal digits
(or fewer if leading zeros are omitted). Thus, subsets of {0, 1, 2, . . . , 31} correspond to
8-digit hexadecimal numbers. For example, the subset {1, 2, 4, 5, 6, 9, 31} corresponds to

4.3. Application: Programming with Sets *

113

0x80000276, which represents the binary number 10000000000000000000010011101102 .
Similarly, 0xFF corresponds to {0, 1, 2, 3, 4, 5, 6, 7} and 0x1101 corresponds to the binary
number 00010001000000012 and to the set {0, 8, 12}.
Now, if you have worked with binary numbers or with hexadecimal numbers, you
know that they have another, more common interpretation. They represent ordinary
integers. Just as 342 represents the integer 3 · 102 + 4 · 101 + 2 · 100 , the binary number
11012 represents the integer 1 · 23 + 1 · 22 + 0 · 21 + 1 · 20 , or 13. When used in this way,
binary numbers are known as base-2 numbers, just as ordinary numbers are base-10
numbers. Hexadecimal numbers can be interpreted as base-16 numbers. For example,
0x3C7 represents the integer 3 · 162 + 12 · 161 + 7 · 160 , or 874. So, does 11012 really
represent the integer 13, or does it represent the set {0, 2, 3} ? The answer is that to
a person, 11012 can represent either. Both are valid interpretations, and the only real
question is which interpretation is useful in a given circumstance. On the other hand, to
the computer, 11012 doesn’t represent anything. It’s just a string of bits, and the computer
manipulates the bits according to its program, without regard to their interpretation.
Of course, we still have to answer the question of whether it is ever useful to interpret
strings of bits in a computer as representing sets.

4.3.2 Computing with sets
If all we could do with sets were to ‘represent’ them, it wouldn’t be very useful. We need
to be able to compute with sets. That is, we need to be able to perform set operations
such as union and complement.
Many programming languages provide operators that perform set operations. In
Java and related languages, the operators that perform union, intersection, and complement are written as | , &, and ~. For example, if x and y are 32-bit integers representing
two subsets, X and Y, of {0, 1, 2, . . . , 31}, then x | y is a 32-bit integer that represents the
set X ∪ Y. Similarly, x & y represents the set X ∩ Y, and ~x represents the complement,
X.
The operators | , &, and ~ are called bitwise logical operators because of the way
they operate on the individual bits of the numbers to which they are applied. If 0 and
1 are interpreted as the logical values false and true, then the bitwise logical operators
perform the logical operations ∨, ∧, and ¬ on individual bits. To see why this is true,
let’s look at the computations that these operators have to perform.
Let k be one of the members of {0, 1, 2, . . . , 31}. In the binary numbers x, y, x | y,
x & y, and ~x, the number k corresponds to the bit in position k. That is, k is in the set
represented by a binary number if and only if the bit in position k in that binary number
is 1. Considered as sets, x & y is the intersection of x and y, so k is a member of the set
represented by x & y if and only if k is a member of both of the sets represented by x and
y. That is, bit k is 1 in the binary number x & y if and only if bit k is 1 in x and bit k is 1
in y. When we interpret 1 as true and 0 as false, we see that bit k of x & y is computed by
applying the logical ‘and’ operator, ∧, to bit k of x and bit k of y. Similarly, bit k of x | y is

4

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computed by applying the logical ‘or’ operator, ∨, to bit k of x and bit k of y. And bit k
of ~x is computed by applying the logical ‘not’ operator, ¬, to bit k of x. In each case, the
logical operator is applied to each bit position separately. (Of course, this discussion is
just a translation to the language of bits of the definitions of the set operations in terms
of logical operators: A ∩ B = { x | x ∈ A ∧ x ∈ B}, A ∪ B = { x | x ∈ A ∨ x ∈ B}, and
A = { x ∈ U | ¬( x ∈ A)}.)
For example, consider the binary numbers 10110102 and 101112 , which represent the
sets {1, 3, 4, 6} and {0, 1, 2, 4}. Then 10110102 & 101112 is 100102 . This binary number
represents the set {1, 4}, which is the intersection {1, 3, 4, 6} ∩ {0, 1, 2, 4}. It’s easier to
see what’s going on if we write out the computation in columns, the way you probably
first learned to do addition:

4

&

1011010
0010111
0010010

{
{
{

6,

4,
4,

3,
2,

4,

1
1, 0
1

}
}
}

Note that in each column in the binary numbers, the bit in the bottom row is computed
as the logical ‘and’ of the two bits that lie above it in the column. I’ve written out the sets
that correspond to the binary numbers to show how the bits in the numbers correspond
to the presence or absence of elements in the sets. Similarly, we can see how the union
of two sets is computed as a bitwise ‘or’ of the corresponding binary numbers.

|

1011010
0010111
1011111

{
{
{

6,
6,

4,
4,

3,

4,

3,

2,

1
1, 0

2,

1, 0

}
}
}

The complement of a set is computed using a bitwise ‘not’ operation. Since we are working with 32-bit binary numbers, the complement is taken with respect to the universal
set {0, 1, 2, . . . , 31}. So, for example,
~10110102 = 111111111111111111111111101001012
Of course, we can apply the operators &, | , and ~ to numbers written in hexadecimal
form, or even in ordinary, base-10 form. When doing such calculations by hand, it is
probably best to translate the numbers into binary form. For example,
0xAB7 & 0x168E = 1010 1011 01112 & 1 0110 1000 11102

= 0 0010 1000 01102
= 0x286
When computing with sets, it is sometimes necessary to work with individual elements. Typical operations include adding an element to a set, removing an element
from a set, and testing whether an element is in a set. However, instead of working
with an element itself, it’s convenient to work with the set that contains that element as

4.3. Application: Programming with Sets *

115

its only member. For example, testing whether 5 ∈ A is the same as testing whether
{5} ∩ A ̸= ∅. The set {5} is represented by the binary number 1000002 or by the hexadecimal number 0x20. Suppose that the set A is represented by the number x. Then,
testing whether 5 ∈ A is equivalent to testing whether 0x20 & x ̸= 0. Similarly, the
set A ∪ {5}, which is obtained by adding 5 to A, can be computed as x | 0x20. The set
A ∖ {5}, which is the set obtained by removing 5 from A if it occurs in A, is represented
by x & ~0x20.
The sets {0}, {1}, {2}, {3}, {4}, {5}, {6}, …, {31} are represented by the hexadecimal
numbers 0x1, 0x2, 0x4, 0x8, 0x10, 0x20, …, 0x80000000. In typical computer applications,
some of these numbers are given names, and these names are thought of as names for
the possible elements of a set (although, properly speaking, they are names for sets containing those elements). Suppose, for example, that a, b, c, and d are names for four
of the numbers from the above list. Then a | c is the set that contains the two elements
corresponding to the numbers a and c. If x is a set, then x & ~d is the set obtained by
removing d from x. And we can test whether b is in x by testing if x & b ̸= 0.
Here is an actual example, which is used in the Apple Mac operating systems (macOS). Characters can be printed or displayed on the screen in various sizes and styles. A
font is a collection of pictures of characters in a particular size and style. On the Mac, a
basic font can be modified by specifying any of the following style attributes: bold, italic,
underline, outline, shadow, condense, and extend. The style of a font is a subset of this set
of attributes. A style set can be specified by or-ing together individual attributes. For example, an underlined, bold, italic font has style set underline | bold | italic. For a plain font,
with none of the style attributes set, the style set is the empty set, which is represented
by the number zero.
The Java programming language uses a similar scheme to specify style attributes for
fonts, but currently there are only two basic attributes, Font.BOLD and Font.ITALIC. A
more interesting example in Java is provided by event types. An event in Java represents
some kind of user action, such as pressing a key on the keyboard. Events are associated
with ‘components’ such as windows, push buttons, and scroll bars. Components can
be set to ignore a given type of event. We then say that that event type is disabled for
that component. If a component is set to process events of a given type, then that event
type is said to be enabled. Each component keeps track of the set of event types that are
currently enabled. It will ignore any event whose type is not in that set. Each event type
has an associated constant with a name such as AWTEvent.MOUSE_EVENT_MASK. These
constants represent the possible elements of a set of event types. A set of event types
can be specified by or-ing together a number of such constants. If c is a component and
x is a number representing a set of event types, then the command ‘c.enableEvents(x)’
enables the events in the set x for the component c. If y represents the set of event types
that were already enabled for c, then the effect of this command is to replace y with the
union, y | x. Another command, “c.disableEvents(x)”, will disable the event types in x for
the component c. It does this by replacing the current set, y, with y & ~x.

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Exercises
1. Suppose that the numbers x and y represent the sets A and B. Show that the set A ∖ B is
represented by x & (~y).
2. Write each of the following binary numbers in hexadecimal:
a) 101101102
b) 102
c) 1111000011112

d) 1010012

3. Write each of the following hexadecimal numbers in binary:
a) 0x123
b) 0xFADE
c) 0x137F
d) 0xFF11
4. Give the value of each of the following expressions as a hexadecimal number:
a) 0x73 | 0x56A
b) ~0x3FF0A2FF
c) (0x44 | 0x95) & 0xE7
d) 0x5C35A7 & 0xFF00
e) 0x5C35A7 & ~0xFF00
f) ~(0x1234 & 0x4321)

4

5. Find a calculator (or a calculator program on a computer) that can work with hexadecimal
numbers. Write a short report explaining how to work with hexadecimal numbers on that
calculator. You should explain, in particular, how the calculator can be used to do the previous
problem.
6. This question assumes that you know how to add binary numbers. Suppose x and y are binary
numbers. Under what circumstances will the binary numbers x + y and x | y be the same?
7. In addition to hexadecimal numbers, the programming languages Java, C, and C++ support
octal numbers. Look up and report on octal numbers in Java, C, or C++. Explain what octal
numbers are, how they are written, and how they are used.
8. In the Unix (and Linux, macOS, …) operating system, every file has an associated set of permissions, which determine who can use the file and how it can be used. The set of permissions
for a given file is represented by a nine-bit binary number. This number is sometimes written
as an octal number. Research and report on the Unix systems of permissions. What set of
permissions is represented by the octal number 752? by the octal number 622? Explain what
is done by the Unix commands chmod g+rw filename and chmod o-w filename in terms of
sets.
9. Most modern programming languages have a boolean data type that has the values true and
false. The usual logical and, or, and not operators on boolean values are represented in Java,
C and C++ by the operators &&, | |, and !. C and C++ (but not Java) allow integer values to be
used in places where boolean values are expected. In this case, the integer zero represents the
boolean value false while any non-zero integer represents the boolean value true. This means
that if x and y are integers, then both x & y and x && y are valid expressions, and both can be
considered to represent boolean values. Do the expressions x & y and x && y always represent
the same boolean value, for any integers x and y? Do the expressions x | y and x | | y always
represent the same boolean values? Explain your answers.
10. Suppose that you, as a programmer, want to write a subroutine that will open a window on
the computer’s screen. The window can have any of the following options: a close box, a zoom
box, a resize box, a minimize box, a vertical scroll bar, a horizontal scroll bar. Design a scheme
whereby the options for the window can be specified by a single parameter to the subroutine.
The parameter should represent a set of options. How would you use your subroutine to
open a window that has a close box and both scroll bars and no other options? Inside your
subroutine, how would you determine which options have been specified for the window?

4.4. Functions

117

4.4 Functions
Both the real world and the world of mathematics are full of what are called, in mathematics, ‘functional relationships’. A functional relationship is a relationship between
two sets, which associates exactly one element from the second set to each element of
the first set.
For example, each item for sale in a store has a price. The first set in this relationship
is the set of items in the store. For each item in the store, there is an associated price,
so the second set in the relationship is the set of possible prices. The relationship is a
functional relationship because each item has a price. That is, the question “What is the
price of this item?” has a single, definite answer for each item in the store.
Similarly, the question “Who is the (biological) mother of this person?” has a single,
definite answer for each person. So, the relationship ‘mother of’ defines a functional
relationship. In this case, the two sets in the relationship are the same set, namely the set
of people. On the other hand, the relationship ‘child of’ is not a functional relationship.
The question “Who is the child of this person?” does not have a single, definite answer
for each person. A given person might not have any child at all. And a given person
might have more than one child. Either of these cases—a person with no child or a
person with more than one child—is enough to show that the relationship ‘child of’ is
not a functional relationship.
Or consider an ordinary map, such as a map of Zuid-Holland or a street map of Rome.
The whole point of the map, if it is accurate, is that there is a functional relationship
between points on the map and points on the surface of the Earth. Perhaps because of
this example, a functional relationship is sometimes called a mapping.
There are also many natural examples of functional relationships in mathematics.
For example, every rectangle has an associated area. This fact expresses a functional
relationship between the set of rectangles and the set of numbers. Every natural number
n has a square, n2 . The relationship ‘square of’ is a functional relationship from the set
of natural numbers to itself.

4.4.1 Formalising the notion of functions
In mathematics, of course, we need to work with functional relationships in the abstract.
To do this, we introduce the idea of function. You should think of a function as a mathematical object that expresses a functional relationship between two sets. The notation
f : A → B expresses the fact that f is a function from the set A to the set B. That is, f
is a name for a mathematical object that expresses a functional relationship from the set
A to the set B. The notation f : A → B is read as “ f is a function from A to B” or more
simply as “ f maps A to B”.



Mathematical functions are different to functions in a programming language in Java. We’ll come back to this in the next section.

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Chapter 4. Sets, Functions, and Relations

If f : A → B and if a ∈ A, the fact that f is a functional relationship from A to B
means that f associates some element of B to a. That element is denoted f ( a). That is,
for each a ∈ A, f ( a) ∈ B and f ( a) is the single, definite answer to the question “What
element of B is associated to a by the function f ?” The fact that f is a function from A to
B means that this question has a single, well-defined answer. Given a ∈ A, f ( a) is called
the value of the function f at a.


4

For example, if I is the set of items for sale in a given store and M is the set of
possible prices, then there is function c : I → M which is defined by the fact
that for each x ∈ I, c( x ) is the price of the item x. Similarly, if P is the set of
people, then there is a function m : P → P such that for each person p, m( p)
is the mother of p. And if N is the set of natural numbers, then the formula
s(n) = n2 specifies a function s : N → N.

It is in the form of formulas such as s(n) = n2 or f ( x ) = x3 − 3x + 7 that most people
first encounter functions. But you should note that a formula by itself is not a function,
although it might well specify a function between two given sets of numbers. Functions
are much more general than formulas, and they apply to all kinds of sets, not just to sets
of numbers.

4.4.2

Operations on functions

Suppose that f : A → B and g : B → C are functions. Given a ∈ A, there is an associated
element f ( a) ∈ B. Since g is a function from B to C, and since f ( a) ∈ B, g associates
some element of C to f ( a). That element is g( f ( a)). Starting with an element a of A, we
have produced an associated element g( f ( a)) of C. This means that we have defined a
new function from the set A to the set C. This function is called the composition of g
with f , and it is denoted by g ◦ f . That is, if f : A → B and g : B → C are functions, then
g ◦ f : A → C is the function which is defined by

( g ◦ f )( a) = g( f ( a))
for each a ∈ A. For example, suppose that p is the function that associates to each
item in a store the price of the item, and suppose that t is a function that associates the
amount of tax on a price to each possible price. The composition, t ◦ p, is the function
that associates to each item the amount of tax on that item. Or suppose that s : N → N
and r : N → N are the functions defined by the formulas s(n) = n2 and r (n) = 3n + 1,
for each n ∈ N. Then r ◦ s is a function from N to N, and for n ∈ N, (r ◦ s)(n) =
r (s(n)) = r (n2 ) = 3n2 + 1. In this case, we also have the function s ◦ r, which satisfies

4.4. Functions

119

(s ◦ r )(n) = s(r (n)) = s(3n + 1) = (3n + 1)2 = 9n2 + 6n + 1. Note in particular that r ◦ s
and s ◦ r are not the same function. The operation ◦ is not commutative.
If A is a set and f : A → A, then f ◦ f , the composition of f with itself, is defined.
For example, using the function s from the preceding example, s ◦ s is the function from
N to N given by the formula (s ◦ s)(n) = s(s(n)) = s(n2 ) = (n2 )2 = n4 . If m is the
function from the set of people to itself which associates to each person that person’s
mother, then m ◦ m is the function that associates to each person that person’s maternal
grandmother.
If a and b are entities, then ( a, b) denotes the ordered pair containing a and b. The
ordered pair ( a, b) differs from the set { a, b} because a set is not ordered. That is, { a, b}
and {b, a} denote the same set, but if a ̸= b, then ( a, b) and (b, a) are different ordered
pairs. More generally, two ordered pairs ( a, b) and (c, d) are equal if and only if both a = c
and b = d. If ( a, b) is an ordered pair, then a and b are referred to as the coordinates of
the ordered pair. In particular, a is the first coordinate and b is the second coordinate.



In high school you would also have to write (x,y)-coordinates using this
ordered pair notation. For instance you would say that the line y = ax + b
b
intersects the y-axis at (0, b) and the x-axis at (− , 0).
a

If A and B are sets, then we can form the set A × B which is defined by
A × B = {( a, b) | a ∈ A and b ∈ B}.
This set is called the cross product or Cartesian product of the sets A and B. The set
A × B contains every ordered pair whose first coordinate is an element of A and whose
second coordinate is an element of B. For example, if X = {c, d} and Y = {1, 2, 3}, then
X × Y = {(c, 1), (c, 2), (c, 3), (d, 1), (d, 2), (d, 3)}. It is possible to extend this idea to the
cross product of more than two sets. The cross product of the three sets A, B, and C is
denoted A × B × C. It consists of all ordered triples ( a, b, c) where a ∈ A, b ∈ B, and
c ∈ C. The definition for four or more sets is similar. The general term for a member of a
cross product is tuple (recall page 42) or, more specifically, ordered n-tuple. For example,
( a, b, c, d, e) is an ordered 5-tuple.
Given a function f : A → B, consider the set {( a, b) ∈ A × B | a ∈ A and b = f ( a)}.
This set of ordered pairs consists of all pairs ( a, b) such that a ∈ A and b is the element of
B that is associated to a by the function f . The set {( a, b) ∈ A × B | a ∈ A and b = f ( a)}
is called the graph of the function f . Since f is a function, each element a ∈ A occurs
once and only once as a first coordinate among the ordered pairs in the graph of f . Given
a ∈ A, we can determine f ( a) by finding that ordered pair and looking at the second
coordinate. In fact, it is convenient to consider the function and its graph to be the same
thing, and to use this as our official mathematical definition.3
3 This

is a convenient definition for the mathematical world, but as is often the case in mathematics, it

4

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Chapter 4. Sets, Functions, and Relations

100
80

a

b

c

1

2

3

60
40
20
0

−10

−5

0

5

10

(a) Graph for the formula f (n) = n2 for −10 ≤
n ≤ 10

(b) Graph for the function g : A → B with
A = { a, b, c} and B = {1, 2, 3}, such that
g = {( a, 2), (b, 3), (c, 1)}.

Figure 4.7: Two different graphs representing functions.

4

You are already familiar with the process of graphing numerical functions from high
school (for instance you can graph the formula f (n) = n2 as depicted in Figure 4.7a),
but we can also graph non-numerical functions as indicated in Figure 4.7b. To do so, we
draw all elements from the set A and connect them to the elements of B they map to.
Definition 4.2. Let A and B be sets. A function from A to B is a subset of A × B which
has the property that for each a ∈ A, the set contains one and only one ordered pair
whose first coordinate is a. If ( a, b) is that ordered pair, then b is called the value of the
function at a and is denoted f ( a). If b = f ( a), then we also say that the function f maps
a to b. The fact that f is a function from A to B is indicated by the notation f : A → B.



For example, if X = { a, b} and Y = {1, 2, 3}, then the set {( a, 2), (b, 1)} is a
function from X to Y, and {(1, a), (2, a), (3, b)} is a function from Y to X. On
the other hand, {(1, a), (2, b)} is not a function from Y to X, since it does not
specify any value for 3. And {( a, 1), ( a, 2), (b, 3)} is not a function from X to
Y because it specifies two different values, 1 and 2, associated with the same
element, a, of X.

Even though the technical definition of a function is a set of ordered pairs, it’s usually
better to think of a function from A to B as something that associates some element of B to
every element of A. The set of ordered pairs is one way of expressing this association. If
the association is expressed in some other way, it’s easy to write down the set of ordered
pairs. For example, the function s : N → N which is specified by the formula s(n) = n2
can be written as the set of ordered pairs {(n, n2 ) | n ∈ N}.
neglects much of the real world. Functional relationships in the real world are meaningful, but we model them
in mathematics with meaningless sets of ordered pairs. We do this for the usual reason: to have something
precise and rigorous enough that we can make logical deductions and prove things about it.

4.4. Functions

121

4.4.3 Properties of functions
Suppose that f : A → B is a function from the set A to the set B. We say that A is the
domain of the function f and that B is the range of the function. We define the image
of the function f to be the set {b ∈ B | ∃ a ∈ A (b = f ( a))}. Put more simply, the image
of f is the set { f ( a) | a ∈ A}. That is, the image is the set of all values, f ( a), of the
function, for all a ∈ A. For example, for the function s : N → N that is specified by
s(n) = n2 , both the domain and the range are N, and the image is the set {n2 | n ∈ N},
or {0, 1, 4, 9, 16, . . . }.



In some cases—particularly in courses like Calculus—the term ‘range’ is used
to refer to what I am calling the image.

Note that the image of a function is a subset of its range. It can be a proper subset, as
in the above example, but it is also possible for the image of a function to be equal to the
range. In that case, the function is said to be onto. Sometimes, the fancier term surjective
is used instead. Formally, a function f : A → B is said to be onto (or surjective) if every
element of B is equal to f ( a) for some element of A. In terms of logic, f is onto if and
only if
(
)
∀b ∈ B ∃ a ∈ A (b = f ( a)) .



For example, let X = { a, b} and Y = {1, 2, 3}, and consider the function
from Y to X specified by the set of ordered pairs {(1, a), (2, a), (3, b)}. This
function is onto because its image, { a, b}, is equal to the range, X. However,
the function from X to Y given by {( a, 1), (b, 3)} is not onto, because its
image, {1, 3}, is a proper subset of its range, Y.
As a further example, consider the function f from Z to Z given by f (n) =
n − 52. To show that f is onto, we need to pick an arbitrary b in the range Z
and show that there is some number a in the domain Z such that f ( a) = b.
So let b be an arbitrary integer; we want to find an a such that a − 52 = b.
Clearly this equation will be true when a = b + 52. So every element b is
the image of the number a = b + 52, and f is therefore onto. Note that if
f had been specified to have domain N, then f would not be onto, as for
some b ∈ Z the number a = b + 52 is not in the domain N (for example, the
integer −73 is not in the image of f , since −21 is not in N.)

If f : A → B and if a ∈ A, then a is associated to only one element of B. This is part
of the definition of a function. However, no such restriction holds for elements of B. If
b ∈ B, it is possible for b to be associated to zero, one, two, three, …, or even to an infinite

4

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number of elements of A. In the case where each element of the range is associated to at
most one element of the domain, the function is said to be one-to-one. Sometimes, the
term injective is used instead. The function f is one-to-one (or injective) if for any two
distinct elements x and y in the domain of f , f ( x ) and f (y) are also distinct. In terms of
logic, f : A → B is one-to-one if and only if
(
)
∀ x ∈ A ∀y ∈ A x ̸= y → f ( x ) ̸= f (y) .
Since a proposition is equivalent to its contrapositive, we can write this condition equivalently as
(
)
∀ x ∈ A ∀y ∈ A f ( x ) = f (y) → x = y .

4

Sometimes, it is easier to work with the definition of one-to-one when it is expressed in
this form.



The function that associates every person to his or her mother is not one-toone because it is possible for two different people to have the same mother.
The function s : N → N specified by s(n) = n2 is one-to-one. However, we
can define a function r : Z → Z by the same formula: r (n) = n2 , for n ∈ Z.
The function r is not one-to-one since two different integers can have the same
square. For example, r (−2) = r (2).

A function that is both one-to-one and onto is said to
be bijective.4 The function that associates each point in a
map of Zuid-Holland to a point in the state itself is presumably bijective. For each point on the map, there is a corresponding point in the province, and vice versa. If we specify
the function f from the set {1, 2, 3} to the set { a, b, c} as the
set of ordered pairs {(1, b), (2, a), (3, c)}, then f is a bijective function. Or consider the function from Z to Z given by
f (n) = n − 52. We have already shown that f is onto. We
can show that it is also one-to-one.
Proof. Pick an arbitrary x and y in Z and assume that
f ( x ) = f (y). This means that x − 52 = y − 52, and adding 52 to both sides of the equation gives x = y. Since x and y were arbitrary, we have proved ∀ x ∈ Z ∀y ∈ Z ( f ( x ) =
f (y) → x = y), that is, that f is one-to-one.
Altogether, then, f is a bijection.
4 Image:

commons.wikimedia.org/wiki/File:Bijection.svg.

4.4. Functions

123

4.4.4 First-class objects
One difficulty that people sometimes have with mathematics is its generality. A set is
a collection of entities, but an ‘entity’ can be anything at all, including other sets. Once
we have defined ordered pairs, we can use ordered pairs as elements of sets. We could
also make ordered pairs of sets. Now that we have defined functions, every function is
itself an entity. This means that we can have sets that contain functions. We can even
have a function whose domain and range are sets of functions. Similarly, the domain or
range of a function might be a set of sets, or a set of ordered pairs. Computer scientists
have a good name for this. They would say that sets, ordered pairs, and functions are
first-class objects or first-class citizens. Once a set, ordered pair, or function has been
defined, it can be used just like any other entity. If they were not first-class objects, there
could be restrictions on the way they can be used. For example, it might not be possible
to use functions as members of sets. (This would make them ‘second class.’)



One way that programming languages differ is by what they allow as firstclass objects. For example, Java added a ‘lambda syntax’ to help writing
‘closures’ in version 8.

For example, suppose that A, B, and C are sets. Then since A × B is a set, we might
have a function f : A × B → C. If ( a, b) ∈ A × B, then the value of f at ( a, b) would be
denoted f (( a, b)). In practice, though, one set of parentheses is usually dropped, and
the value of f at ( a, b) is denoted f ( a, b). As a particular example, we might define a
function p : N × N → N with the formula p(n, m) = nm + 1. Similarly, we might define
a function q : N × N × N → N × N by q(n, m, k ) = (nm − k, nk − n).
Suppose that A and B are sets. There are, in general, many functions that map A to B.
We can gather all those functions into a set. This set, whose elements are all the functions
from A to B, is denoted B A . (We’ll see later why this notation is reasonable.) Using this
notation, saying f : A → B is exactly the same as saying f ∈ B A . Both of these notations
assert that f is a function from A to B.(Of course,
we can also form an unlimited number
)
of other sets, such as the power set P B A , the cross product B A × A, or the set A A× A ,
which contains all the functions from the set A × A to the set A. And of course, any of
these sets can be the domain or range of a function. An example of this is the function
E : B A × A → B defined by the formula E( f , a) = f ( a). Let’s see if we can make sense of
this notation. Since the domain of E is B A × A, an element in the domain is an ordered
pair in which the first coordinate is a function from A to B and the second coordinate is
an element of A. Thus, E( f , a) is defined for a function f : A → B and an element a ∈ A.
Given such an f and a, the notation f ( a) specifies an element of B, so the definition of
E( f , a) as f ( a) makes sense. The function E is called the ‘evaluation function’ since it
captures the idea of evaluating a function at an element of its domain.

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Exercises
1. Let A = {1, 2, 3, 4} and let B = { a, b, c}. Find the sets A × B and B × A.
2. Let A be the set { a, b, c, d}. Let f be the function from A to A given by the set of ordered
pairs {( a, b), (b, b), (c, a), (d, c)}, and let g be the function given by the set of ordered pairs
{( a, b), (b, c), (c, d), (d, d)}. Find the set of ordered pairs for the composition g ◦ f .
3. Let A = { a, b, c} and let B = {0, 1}. Find all possible functions from A to B. Give each function
as a set of ordered pairs. (Hint: Every such function corresponds to one of the subsets of A.)
4. Consider the functions from Z to Z which are defined by the following formulas. Decide
whether each function is onto and whether it is one-to-one; prove your answers.
a) f (n) = 2n
b) g(n) = n + 1
c) h(n) = n2 + n + 1
{
n/2,
if n is even
d) s(n) =
(n+1)/2, if n is odd

4

5. Prove that composition of functions is an associative operation. That is, prove that for functions
f : A → B, g : B → C, and h : C → D, the compositions (h ◦ g) ◦ f and h ◦ ( g ◦ f ) are equal.
6. Suppose that f : A → B and g : B → C are functions and that g ◦ f is one-to-one.
a) Prove that f is one-to-one. (Hint: use a proof by contradiction.)
b) Find a specific example that shows that g is not necessarily one-to-one.
7. Suppose that f : A → B and g : B → C, and suppose that the composition g ◦ f is an onto
function.
a) Prove that g is an onto function.
b) Find a specific example that shows that f is not necessarily onto.

4.5

Application: Programming with Functions

Functions are fundamental in computer programming, although not everything in programming that goes by the name of ‘function’ is a function according to the mathematical
definition.



In this section we go into detail about functions in computer programming.
You won’t be examined on this in Reasoning & Logic! You will find that there
is again quite some overlap with your study materials from Object-Oriented
Programming and later on in your curriculum with Concepts of Programming
Languages.

In computer programming, a function is a routine that is given some data as input
and that will calculate and return an answer based on that data. For example, Java, a
function that calculates the square of an integer could be written
2

int square(int n) {
return n*n;
}

4.5. Application: Programming with Functions *

125

In Java, int is a data type. From the mathematical point of view, a data type is a set. The
data type int is the set of all integers that can be represented as 32-bit binary numbers.
Mathematically, then, int ⊆ Z. (You should get used to the fact that sets and functions
can have names that consist of more than one character, since it’s done all the time in computer programming.) The first line of the above function definition, int square(int n),
says that we are defining a function named square whose range is int and whose domain
is int. In the usual notation for functions, we would express this as square : int → int, or
possibly as square ∈ intint , where intint is the set of all functions that map the set int to
the set int.
The first line of the function, int square(int n), is called the prototype of the function. The prototype specifies the name, the domain, and the range of the function and so
carries exactly the same information as the notation f : A → B. The ‘n’ in int square(int n)
is a name for an arbitrary element of the data type int. We call n a parameter of the
function. The rest of the definition of square tells the computer to calculate the value
of square(n) for any n ∈ int by multiplying n times n. The statement return n*n says
that n ∗ n is the value that is computed, or ‘returned’, by the function. (The ∗ stands for
multiplication.)
Java has many data types in addition to int. There is a boolean data type named
boolean. The values of type boolean are true and false. Mathematically, boolean is a name
for the set {true, false}. The type double consists of real numbers, which can include a
decimal point. Of course, on a computer, it’s not possible to represent the entire infinite
set of real numbers, so double represents some subset of the mathematical set of real
numbers. There is also a data type whose values are strings of characters, such as “Hello
world” or “xyz152QQZ”. The name for this data type in Java is string. All these types,
and many others, can be used in functions. For example, in Java, m % n is the remainder
when the integer m is divided by the integer n. We can define a function to test whether
an integer is even as follows:

4

1

3

boolean even(int k) {
return k % 2 == 0;
}

You don’t need to worry about all the details here, but you should understand that the
prototype, boolean even(int k), says that even is a function from the set int to the set
boolean. That is, even : int → boolean. Given an integer N, even( N ) has the value true if N
is an even integer, and it has the value false if N is an odd integer.
A function can have more than one parameter. For example, we might define a function with prototype int index(string str, string sub). If s and t are strings, then
index(s, t) would be the int that is the value of the function at the ordered pair (s, t).
We see that the domain of index is the cross product string × string, and we can write
index : string × string → int or, equivalently, index ∈ intstring×string .
Not every Java function is actually a function in the mathematical sense. In mathematics, a function must associate a single value in its range to each value in its domain.

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There are two things that can go wrong: the value of the function might not be defined for
every element of the domain, and the function might associate several different values
to the same element of the domain. Both of these things can happen with Java functions.
In computer programming, it is very common for a ‘function’ to be undefined for
some values of its parameter. In mathematics, a partial function from a set A to a set B
is defined to be a function from a subset of A to B. A partial function from A to B can be
undefined for some elements of A, but when it is defined for some a ∈ A, it associates
just one element of B to a. Many functions in computer programs are actually partial
functions. (When dealing with partial functions, an ordinary function, which is defined
for every element of its domain, is sometimes referred to as a total function. Note that—
with the mind-boggling logic that is typical of mathematicians—a total function is a type
of partial function, because a set is a subset of itself.)
It’s also very common for a ‘function’ in a computer program to produce a variety
of values for the same value of its parameter. A common example is a function with
prototype int random(int N), which returns a random integer between 1 and N. The
value of random(5) could be 1, 2, 3, 4, or 5. This is not the behaviour of a mathematical
function—but it’s very useful when programming!

4

Even though many functions in computer programs are not really mathematical functions, I will continue to refer to them as functions in this section. Mathematicians will
just have to stretch their definitions a bit to accommodate the realities of computer programming.

4.5.1

Functions as first-class objects

In most programming languages, functions are not first-class objects. That is, a function
cannot be treated as a data value in the same way as a string or an int. However, newer
versions of Java do take a step in this direction with ‘lambda expressions’. In Java it is not
yet possible for a function to be a parameter to another function. For example, suppose
in Java we could write the function prototype
1

double sumten(Function <Integer ,Double > f)

This is a prototype for a function named sumten whose parameter is a function. The
parameter is specified by the prototype Function<Integer,Double> f. This means that
the parameter must be a function from int to double. The parameter name, f , stands for an
arbitrary such function. Mathematically, f ∈ doubleint , and so sumten : doubleint → double.
My idea is that sumten( f ) would compute f (1) + f (2) + · · · + f (10). A more useful
function would be able to compute f ( a) + f ( a + 1) + · · · + f (b) for any integers a and b.
This just means that a and b should be parameters to the function. The prototype for the
improved function would look like
1

double sum(Function <Integer ,Double > f, int a, int b)

4.5. Application: Programming with Functions *

127

The parameters to sum form an ordered triple in which the first coordinate is a function
and the second and third coordinates are integers. So, we could write
sum : doubleint × int × int → double
It’s interesting that computer programmers deal routinely with such complex objects.
Languages where functions are first-class objects are for example Python and Scala.
These languages support what is called functional programming.
One of the most accessible languages that supports functional programming is JavaScript, a language that is used on webpages. (Although the names are similar, JavaScript
and Java are only distantly related. You probably knew that.) In JavaScript, the function
that computes the square of its parameter could be defined as
1

3

function square(n) {
return n*n;
}

This is similar to the Java definition of the same function, but you’ll notice that no type
is specified for the parameter n or for the value computed by the function. Given this
definition of square, square(x) would be legal for any x of any type. (Of course, the value
of square(x) would be undefined for most types, so square is a very partial function, like
most functions in JavaScript.) In effect, all possible data values in JavaScript are bundled
together into one set, which I will call data. We then have square : data → data.5
In JavaScript, a function really is a first-class object. We can begin to see this by looking at an alternative definition of the function square:
1

square = function(n) { return n*n; }

Here, the notation function(n) { return n*n; } creates a function that computes the
square of its parameter, but it doesn’t give any name to this function. This function object
is then assigned to a variable named square. The value of square can be changed later, with
another assignment statement, to a different function or even to a different type of value.
This notation for creating function objects can be used in other places besides assignment
statements. Suppose, for example, that a function with prototype function sum(f,a,b)
has been defined in a JavaScript program to compute f ( a) + f ( a + 1) + · · · + f (b). Then
we could compute 12 + 22 + · · · + 1002 by saying
1

sum(function(n) { return n*n; }, 1, 100)

Here, the first parameter is the function that computes squares. We have created and
used this function without ever giving it a name.
It is even possible in JavaScript for a function to return another function as its value.
For example,
5 Not all functional programming languages lump data types together in this way. There is a functional
programming language, Haskell, for example, that is as strict about types as C++. For information about
Haskell, see www.haskell.org.

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1

3

Chapter 4. Sets, Functions, and Relations

function monomial(a, n) {
return (function(x) { a*Math.pow(x,n); });
}

Here, Math.pow(x,n) computes x n , so for any numbers a and n, the value of monomial(a,n)
is a function that computes ax n . Thus,
1

f = monomial (2,3);

would define f to be the function that satisfies f ( x ) = 2x3 , and if sum is the function
described above, then
1

sum( monomial (8,4), 3, 6 )

would compute 8 ∗ 34 + 8 ∗ 44 + 8 ∗ 54 + 8 ∗ 64 . In fact, monomial can be used to create an unlimited number of new functions from scratch. It is even possible to write
monomial(2,3)(5) to indicate the result of applying the function monomial(2,3) to the
value 5. The value represented by monomial(2,3)(5) is 2 ∗ 53 , or 250. This is real functional programming and might give you some idea of its power.

4

Exercises
1. For each of the following Java-like function prototypes, translate the prototype into a standard
mathematical function specification, such as func : float → int.
a) int strlen(string s)
b) double pythag(double x, double y)
c) int round(double x)
d) string sub(string s, int n, int m)
e) string unlikely(Function<String,Integer> f )
f) int h( Function<Integer,Integer> f, Function<Integer,Integer> g )
2. Write a Java-like function prototype for a function that belongs to each of the following sets.
a) stringstring
b) booleanfloat×float
int

c) floatint

3. It is possible to define new types in Java by using classes. For example, the definition
1

3

Class point {
double x;
double y;
}

defines a new type named point. A value of type point contains two values of type double. What
mathematical operation corresponds to the construction of this data type? Why?
4. Let square, sum and monomial be the JavaScript functions described in this section. What is the
value of each of the following?

4.6. Counting Past Infinity

129

a) sum(square, 2, 4)
b) sum(monomial(5,2), 1, 3)
c) monomial(square(2), 7)
d) sum(function(n) { return 2 ∗ n; }, 1, 5)
e) square(sum(monomial(2,3), 1, 2))
5. Write a JavaScript function named compose that computes the composition of two functions.
That is, compose( f ,g) is f ◦ g, where f and g are functions of one parameter. Recall that f ◦ g is
the function defined by ( f ◦ g)( x ) = f ( g( x )).

4.6 Counting Past Infinity
As children, we all learned to answer the question “How many?” by counting with
numbers: 1, 2, 3, 4, …. But the question of “How many?” was asked and answered long
before the abstract concept of number was invented. The answer can be given in terms
of “as many as.” How many cousins do you have? As many cousins as I have fingers
on both hands. How many sheep do you own? As many sheep as there are notches on
this stick. How many baskets of wheat must I pay in taxes? As many baskets as there
are stones in this box. The question of how many things are in one collection of objects
is answered by exhibiting another, more convenient, collection of objects that has just as
many members.
In set theory, the idea of one set having just as many members as another set is expressed in terms of one-to-one correspondence. A one-to-one correspondence between
two sets A and B pairs each element of A with an element of B in such a way that every
element of B is paired with one and only one element of A. The process of counting, as
it is learned by children, establishes a one-to-one correspondence between a set of n objects and the set of numbers from 1 to n. The rules of counting are the rules of one-to-one
correspondence: Make sure you count every object, make sure you don’t count the same
object more than once. That is, make sure that each object corresponds to one and only
one number. Earlier in this chapter, we used the fancy name ‘bijective function’ to refer
to this idea, but we can now see it as as an old, intuitive way of answering the question
“How many?”

4.6.1 Cardinality
In counting, as it is learned in childhood, the set {1, 2, 3, . . . , n} is used as a typical set
that contains n elements. In mathematics and computer science, it has become more
common to start counting with zero instead of with one, so we define the following sets
to use as our basis for counting:

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Chapter 4. Sets, Functions, and Relations
N0
N1
N2
N3
N4

= ∅,
= {0},
= {0, 1},
= {0, 1, 2},
= {0, 1, 2, 3},

a set with 0 elements
a set with 1 element
a set with 2 elements
a set with 3 elements
a set with 4 elements

and so on. In general, Nn = {0, 1, 2, . . . , n − 1} for each n ∈ N. For each natural number
n, Nn is a set with n elements. Note that if n ̸= m, then there is no one-to-one correspondence between Nn and Nm . This is obvious, but like many obvious things is not all
that easy to prove rigorously, and we omit the argument here.
Theorem 4.6. For each n ∈ N, let Nn be the set Nn = {0, 1, . . . , n − 1}. If n ̸= m, then there
is no bijective function from Nm to Nn .

4

We can now make the following definitions:
Definition 4.3. A set A is said to be finite if there is a one-to-one correspondence between
A and Nn for some natural number n. We then say that n is the cardinality of A. The
notation | A| is used to indicate the cardinality of A. That is, if A is a finite set, then
| A| is the natural number n such that there is a one-to-one correspondence between A
and Nn . In layman’s terms: | A| is the number items in A. A set that is not finite is
said to be infinite. That is, a set B is infinite if for every n ∈ N, there is no one-to-one
correspondence between B and Nn .
Fortunately, we don’t always have to count every element in a set individually to
determine its cardinality. Consider, for example, the set A × B, where A and B are finite
sets. If we already know | A| and | B|, then we can determine | A × B| by computation,
without explicit counting of elements. In fact, | A × B| = | A| · | B|. The cardinality of
the cross product A × B can be computed by multiplying the cardinality of A by the
cardinality of B. To see why this is true, think of how you might count the elements of
A × B. You could put the elements into piles, where all the ordered pairs in a pile have
the same first coordinate. There are as many piles as there are elements of A, and each
pile contains as many ordered pairs as there are elements of B. That is, there are | A| piles,
with | B| items in each. By the definition of multiplication, the total number of items in all
the piles is | A| · | B|. A similar result holds for the cross product of more that two finite
sets. For example, | A × B × C | = | A| · | B| · |C |.
It’s also easy to compute | A ∪ B| in the case where A and B are disjoint finite sets.
(Recall that two sets A and B are said to be disjoint if they have no members in common,
that is, if A ∩ B = ∅.) Suppose | A| = n and | B| = m. If we wanted to count the elements
of A ∪ B, we could use the n numbers from 0 to n − 1 to count the elements of A and then
use the m numbers from n to n + m − 1 to count the elements of B. This amounts to a oneto-one correspondence between A ∪ B and the set Nn+m . We see that | A ∪ B| = n + m.
That is, for disjoint finite sets A and B, | A ∪ B| = | A| + | B|.
What about A ∪ B, where A and B are not disjoint? We have to be careful not to
count the elements of A ∩ B twice. After counting the elements of A, there are only

4.6. Counting Past Infinity

131

| B| − | A ∩ B| new elements in B that still need to be counted. So we see that for any two
finite sets A and B, | A ∪ B| = | A| + | B| − | A ∩ B|.
What about the number of subsets of a finite set A? What is the relationship between
| A| and |P( A)|? The answer is provided by the following theorem.
Theorem 4.7. A finite set with cardinality n has 2n subsets.
Proof. Let P(n) be the statement “Any set with cardinality n has 2n subsets”. We will use
induction to show that P(n) is true for all n ∈ N.
Base case: For n = 0, P(n) is the statement that a set with cardinality 0 has 20 subsets.
The only set with 0 elements is the empty set. The empty set has exactly 1 subset, namely
itself. Since 20 = 1, P(0) is true.
Inductive case: Let k be an arbitrary element of N, and assume that P(k) is true.
That is, assume that any set with cardinality k has 2k elements. (This is the induction
hypothesis.) We must show that P(k + 1) follows from this assumption. That is, using
the assumption that any set with cardinality k has 2k subsets, we must show that any set
with cardinality k + 1 has 2k+1 subsets.
Let A be an arbitrary set with cardinality k + 1. We must show that |P( A)| = 2k+1 .
Since | A| > 0, A contains at least one element. Let x be some element of A, and let
B = A ∖ { x }. The cardinality of B is k, so we have by the induction hypothesis that
|P( B)| = 2k . Now, we can divide the subsets of A into two classes: subsets of A that do
not contain x and subsets of A that do contain x. Let Y be the collection of subsets of
A that do not contain x, and let X be the collection of subsets of A that do contain x. X
and Y are disjoint, since it is impossible for a given subset of A both to contain and to
not contain x. It follows that |P( A)| = | X ∪ Y | = | X | + |Y |.
Now, a member of Y is a subset of A that does not contain x. But that is exactly
the same as saying that a member of Y is a subset of B. So Y = P( B), which we know
contains 2k members. As for X, there is a one-to-one correspondence between P( B) and
X. Namely, the function f : P( B) → X defined by f (C ) = C ∪ { x } is a bijective function.
(The proof of this is left as an exercise.) From this, it follows that | X | = |P( B)| = 2k .
Putting these facts together, we see that |P( A)| = | X | + |Y | = 2k + 2k = 2 · 2k = 2k+1 .
This completes the proof that P(k ) → P(k + 1).
We have seen that the notation A B represents the set of all functions
  from B to A.
Suppose A and B are finite, and that | A| = n and | B| = m. Then  A B  = nm = | A|| B| .
(This fact is one of the reasons why the notation A B is reasonable.) One way to see
this is to note that there is a one-to-one correspondence between A B and a cross product
A × A × · · · A, where the number of terms in the cross product is m. (This will be shown
in one of the exercises at the end of this section.) It follows that  A B  = | A| · | A| · · · | A| =
n · n · · · n, where the factor n occurs m times in the product. This product is, by definition, nm .
This discussion about computing cardinalities is summarized in the following theorem:

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Chapter 4. Sets, Functions, and Relations

Theorem 4.8. Let A and B be finite sets. Then
• | A × B | = | A | · | B |.
• | A ∪ B | = | A | + | B | − | A ∩ B |.
• If A and B are disjoint then | A ∪ B| = | A| + | B|.
 
•  A B  = | A || B| .
• |P( A)| = 2| A| .

4

When it comes to counting and computing cardinalities, this theorem is only the beginning of the story. There is an entire large and deep branch of mathematics known
as combinatorics that is devoted mostly to the problem of counting. But the theorem is
already enough to answer many questions about cardinalities.
For example, suppose that | A| = n and | B| = m. We can form the set P( A × B), which
consists of all subsets of A × B. Using the theorem, we can compute that |P( A × B)| =
2| A× B | = 2| A|·| B| = 2nm . If we assume that A and B are disjoint, then we can compute
that  A A∪ B  = | A|| A∪ B| = nn+m .
To be more concrete, let X = { a, b, c, d, e} and let Y = {c, d, e, f } where a,
b, c, d, e, and f are distinct.
| X × Y | = 5 · 4 = 20 while | X ∪ Y | =
 Then

5 + 4 − |{c, d, e}| = 6 and  XY  = 54 = 625.
We can also answer some simple practical questions. Suppose that in a restaurant you can



choose one appetizer and one main course. What is the number of possible meals? If A
is the set of possible appetizers and C is the set of possible main courses, then your meal
is an ordered pair belonging to the set A × C. The number of possible meals is | A × C |,
which is the product of the number of appetizers and the number of main courses.
Or suppose that four different prizes are to be awarded, and that the set of people
who are eligible for the prizes is A. Suppose that | A| = n. How many different ways are
there to award the prizes? One way to answer this question is to view a way of awarding
the prizes as a function from the set of prizes to the set of people.
if P is the set of
 Then,

prizes, the number of different ways of awarding the prizes is  A P . Since | P| = 4 and
| A| = n, this is n4 . Another way to look at it is to note that the people who win the prizes
form an ordered tuple ( a, b, c, d), which is an element of A × A × A × A. So the number
of different ways of awarding the prizes is | A × A × A × A|, which is | A| · | A| · | A| · | A|.
This is | A|4 , or n4 , the same answer we got before.6
6 This discussion assumes that one person can receive any number of prizes. What if the prizes have to go
to four different people? This question takes us a little farther into combinatorics than I would like to go, but
the answer is not hard. The first award can be given to any of n people. The second prize goes to one of the
remaining n − 1 people. There are n − 2 choices for the third prize and n − 3 for the fourth. The number of
different ways of awarding the prizes to four different people is the product n(n − 1)(n − 2)(n − 3).

4.6. Counting Past Infinity

133

4.6.2 Counting to infinity
So far, we have only discussed finite sets. N, the set of natural numbers {0, 1, 2, 3, . . . }, is
an example of an infinite set. There is no one-to-one correspondence between N and any
of the finite sets Nn . Another example of an infinite set is the set of even natural numbers,
E = {0, 2, 4, 6, 8, . . . }. There is a natural sense in which the sets N and E have the same
number of elements. That is, there is a one-to-one correspondence between them. The
function f : N → E defined by f (n) = 2n is bijective. We will say that N and E have
the same cardinality, even though that cardinality is not a finite number. Note that E is
a proper subset of N. That is, N has a proper subset that has the same cardinality as N.
We will see that not all infinite sets have the same cardinality. When it comes to
infinite sets, intuition is not always a good guide. Most people seem to be torn between
two conflicting ideas. On the one hand, they think, it seems that a proper subset of a set
should have fewer elements than the set itself. On the other hand, it seems that any two
infinite sets should have the same number of elements. Neither of these is true, at least if
we define having the same number of elements in terms of one-to-one correspondence.
A set A is said to be countably infinite if there is a one-to-one correspondence between
N and A. A set is said to be countable if it is either finite or countably infinite. An infinite set that is not countably infinite is said to be uncountable. If X is an uncountable set,
then there is no one-to-one correspondence between N and X.
The idea of ‘countable infinity’ is that even though a countably infinite set cannot
be counted in a finite time, we can imagine counting all the elements of A, one-by-one,
in an infinite process. A bijective function f : N → A provides such an infinite listing:
( f (0), f (1), f (2), f (3), . . . ). Since f is onto, this infinite list includes all the elements of
A. In fact, making such a list effectively shows that A is countably infinite, since the list
amounts to a bijective function from N to A. For an uncountable set, it is impossible to
make a list, even an infinite list, that contains all the elements of the set.
Before you start believing in uncountable sets, you should ask for an example. In
Chapter 3, we worked with the infinite sets Z (the integers), Q (the rationals), R (the
reals), and R ∖ Q (the irrationals). Intuitively, these are all ‘bigger’ than N, but as we
have already mentioned, intuition is a poor guide when it comes to infinite sets. Are any
of Z, Q, R, and R ∖ Q in fact uncountable?
It turns out that both Z and Q are only countably infinite. The proof that Z is countable is left as an exercise; we will show here that the set of non-negative rational numbers
is countable. (The fact that Q itself is countable follows easily from this.) The reason is
that it’s possible to make an infinite list containing all the non-negative rational numbers.
Start the list with all the non-negative rational numbers n/m such that n + m = 1. There
is only one such number, namely 0/1. Next come numbers with n + m = 2. They are
0/2 and 1/1, but we leave out 0/2 since it’s just another way of writing 0/1, which is
already in the list. Now, we add the numbers with n + m = 3, namely 0/3, 1/2, and
2/1. Again, we leave out 0/3, since it’s equal to a number already in the list. Next come
numbers with n + m = 4. Leaving out 0/4 and 2/2 since they are already in the list, we

4

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Chapter 4. Sets, Functions, and Relations

add 1/3 and 3/1 to the list. We continue in this way, adding numbers with n + m = 5,
then numbers with n + m = 6, and so on. The list looks like
)
(
0 1 1 2 1 3 1 2 3 4 1 5 1 2
, , , , , , , , , , , , , ,...
1 1 2 1 3 1 4 3 2 1 5 1 6 5
This process can be continued indefinitely, and every non-negative rational number will
eventually show up in the list. So we get a complete, infinite list of non-negative rational
numbers. This shows that the set of non-negative rational numbers is in fact countable.

4.6.3

4

Uncountable sets

On the other hand, R is uncountable. It is not possible to make an infinite list that contains every real number. It is not even possible to make a list that contains every real
number between zero and one. Another way of saying this is that every infinite list of
real numbers between zero and one, no matter how it is constructed, leaves something
out. To see why this is true, imagine such a list, displayed in an infinitely long column.
Each row contains one number, which has an infinite number of digits after the decimal
point. Since it is a number between zero and one, the only digit before the decimal point
is zero. For example, the list might look like this:
0.90398937249879561297927654857945…
0.12349342094059875980239230834549…
0.22400043298436234709323279989579…
0.50000000000000000000000000000000…
0.77743449234234876990120909480009…
0.77755555588888889498888980000111…
0.12345678888888888888888800000000…
0.34835440009848712712123940320577…
0.93473244447900498340999990948900…
..
.
This is only (a small part of) one possible list. How can we be certain that every such
list leaves out some real number between zero and one? The trick is to look at the digits
shown in bold face. We can use these digits to build a number that is not in the list.
Since the first number in the list has a 9 in the first position after the decimal point,
we know that this number cannot equal any number of, for example, the form 0.4….
Since the second number has a 2 in the second position after the decimal point, neither
of the first two numbers in the list is equal to any number that begins with 0.44…. Since
the third number has a 4 in the third position after the decimal point, none of the first
three numbers in the list is equal to any number that begins 0.445…. We can continue to
construct a number in this way, and we end up with a number that is different from every
number in the list. The nth digit of the number we are building must differ from the nth

4.6. Counting Past Infinity

135

digit of the nth number in the list. These are the digits shown in bold face in the above
list. To be definite, I use a 5 when the corresponding boldface number is 4, and otherwise
I use a 4. For the list shown above, this gives a number that begins 0.44544445…. The
number constructed in this way is not in the given list, so the list is incomplete. The same
construction clearly works for any list of real numbers between zero and one. No such
list can be a complete listing of the real numbers between zero and one, and so there can
be no complete listing of all real numbers. We conclude that the set R is uncountable.
The technique used in this argument is called diagonalization. It is named after the
fact that the bold face digits in the above list lie along a diagonal line.



Proofs by diagonalisation are not a part of Reasoning & Logic. You should be
able to prove a set is countably infinite (by finding a bijection from N to the
set), but you will not be asked to prove a set is uncountable. We leave that for
the course Automata, Languages, and Computability later in your curriculum.

This proof was discovered by a mathematician named Georg Cantor, who caused
quite a fuss in the nineteenth century when he came up with the idea that there are
different kinds of infinity. Since then, his notion of using one-to-one correspondence to
define the cardinalities of infinite sets has been accepted. Mathematicians now consider
it almost intuitive that N, Z, and Q have the same cardinality while R has a strictly larger
cardinality.



To say that Georg Cantor (1845–1918) caused a fuss in
mathematics is the least you can say. Cantor’s theory
of transfinite numbers was originally regarded as so
counter-intuitive—even shocking—that it encountered
resistance from mathematical contemporaries. Like
Frege, Cantor was a German mathematician contributing to logic and set theory. Like Peirce, Cantor’s ideas
did not find acceptance until later in his life. In 1904,
the Royal Society awarded Cantor its Sylvester Medal,
the highest honour it can confer for work in mathematics.
Source: en.wikipedia.org/wiki/Georg_Cantor.

Theorem 4.9. Suppose that X is an uncountable set, and that K is a countable subset of X. Then
the set X ∖ K is uncountable.

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Chapter 4. Sets, Functions, and Relations

Proof. Let X be an uncountable set. Let K ⊆ X, and suppose that K is countable. Let
L = X ∖ K. We want to show that L is uncountable. Suppose that L is countable. We
will show that this assumption leads to a contradiction.
Note that X = K ∪ ( X ∖ K ) = K ∪ L. You will show in Exercise 11 of this section that
the union of two countable sets is countable. Since X is the union of the countable sets K
and L, it follows that X is countable. But this contradicts the fact that X is uncountable.
This contradiction proves the theorem.

4

In the proof, both q and ¬q are shown to follow from the assumptions, where q is
the statement “X is countable”. The statement q is shown to follow from the assumption
that X ∖ K is countable. The statement ¬q is true by assumption. Since q and ¬q cannot
both be true, at least one of the assumptions must be false. The only assumption that can
be false is the assumption that X ∖ K is countable.
This theorem, by the way, has the following easy corollary. (A corollary is a theorem
that follows easily from another, previously proved theorem.)
Corollary 4.10. The set of irrational real numbers is uncountable.
Proof. Let I be the set of irrational real numbers. By definition, I = R ∖ Q. We have
already shown that R is uncountable and that Q is countable, so the result follows immediately from the previous theorem.
You might still think that R is as big as things get, that is, that any infinite set is in
one-to-one correspondence with R or with some subset of R. In fact, though, if X is
any set then it’s possible to find a set that has strictly larger cardinality than X. In fact,
P( X ) is such a set. A variation of the diagonalization technique can be used to show that
there is no one-to-one correspondence between X and P( X ). Note that this is obvious
for finite sets, since for a finite set X, |P( X )| = 2| X | , which is larger than | X |. The point
of the theorem is that it is true even for infinite sets.
Theorem 4.11. Let X be any set. Then there is no one-to-one correspondence between X and
P( X ).
Proof. Given an arbitrary function f : X → P( X ), we can show that f is not onto. Since
a one-to-one correspondence is both one-to-one and onto, this shows that f is not a oneto-one correspondence.
Recall that P( X ) is the set of subsets of X. So, for each x ∈ X, f ( x ) is a subset of X.
We have to show that no matter how f is defined, there is some subset of X that is not in
the image of f .
Given f , we define A to be the set A = { x ∈ X | x ̸∈ f ( x )}. The test “x ̸∈ f ( x )”
makes sense because f ( x ) is a set. Since A ⊆ X, we have that A ∈ P( X ). However, A
is not in the image of f . That is, for every y ∈ X, A ̸= f (y).7 To see why this is true,
7 In fact, we have constructed A so that the sets A and f ( y ) differ in at least one element, namely y itself.
This is where the ‘diagonalization’ comes in.

4.6. Counting Past Infinity

137

let y be any element of X. There are two cases to consider. Either y ∈ f (y) or y ̸∈ f (y).
We show that whichever case holds, A ̸= f (y). If it is true that y ∈ f (y), then by the
definition of A, y ̸∈ A. Since y ∈ f (y) but y ̸∈ A, f (y) and A do not have the same
elements and therefore are not equal. On the other hand, suppose that y ̸∈ f (y). Again,
by the definition of A, this implies that y ∈ A. Since y ̸∈ f (y) but y ∈ A, f (y) and A do
not have the same elements and therefore are not equal. In either case, A ̸= f (y). Since
this is true for any y ∈ X, we conclude that A is not in the image of f and therefore f is
not a one-to-one correspondence.
From this theorem, it follows that there is no one-to-one correspondence between
R and P(R). The cardinality of P(R) is strictly bigger than the cardinality of R. But
it doesn’t stop there. P(P(R)) has an even bigger cardinality, and the cardinality of
P(P(P(R))) is bigger still. We could go on like this forever, and we still won’t have
exhausted all the possible cardinalities. If we let X be the infinite union R ∪ P(R) ∪
P(P(R)) ∪ · · · , then X has larger cardinality than any of the sets in the union. And then
there’s P(X), P(P(X)), X ∪ P(X) ∪ P(P(X)) ∪ · · · . There is no end to this. There is no
upper limit on possible cardinalities, not even an infinite one! We have counted past
infinity.

4.6.4 A final note on infinities
We have seen that |R| is strictly larger than |N|. We end this section with what might look
like a simple question: Is there a subset of R that is neither in one-to-one correspondence
with N nor with R? That is, is the cardinality of R the next largest cardinality after
the cardinality of N, or are there other cardinalities intermediate between them? This
problem was unsolved for quite a while, and the solution, when it was found, proved to
be completely unexpected. It was shown that both ‘yes’ and ‘no’ are consistent answers
to this question! That is, the logical structure built on the system of axioms that had
been accepted as the basis of set theory was not extensive enough to answer the question.
The question is ‘undecidable’ in that system. We will come back to ‘undecidability’ in
Chapter 5.
It is possible to extend the system of axioms underlying set theory in various ways. In
some extensions, the answer is yes. In others, the answer is no. You might object, “Yes,
but which answer is true for the real real numbers?” Unfortunately, it’s not even clear
whether this question makes sense, since in the world of mathematics, the real numbers
are just part of a structure built from a system of axioms. And it’s not at all clear whether
the ‘real numbers’ exist in some sense in the real world. If all this sounds like it’s a bit
of a philosophical muddle, it is. That’s the state of things today at the foundation of
mathematics, and it has implications for the foundations of computer science, as we’ll
see in the next chapter.

4

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Chapter 4. Sets, Functions, and Relations

Exercises

4

1. Suppose that A, B, and C are finite sets which are pairwise disjoint. (That is, A ∩ B = A ∩ C =
B ∩ C = ∅.) Express the cardinality of each of the following sets in terms of | A|, | B|, and |C |.
Which of your answers depend on the fact that the sets are pairwise disjoint?
a) P( A ∪ B)
b) A × ( BC )
c) P( A) × P(C )
d) A B×C
e) ( A × B)C
f) P( A B )
C
g) ( A ∪ B)
h) ( A ∪ B) × A
i) A × A × B × B
2. Suppose that A and B are finite sets which are not necessarily disjoint. What are all the possible
values for | A ∪ B| ?
3. Let’s say that an ‘identifier’ consists of one or two characters. The fist character is one of the
twenty-six letters (A, B, …, C). The second character, if there is one, is either a letter or one of
the ten digits (0, 1, …, 9). How many different identifiers are there? Explain your answer in
terms of unions and cross products.
4. Suppose that there are five books that you might bring along to read on your vacation. In how
many different ways can you decide which books to bring, assuming that you want to bring at
least one? Why?
5. Show that the cardinality of a finite set is well-defined. That is, show that if f is a bijective
function from a set A to Nn , and if g is a bijective function from A to Nm , then n = m.
6. Finish the proof of Theorem 4.7 by proving the following statement: Let A be a non-empty set,
and let x ∈ A. Let B = A ∖ { x }. Let X = {C ⊆ A | x ∈ C }. Define f : P( B) → X by the
formula f (C ) = C ∪ { x }. Show that f is a bijective function.
 
7. Use induction on the cardinality of B to show that for any finite sets A and B,  A B  = | A|| B| .
(Hint: For the case where B ̸= ∅, choose x ∈ B, and divide A B into classes according to the
value of f ( x ).)
8. Let A and B be finite sets with | A| = n and | B| = m. List the elements of B as B = {b0 , b1 , . . . , bm−1 }.
Define the (function F : A B → A × A ×
) · · · × A, where A occurs m times in the cross product,
by F ( f ) = f (b0 ), f (b1 ), . . . , f (bm−1 ) . Show that F is a one-to-one correspondence.
9. Show that Z, the set of integers, is countable by finding a one-to-one correspondence between
N and Z.
10. Show that the set N × N is countable.
11. Complete the proof of Theorem 2.9 as follows:
a) Suppose that A and B are countably infinite sets. Show that A ∪ B is countably infinite.
b) Suppose that A and B are countable sets. Show that A ∪ B is countable.
12. Prove that each of the following statements is true. In each case, use a proof by contradiction.
a) Let X be a countably infinite set, and let N be a finite subset of X. Then X ∖ N is countably infinite.
b) Let A be an infinite set, and let X be a subset of A. Then at least one of the sets X and
A ∖ X is infinite.
c) Every subset of a finite set is finite.
13. Let A and B be sets and let ⊥ be an entity that is not a member of B. Show that there is a
one-to-one correspondence between the set of functions from A to B ∪ {⊥} and the set of
partial functions from A to B. (Partial functions were defined in Section 4.5. The symbol ‘⊥’
is sometimes used in theoretical computer science to represent the value ‘undefined.’)

4.7. Relations

139

4.7 Relations
In Section 4.4, we saw that ‘mother of’ is a functional relationship because every person
has one and only one mother, but that ‘child of’ is not a functional relationship, because a
person can have no children or more than one child. However, the relationship expressed
by ‘child of’ is certainly one that we have a right to be interested in and one that we should
be able to deal with mathematically.
There are many examples of relationships that are not functional relationships. The
relationship that holds between two natural numbers n and m when n ≤ m is an example
in mathematics. The relationship between a person and a book that that person has
on loan from the library is another. Some relationships involve more than two entities,
such as the relationship that associates a name, an address, and a phone number in
an address book or the relationship that holds among three real numbers x, y, and z if
x2 + y2 + z2 = 1. Each of these relationships can be represented mathematically by what
is called a ‘relation’.
A relation on two sets, A and B, is defined to be a subset of A × B. Since a function
from A to B is defined, formally, as a subset of A × B that satisfies certain properties,
a function is a relation. However, relations are more general than functions, since any
subset of A × B is a relation. We also define a relation among three or more sets to be
a subset of the cross product of those sets. In particular, a relation on A, B, and C is a
subset of A × B × C.



For example, if P is the set of people and B is the set of books owned by a
library, then we can define a relation R on the sets P and B to be the set R =
{( p, b) ∈ P × B | p has b out on loan}. The fact that a particular ( p, b) ∈ R
is a fact about the world that the library will certainly want to keep track of.
When a collection of facts about the world is stored on a computer, it is called
a database. We’ll see in the next section that relations are the most common
means of representing data in databases.

If A is a set and R is a relation on the sets A and A (that is, on two copies of A), then
R is said to be a binary relation on A. That is, a binary relation on the set A is a subset of
A × A. The relation consisting of all ordered pairs (c, p) of people such that c is a child
of p is a binary relation on the set of people. The set {(n, m) ∈ N × N | n ≤ m} is a
binary relation on N. Similarly, we define a ternary relation on a set A to be a subset of
A × A × A. The set {( x, y, z) ∈ R × R × R | x2 + y2 + z2 = 1} is a ternary relation on R.
For complete generality, we can define an n-ary relation on A, for any positive integer
n, to be a subset of A × A × · · · × A, where A occurs n times in the cross product.
For the rest of this section, we will be working exclusively with binary relations. Suppose that R ⊆ A × A. That is, suppose that R is a binary relation on a set A. If ( a, b) ∈ R,
then we say that a is related to b by R. Instead of writing “( a, b) ∈ R”, we will often write

4

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Chapter 4. Sets, Functions, and Relations

“a R b”. This notation is used in analogy to the notation n ≤ m to express the relation
that n is less than or equal to m. Remember that a R b is just an alternative way of writing
( a, b) ∈ R. In fact, we could consider the relation ≤ to be a set of ordered pairs and write
(n, m) ∈ ≤ in place of the notation n ≤ m.

4.7.1

4

Properties of relations

In many applications, attention is restricted to relations that satisfy some property or set
of properties. (This is, of course, just what we do when we study functions.) We begin
our discussion of binary relations by considering several important properties. In this
discussion, let A be a set and let R be a binary relation on A, that is, a subset of A × A.
R is said to be reflexive if ∀ a ∈ A ( a R a). That is, a binary relation on a set is reflexive
if every element of the set is related to itself. This is true, for example, for the relation
≤ on the set N, since n ≤ n for every n ∈ N. On the other hand, it is not true for the
relation < on N, since, for example, the statement 17( < 17 is false.8
)
R is called transitive if ∀ a ∈ A, ∀b ∈ A, ∀c ∈ A ( a R b ∧ b R c) → ( a R c) . Transitivity allows us to ‘chain together’ two true statements a R b and b R c, which are ‘linked’
by the b that occurs in each statement, to deduce that a R c. For example, suppose P is
the set of people, and define the relation C on P such that x P y if and only if x is a child
of y. The relation P is not transitive because the child of a child of a person is not a child
of that person. Suppose, on the other hand, that we define a relation D on P such that
x D y if and only if x is a descendent of y. Then D is a transitive relation on the set of
people, since a descendent of a descendent of a person is a descendent of that person.
That is, from the facts that Elizabeth is a descendent of Victoria and Victoria is a descendent of James, we can deduce that Elizabeth is a descendent of James. In the mathematical
world, the relations ≤ and < on the set N are both transitive.
R is said to be symmetric if ∀ a ∈ A, ∀b ∈ B ( a R b → b R a). That is, whenever a is
related to b, it follows that b is related to a. The relation “is a first cousin of” on the set of
people is symmetric, since whenever x is a first cousin of y, we have automatically that y
is a first cousin of x. On the other hand, the “child of” relation is certainly not symmetric.
The relation ≤ on N is not symmetric. From the fact that n ≤ m, we cannot conclude
that m ≤ n. It is true for some n and m in N that n ≤ m → m ≤ n, but it is not true for all
n and m in N.
(
)
Finally, R is antisymmetric if ∀ a ∈ A, ∀b ∈ B ( a R b ∧ b R a) → a = b . The relation
R is antisymmetric if for any two distinct elements x and y of A, we can’t have both x R y
and y R x. The relation ≤ on N is antisymmetric because from the facts that n ≤ m
and m ≤ n, we can deduce that n = m. The relation ‘child of’ on the set of people is
antisymmetric since it’s impossible to have both that x is a child of y and y is a child of x.
8 Note that to show that the relation R is not reflexive, you only need to find one a such that a R a is false.
(
)
(
)
This follows from the fact that ¬ ∀ a ∈ A ( a R a) ≡ ∃ a ∈ A ¬( a R a) . A similar remark holds for each of the
properties of relations that are discussed here.

4.7. Relations



141

During lectures, we’ll think about how to draw relations graphically. See the
figure in Section 4.4.3 for one kind of graphical depiction.

There are a few combinations of properties that define particularly useful types of
binary relations. The relation ≤ on the set N is reflexive, antisymmetric, and transitive.
These properties define what is called a partial order: a partial order on a set A is a
binary relation on A that is reflexive, antisymmetric, and transitive.
Another example of a partial order is the subset relation, ⊆, on the power set of any
set. If X is a set, then of course P( X ) is a set in its own right, and ⊆ can be considered to
be a binary relation on this set. Two elements A and B of P( X ) are related by ⊆ if and
only if A ⊆ B. This relation is reflexive since every set is a subset of itself. The fact that
it is antisymmetric follows from Theorem 4.1. The fact that it is transitive was Exercise
11 in Section 4.1.
The ordering imposed on N by ≤ has one important property that the ordering of
subsets by ⊆ does not share. If n and m are natural numbers, then at least one of the
statements n ≤ m and m ≤ n must be true. However, if A and B are subsets of a set X, it
is certainly possible that both A ⊆ B and B ⊆ A are false. A binary relation R on a set A
is said to be a total order if it is a partial order and furthermore for any two elements a
and b of A, either a R b or b R a. The relation ≤ on the set N is a total order. The relation ⊆
on P( X ) is not. (Note once again the slightly odd mathematical language: A total order
is a kind of partial order—not, as you might expect, the opposite of a partial order.)
For another example of ordering, let L be the set of strings that can be made from
lowercase letters. L contains both English words and nonsense strings such as “sxjja”.
There is a commonly used total order on the set L, namely alphabetical order.

4.7.2 Equivalence relations
We’ll approach another important kind of binary relation indirectly, through what might
at first appear to be an unrelated idea. Let A be a set. A partition of A is defined to be
a collection of non-empty subsets of A such that each pair of distinct subsets in the collection is disjoint and the union of all the subsets in the collection is A. A partition of A
is just a division of all the elements of A into non-overlapping subsets. For example, the
sets {1, 2, 6}, {3, 7}, {4, 5, 8, 10}, and {9} form a partition of the set {1, 2, . . . , 10}. Each
element of {1, 2, . . . , 10} occurs in exactly one of the sets that make up the partition. As
another example, we can partition the set of all people into two sets, the set of males and
the set of females. Biologists try to partition the set of all organisms into different species. Librarians try to partition books into various categories such as fiction, biography,
and poetry. In the real world, classifying things into categories is an essential activity,
although the boundaries between categories are not always well-defined. The abstract
mathematical notion of a partition of a set models the real-world notion of classification.
In the mathematical world, though, the categories are sets and the boundary between

4

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4

Chapter 4. Sets, Functions, and Relations

two categories is sharp.
In the real world, items are classified in the same category because they are related
in some way. This leads us from partitions back to relations. Suppose that we have a
partition of a set A. We can define a relation R on A by declaring that for any a and b in
A, a R b if and only if a and b are members of the same subset in the partition. That is,
two elements of A are related if they are in the same category. It is clear that the relation
defined in this way is reflexive, symmetric, and transitive.
An equivalence relation is defined to be a binary relation that is reflexive, symmetric,
and transitive. Any relation defined, as above, from a partition is an equivalence relation.
Conversely, we can show that any equivalence relation defines a partition. Suppose that
R is an equivalence relation on a set A. Let a ∈ A. We define the equivalence class of a
under the equivalence relation R to be the subset [ a]R defined as [ a]R = {b ∈ A | b R a}.
That is, the equivalence class of a is the set of all elements of A that are related to a. In
most cases, we’ll assume that the relation in question is understood, and we’ll write [ a]
instead of [ a]R . Note that each equivalence class is a subset of A. The following theorem
shows that the collection of equivalence classes form a partition of A.
Theorem 4.12. Let A be a set and let R be an equivalence relation on A. Then the collection of
all equivalence classes under R is a partition of A.
Proof. To show that a collection of subsets of A is a partition, we must show that each
subset is non-empty, that the intersection of two distinct subsets is empty, and that the
union of all the subsets is A.
If [ a] is one of the equivalence classes, it is certainly non-empty, since a ∈ [ a]. (This
follows from the fact that R is reflexive, and hence a R a.) To show that A is the union of
all the equivalence classes, we just have to show that each element of A is a member of
one of the equivalence classes. Again, the fact that a ∈ [ a] for each a ∈ A shows that this
is true.
Finally, we have to show that the intersection of two distinct equivalence classes is
empty. Suppose that a and b are elements of A and consider the equivalence classes [ a]
and [b]. We have to show that if [ a] ̸= [b], then [ a] ∩ [b] = ∅. Equivalently, we can show
the converse: If [ a] ∩ [b] ̸= ∅ then [ a] = [b]. So, assume that [ a] ∩ [b] ̸= ∅. Saying that
a set is not empty just means that the set contains some element, so there must be an
x ∈ A such that x ∈ [ a] ∩ [b]. Since x ∈ [ a], x R a. Since R is symmetric, we also have
a R x. Since x ∈ [b], x R b. Since R is transitive and since ( a R x ) ∧ ( x R b), it follows that
a R b.
Our object is to deduce that [ a] = [b]. Since [ a] and [b] are sets, they are equal if and
only if [ a] ⊆ [b] and [b] ⊆ [ a]. To show that [ a] ⊆ [b], let c be an arbitrary element of
[ a]. We must show that c ∈ [b]. Since c ∈ [ a], we have that c R a. And we have already
shown that a R b. From these two facts and the transitivity of R, it follows that c R b. By
definition, this means that c ∈ [b]. We have shown that any member of [ a] is a member
of [b] and therefore that [ a] ⊆ [b]. The fact that [b] ⊆ [ a] can be shown in the same way.
We deduce that [ a] = [b], which proves the theorem.

4.7. Relations

143

The point of this theorem is that if we can find a binary relation that satisfies certain
properties, namely the properties of an equivalence relation, then we can classify things
into categories, where the categories are the equivalence classes.
For example, suppose that U is a possibly infinite set. Define a binary relation ∼ on
P(U ) as follows: For X and Y in P(U ), X ∼ Y if and only if there is a bijective function
from the set X to the set Y. In other words, X ∼ Y means that X and Y have the same
cardinality. Then ∼ is an equivalence relation on P(U ). (The symbol ∼ is often used to
denote equivalence relations. It is usually read ‘is equivalent to’.) If X ∈ P(U ), then the
equivalence class [ X ]∼ consists of all the subsets of U that have the same cardinality as
X. We have classified all the subsets of U according to their cardinality—even though
we have never said what an infinite cardinality is. (We have only said what it means to
have the same cardinality.)
You might know the popular puzzle called Rubik’s
Cube9 , a cube made of smaller cubes with coloured sides
that could be manipulated by twisting layers of little cubes.
The object was to manipulate the cube so that the colours of
the little cubes formed a certain configuration. Define two
configurations of the cube to be equivalent if it’s possible to
manipulate one configuration into the other by a sequence
of twists. This is, in fact, an equivalence relation on the set
of possible configurations. (Symmetry follows from the fact
that each move is reversible.) It has been shown that this
equivalence relation has exactly twelve equivalence classes.
The interesting fact is that it has more than one equivalence
class: If the configuration that the cube is in and the configuration that you want to
achieve are not in the same equivalence class, then you are doomed to failure.



Fortunately if you buy the cube in the configuration you want it to be in (i.e.,
with 6 faces all having just one colour), you can still be successful. Since all
configurations you can get to using legal moves are in the same class, you
can only move to another class of configurations with an illegal move. So it
is only by taking the cube apart and incorrectly putting it back together that
you can really change this puzzle from difficult to impossible.

Suppose that R is a binary relation on a set A. Even though R might not be transitive,
it is always possible to construct a transitive relation from R in a natural way. If we
think of a R b as meaning that a is related by R to b ‘in one step’, then we consider the
relationship that holds between two elements x and y when x is related by R to y ‘in
one or more steps’. This relationship defines a binary relation on A that is called the
9 Image:

commons.wikimedia.org/wiki/File:Rubiks_cube_scrambled.jpg.

4

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Chapter 4. Sets, Functions, and Relations

transitive closure of R. The transitive closure of R is denoted R∗ . Formally, R∗ is defined
as follows: For a and b in A, a R∗ b if there is a sequence x0 , x1 , . . . xn of elements of A,
where n > 0 and x0 = a and xn = b, such that x0 R x1 , x1 R x2 , …, and xn−1 R xn .



You will revist the notion of transitive closures in the course Automata, Languages, and Computability.

For example, if a R c, c R d, and d R b, then we would have that a R∗ b. Of course, we
would also have that a R∗ c, and a R∗ d.

4



For a practical example, suppose that C is the set of all cities and let A be the
binary relation on C such that for x and y in C, x A y if there is a regularly
scheduled airline flight from x to y. Then the transitive closure A∗ has a
natural interpretation: x A∗ y if it’s possible to get from x to y by a sequence
of one or more regularly-scheduled airline flights. You’ll find a few more
examples of transitive closures in the exercises.

Exercises
1. For a finite set, it is possible to define a binary relation on the set by listing the elements of the
relation, considered as a set of ordered pairs. Let A be the set { a, b, c, d}, where a, b, c, and
d are distinct. Consider each of the following binary relations on A. Is the relation reflexive?
Symmetric? Antisymmetric? Transitive? Is it a partial order? An equivalence relation?
a) R = {( a, b), ( a, c), ( a, d)}.
b) S = {( a, a), (b, b), (c, c), (d, d), ( a, b), (b, a)}.
c) T = {(b, b), (c, c), (d, d)}.
d) C = {( a, b), (b, c), ( a, c), (d, d)}.
e) D = {( a, b), (b, a), (c, d), (d, c)}.
2. Let A be the set {1, 2, 3, 4, 5, 6}. Consider the partition of A into the subsets {1, 4, 5}, {3}, and
{2, 6}. Write out the associated equivalence relation on A as a set of ordered pairs.
3. Consider each of the following relations on the set of people. Is the relation reflexive? Symmetric? Transitive? Is it an equivalence relation?
a) x is related to y if x and y have the same biological parents.
b) x is related to y if x and y have at least one biological parent in common.
c) x is related to y if x and y were born in the same year.
d) x is related to y if x is taller than y.
e) x is related to y if x and y have both visited Indonesia.
4. It is possible for a relation to be both symmetric and antisymmetric. For example, the equality
relation, =, is a relation on any set which is both symmetric and antisymmetric. Suppose that
A is a set and R is a relation on A that is both symmetric and antisymmetric. Show that R is a

4.8. Application: Relational Databases *

145

subset of = (when both relations are considered as sets of ordered pairs). That is, show that
for any a and b in A, ( a R b) → ( a = b).
5. Let ∼ be the relation on R, the set √
of real numbers,
such that for x and y in R, x √
∼ y if and
√
2
−
1
∼
2
+
17
because
the
difference,
(
2 − 1) −
only
if
x
−
y
∈
Z.
For
example,
√
( 2 + 17), is −18, which is an integer. Show that ∼ is an equivalence relation. Show that each
equivalence class [ x ]∼ contains exactly one number a which satisfies 0 ≤ a < 1. (Thus, the
set of equivalence classes under ∼ is in one-to-one correspondence with the half-open interval
[0, 1).)
6. Let A and B be any sets, and suppose f : A → B. Define a relation ∼ on B such that for any x
and y in A, x ∼ y if and only if f ( x ) = f (y). Show that ∼ is an equivalence relation on A.
7. Let Z+ be the set of positive integers {1, 2, 3, . . . }. Define a binary relation D on Z+ such that
for n and m in Z+ , n D m if n divides evenly into m, with no remainder. Equivalently, n D m if
n is a factor of m, that is, if there is a k in Z+ such that m = nk. Show that D is a partial order.
8. Consider the set N × N, which consists of all ordered pairs of natural numbers. Since N × N
is a set, it is possible to have binary relations on N × N. Such a relation would be a subset
of (N × N) × (N × N). Define a binary relation ⪯ on N × N such that for (m, n) and (k, ℓ)
in N × N, (m, n) ⪯ (k, ℓ) if and only if either m < k or ((m = k) ∧ (n ≤ ℓ)). Which of the
following are true?
a) (2, 7) ⪯ (5, 1)
b) (8, 5) ⪯ (8, 0)
c) (0, 1) ⪯ (0, 2)
d) (17, 17) ⪯ (17, 17)
Show that ⪯ is a total order on N × N.
9. Let ∼ be the relation defined on N × N such that (n, m) ∼ (k, ℓ) if and only if n + ℓ = m + k.
Show that ∼ is an equivalence relation.
10. Let P be the set of people and let C be the ‘child of’ relation. That is x C y means that x is a child
of y. What is the meaning of the transitive closure C∗ ? Explain your answer.
11. Let R be the binary relation on N such that x R y if and only if y = x + 1. Identify the transitive
closure R∗ . (It is a well-known relation.) Explain your answer.
12. Suppose that R is a reflexive, symmetric binary relation on a set A. Show that the transitive
closure R∗ is an equivalence relation.

4.8 Application: Relational Databases
One of the major uses of computer systems is to store and manipulate collections of data.
A database is a collection of data that has been organized so that it is possible to add and
delete information, to update the data that it contains, and to retrieve specified parts of
the data. A Database Management System, or DBMS, is a computer program that makes
it possible to create and manipulate databases. A DBMS must be able to accept and
process commands that manipulate the data in the databases that it manages. These
commands are called queries, and the languages in which they are written are called
query languages. A query language is a kind of specialized programming language.
There are many different ways that the data in a database could be represented. Different DBMS’s use various data representations and various query languages. However,

4

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Chapter 4. Sets, Functions, and Relations

data is most commonly stored in relations. A relation in a database is a relation in the
mathematical sense. That is, it is a subset of a cross product of sets. A database that stores
its data in relations is called a relational database. The query language for most relational database management systems is some form of the language known as Structured
Query Language, or SQL. In this section, we’ll take a very brief look at SQL, relational
databases, and how they use relations.


4

You’ll learn more about databases in the courses Web & Database Technology
and Information & Data Management.

A relation is just a subset of a cross product of sets. Since we are discussing computer
representation of data, the sets are data types. As in Section 4.5, we’ll use data type
names such as int and string to refer to these sets. A relation that is a subset of the cross
product int × int × string would consist of ordered 3-tuples such as (17, 42, “hike”). In
a relational database, the data is stored in the form of one or more such relations. The
relations are called tables, and the tuples that they contain are called rows or records.
As an example, consider a lending library that wants to store data about its members,
the books that it owns, and which books the members have out on loan. This data could
be represented in three tables, as illustrated in Figure 4.8. The relations are shown as
tables rather than as sets of ordered tuples, but each table is, in fact, a relation. The rows
of the table are the tuples. The Members table, for example, is a subset of int × string ×
string × string, and one of the tuples is (1782, “Smit, Johan”, “107 Main St”, “New York,
NY”). A table does have one thing that ordinary relations in mathematics do not have.
Each column in the table has a name. These names are used in the query language to
manipulate the data in the tables.
The data in the Members table is the basic information that the library needs in order
to keep track of its members, namely the name and address of each member. A member
also has a MemberID number, which is presumably assigned by the library. Two different
members can’t have the same MemberID, even though they might have the same name or
the same address. The MemberID acts as a primary key for the Members table. A given
value of the primary key uniquely identifies one of the rows of the table. Similarly, the
BookID in the Books table is a primary key for that table. In the Loans table, which
holds information about which books are out on loan to which members, a MemberID
unambiguously identifies the member who has a given book on loan, and the BookID
says unambiguously which book that is. Every table has a primary key, but the key can
consist of more than one column. The DBMS enforces the uniqueness of primary keys.
That is, it won’t let users make a modification to the table if it would result in two rows
having the same primary key.
The fact that a relation is a set—a set of tuples—means that it can’t contain the same
tuple more than once. In terms of tables, this means that a table shouldn’t contain two
identical rows. But since no two rows can contain the same primary key, it’s impossible

4.8. Application: Relational Databases *

147

for two rows to be identical. So tables are in fact relations in the mathematical sense.
The library must have a way to add and delete members and books and to make a
record when a book is borrowed or returned. It should also have a way to change the
address of a member or the due date of a borrowed book. Operations such as these
are performed using the DBMS’s query language. SQL has commands named INSERT,
DELETE, and UPDATE for performing these operations. The command for adding Barack
Obama as a member of the library with MemberID 999 would be
INSERT INTO Members
VALUES (999, "Barack Obama",
"1600 Pennsylvania Ave", "Washington, DC")
When it comes to deleting and modifying rows, things become more interesting because
it’s necessary to specify which row or rows will be affected. This is done by specifying a
condition that the rows must fulfill. For example, this command will delete the member
with ID 4277:
DELETE FROM Members
WHERE MemberID = 4277
It’s possible for a command to affect multiple rows. For example,
DELETE FROM Members
WHERE Name = "Smit, Johan"
would delete every row in which the name is “Smit, Johan”. The update command also
specifies what changes are to be made to the row:
UPDATE Members
SET Address="19 South St", City="Hartford, CT"
WHERE MemberID = 4277
Of course, the library also needs a way of retrieving information from the database.
SQL provides the SELECT command for this purpose. For example, the query
SELECT Name, Address
FROM Members
WHERE City = "New York, NY"
asks for the name and address of every member who lives in New York City. The last
line of the query is a condition that picks out certain rows of the “Members” relation,
namely all the rows in which the City is “New York, NY”. The first line specifies which
data from those rows should be retrieved. The data is actually returned in the form of a
table. For example, given the data in Figure 4.8, the query would return this table:

4

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Chapter 4. Sets, Functions, and Relations
Smit, Johan
Jones, Mary
Lee, Joseph
O’Neil, Sally

107 Main St
1515 Center Ave
90 Park Ave
89 Main St

The table returned by a SELECT query can even be used to construct more complex queries. For example, if the table returned by SELECT has only one column, then it can be used
with the IN operator to specify any value listed in that column. The following query will
find the BookID of every book that is out on loan to a member who lives in New York
City:

4

SELECT BookID
FROM Loans
WHERE MemberID IN (SELECT MemberID
FROM Members
WHERE City = "New York, NY")
More than one table can be listed in the FROM part of a query. The tables that are
listed are joined into one large table, which is then used for the query. The large table
is essentially the cross product of the joined tables, when the tables are understood as
sets of tuples. For example, suppose that we want the titles of all the books that are out
on loan to members who live in New York City. The titles are in the Books table, while
information about loans is in the Loans table. To get the desired data, we can join the
tables and extract the answer from the joined table:
SELECT Title
FROM Books, Loans
WHERE MemberID IN (SELECT MemberID
FROM Members
WHERE City = "New York, NY")
In fact, we can do the same query without using the nested SELECT. We need one more
bit of notation: If two tables have columns that have the same name, the columns can
be named unambiguously by combining the table name with the column name. For example, if the Members table and Loans table are both under discussion, then the MemberID
columns in the two tables can be referred to as Members.MemberID and Loans.MemberID.
So, we can say:
SELECT Title
FROM Books, Loans
WHERE City ="New York, NY"
AND Members.MemberID = Loans.MemberID
This is just a sample of what can be done with SQL and relational databases. The
conditions in WHERE clauses can get very complicated, and there are other operations

4.8. Application: Relational Databases *

149

besides the cross product for combining tables. The database operations that are needed
to complete a given query can be complex and time-consuming. Before carrying out a
query, the DBMS tries to optimize it. That is, it manipulates the query into a form that
can be carried out most efficiently. The rules for manipulating and simplifying queries
form an algebra of relations, and the theoretical study of relational databases is in large
part the study of the algebra of relations.

Exercises
1. Using the library database from Figure 4.8, what is the result of each of the following SQL
commands?
a)

SELECT Name, Address
FROM Members
WHERE Name = "Smit, Johan"

b)

DELETE FROM Books
WHERE Author = "Isaac Asimov"

c)

UPDATE Loans
SET DueDate = "20 November"
WHERE BookID = 221

d)

SELECT Title
FROM Books, Loans
WHERE Books.BookID = Loans.BookID

e)

DELETE FROM Loans
WHERE MemberID IN (SELECT MemberID
FROM Members
WHERE Name = "Lee, Joseph")

2. Using the library database from Figure 4.8, write an SQL command to do each of the following
database manipulations:
a) Find the BookID of every book that is due on 1 November 2010.
b) Change the DueDate of the book with BookID 221 to 15 November 2010.
c) Change the DueDate of the book with title “Summer Lightning” to 14 November 2010.
Use a nested SELECT.
d) Find the name of every member who has a book out on loan. Use joined tables in the
FROM clause of a SELECT command.
3. Suppose that a university colleges wants to use a database to store information about its students, the courses that are offered in a given term, and which students are taking which courses.
Design tables that could be used in a relational database for representing this data. Then write
SQL commands to do each of the following database manipulations. (You should design your
tables so that they can support all these commands.)

4

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Chapter 4. Sets, Functions, and Relations
a) Enroll the student with ID number 1928882900 in “English 260”.
b) Remove “Johan Smit” from “Biology 110”.
c) Remove the student with ID number 2099299001 from every course in which that student is enrolled.
d) Find the names and addresses of the students who are taking “Computer Science 229”.
e) Cancel the course “History 101”.

4

4.8. Application: Relational Databases *

151

Members
MemberID

Name

Address

City

1782
2889
378
4277
5704

Smit, Johan
Jones, Mary
Lee, Joseph
Smit, Johan
O’Neil, Sally

107 Main St
1515 Center Ave
90 Park Ave
2390 River St
89 Main St

New York, NY
New York, NY
New York, NY
Newark, NJ
New York, NY

Books
BookID

Title

Author

182
221
38
437
598
629
720

I, Robot
The Sound and the Fury
Summer Lightning
Pride and Prejudice
Left Hand of Darkness
Foundation Trilogy
The Amber Spyglass

Isaac Asimov
William Faulkner
P.G. Wodehouse
Jane Austen
Ursula LeGuin
Isaac Asimov
Philip Pullman

Loans
MemberID

BookID

DueDate

378
221
8 October 2010
2889
182
1 November 2010
4277
221
1 November 2010
1782
38
30 October 2010
Figure 4.8: Tables that could be part of a relational database. Each table has a name,
shown above the table. Each column in the table also has a name, shown in the top row
of the table. The remaining rows hold the data.

4

Chapter 5

Looking Beyond
oming to the right conclusion has been the theme of this book. We learned how to

C represent statements formally in Chapter 2, and how to prove or disprove statements

in Chapter 3. We looked at more important foundational parts of computer science in
Chapter 4: sets, functions and relations.

Last chapter, I said that the foundations of mathematics are in “a bit of a philosophical
muddle”. That was at the end of our discussion about counting past infinity (Section 4.6).
The questions from the work of Cantor, Russell and others became more profound in the
1930s thanks to Kurt Gödel. All this just before practical computers were invented—yes,
the theoretical study of computing began before computers existed!
Since this book is about the foundations of computation, let’s mention Gõdel and his
contemporary, Alan Turing.
Gödel published his two incompleteness theorems in 1931. The first incompleteness
theorem states that for any self-consistent recursive axiomatic system powerful enough
to describe the arithmetic of the natural numbers1 , there are true propositions about the
naturals that cannot be proved from the axioms. In other words, in any formal system of
logic, there will always be statements that you can never prove nor disprove: you don’t
know.
These two theorems ended a half-century of attempts, beginning with the work of
Frege and culminating in the work of Russell and others, to find a set of axioms sufficient
for all mathematics.

1 For example, Peano arithmetic, which is a recursive definition of N: see en.wikipedia.org/wiki/Peano_
axioms.

153

154



Chapter 5. Looking Beyond *

John von Neumann, one of the pioneers of the
first computers, wrote “Kurt Gödel’s achievement in
modern logic is singular and monumental—indeed
it is more than a monument, it is a landmark which
will remain visible far in space and time.” John
von Neumann was a brilliant Hungarian-American
mathematician, physicist and computer scientist.
Among many other things, he invented the von Neumann architecture (familiar from Computer Organisation?) and is called the ‘midwife’ of the modern
computer.
Source: en.wikipedia.org/wiki/Kurt_Gödel and en.wikipedia.org/
wiki/John_von_Neumann.

5

Around the same time, one of the early models of computation was developed by
the British mathematician, Alan Turing. Turing was interested in studying the theoretical abilities and limitations of computation. (This before the first computers! Von Neumann knew Turing and emphasized that “the fundamental conception [of the modern
computer] is owing to Turing” and not to himself.2 ) His model for computation is a very
simple abstract computing machine which has come to be known as a Turing machine.
While Turing machines are not very practical, their use as a fundamental model for computation means that every computer scientist should be familiar with them, at least in a
general way.3 You’ll learn about them in Automata, Languages, and Computability.
We can also use Turing machines to define ‘computable languages’. The idea is that
anything that it is possible to compute, you can compute using a Turing machine (just
very slowly). Turing, with American mathematician Alonzo Church, made a hypothesis
about the nature of computable functions.4 It states that a function on the natural numbers is computable by a human being following an algorithm (ignoring resource limitations), if and only if it is computable by a Turing machine.
So Gödel established that there are some things we can never tell whether they are
true or false, and Turing established a computational model for that the things we can
compute.
2 A quote from von Neumann’s colleague Frankel.
See en.wikipedia.org/wiki/Von_Neumann_
architecture.
3 In fact, the Brainf*ck programming language we mentioned earlier is not much more than a Turing machine.
4 It’s not a theorem because it is not proved, but all theoretical computer scientists believe it to be true.
Between them, Church and Turing did prove that a function is λ-computable if and only if it is Turing computable if and only if it is general recursive. Another thesis that is widely believed but not proved is: for the things
we can compute, there is a difference between those that need only polynomial time (in the size of the input)
for the computation and those that need more than polynomial time. More about that in Complexity Theory.

155
Is there a link between these results? The halting problem is to determine, given a
program and an input to the program, whether the program will eventually halt when
run with that input. Turing proved in 1936 that a general algorithm to solve the halting
problem for all possible program-input pairs cannot exist.
We end the book with the idea of the proof:
Proof. Consider a model of computation, such a Turing machine. For any program f
that might determine if programs halt, construct a ‘pathological’ program g as follows.
When called with an input, g passes its own source and its input to f , and when f returns
an output (halt/not), g then specifically does the opposite of what f predicts g will do.
No f can exist that handles this case.

5

Further Reading
The books by Christian & Griffiths, Hofstadter and Smullyan are on recreational maths.
Hofstadter’s is a winner of the Pulitzer Prize. The other titles are textbooks; Hammack’s
book is an Open Textbook, just like this book is.
[1] Christian, B. and Griffiths, T. (2016). Algorithms to Live By: The Computer Science of
Human Decisions. New York, NY: Henry Holt and Co.
[2] Critchlow, C. and Eck, D. (2011). Foundations of Computation, version
2.3.1. Geneva, NY: Hobart and William Smith Colleges. math.hws.edu/
FoundationsOfComputation/
[3] Epp, S. S. (2011). Discrete Mathematics with Applications, 4th edition. Boston, MA:
Cengage Learning.
[4] Grassmann, W. K. and Tremblay, J.-P. (1996). Logic and Discrete Mathematics. Upper
Saddle River, NJ: Prentice-Hall.
[5] Hammack, R. (2018). Book of Proof, 3rd edition. Richmond, VA: Virginia Commonwealth University. www.people.vcu.edu/~rhammack/BookofProof3/
[6] Hofstadter, D. (1979). Gödel, Escher, Bach: An Eternal Golden Braid. New York, NY:
Basic Books.
[7] Huth, M. and Ryan, M. (2004). Logic in Computer Science, 2nd edition. Cambridge,
UK: Cambridge University Press.
[8] Smullyan, R. M. (1979). What is the Name of This Book? Upper Saddle River, NJ:
Prentice-Hall.

157

Index
n-ary relation, 139

brace symbol, 42, 89

addition, binary, 33
algebra, 15, 106, 149
Algorithm Design, 78
Algorithms & Data Structures, 1, 74, 78, 82
and (logical operator), 6, 102
AND gate, 25
antecedent, 10
antisymmetric relation, 140
argument, 46
associative operator, 7
Automata, Languages, and Computability,
61, 135, 144, 154

Calculus, 1, 121
calling a subroutine, 77
Cantor, George, 135
cardinality, 130, 133, 143
Cartesian product, 119
Church, Alonzo, 154
CNF, see conjunctive normal form
combinatorial logic circuit, 28
combinatorics, 132
complement, 104
Complexity Theory, 154
composition, 118
compound proposition, 7
computability, 154
computable languages, 154
Computer Organisation, 1, 4, 6, 22, 24, 32,
78, 81, 111, 154
Concepts of Programming Languages, 124
conclusion, 3, 10, 46
conditional, 10
conditional operator, 9
conjunction, 6
conjunctive normal form, 34
consequent, 10
contingency, 13
contradiction, 13
contrapositive, 11
converse, 11
coordinates, 119

base case, 71, 75, 78
base-2 number, 113
biconditional operator, 9, see also iff, 11
bijective, 122
bijective function, 129
binary number, 32, 111
binary relation, 139
binary sort tree, 84
binary tree, 81
bit, 111
bitwise logical operator, 113
Boole, George, 15
Boolean algebra, 6, 15, 19
and logic circuits, 31
in predicate logic, 39
in set theory, 106
bound variable, 38

159

corollary, 136
countable set, 133
countably infinite, 133
counterexample, 50, 61
cross product, 119
Cruyff, Johan, 6

existential quantifier, 38
Explosion, Principle of, see Principle of Explosion
extensional definition, 92
false (logical value), 15
feedback in circuits, 27
Feyenoord, 10, 11, 47
Fibonacci sequence, 86
finite set, 130
first-class citizen, see first-class objects
first-class object, 123, 126
font, 115
formal proof, see proof
formal structure, 41
free variable, 38
Frege, Friedrich, 56
Frege, Gottlieb, 56
function, 117, 120
in computer programming, 124
partial, 126
functional programming, 127
functionally complete, 12
funtionally complete, 12

database, 139, 145
Database Management System, 145
DBMS, see Database Management System
deduction, 3, 46, 47
Delft, 4
DeMorgan’s Laws, 22, 43, 104
diagonalization, 135
Digital Systems, 35
direct proofs, 66
disjoint sets, 94
disjunction, 6
disjunctive normal form, 29
distributive law, 15, 22, 23, 103
divisible by m, 62
DNF, see disjunctive normal form
domain, 121
domain of discourse, 36
dominos
Gödel, Kurt, 57, 153
analogy for mathematical induction, 70
Galileo, 14
double negation, 17
Gauss, Carl, 74
duality, 16
graph, 119
e, 62
element (of a set), 89
empty set, 91
entity, 35
variable, 38
equals sign, 15
equivalence class, 142
equivalence relation, 142
Euclid, 56, 68
even, 63
excluded middle, law of, 21
exclusive or operator, 9
existence proof, 60

halting problem, 155
hexadecimal number, 112
hypothesis, 10, 60
identity law, 21
if and only if, 11
iff, see also if and only if, 62
image, 121
implication, 10
incompleteness theorems, 153
indirect proof, 66
induction, 70
second form, 75
160

induction hypothesis, 71
induction, structural, see structural induction
inductive case, 71
inductive hypothesis, 71
infinite set, 130
Information & Data Management, 1, 146
injective, 122
integers, 62
intensional definition, 92
intersection, 93, 102
invalid argument, 50
invariant, 87, 100
inverse, 11
irrational numbers, 62
iteration, 87
Java, 5, 9, 78, 82, 124
JavaScript, 127
Karnaugh-map, 32
Khan Academy, 70
laws of Boolean Algebra, 16
leaf node, 84
left subtree, 82
lemma, 68
Linear Algebra, 1
logic circuit, 25
and compound propositions, 26
for an input/output table, 30
simplifying, 31
logic gate, 24–25
logical connective, 5
logical deduction, see deduction
logical equivalence, 8, 13, 15, 18, 19
and logical deduction, 48
in predicate logic, 42, 44
logical operator, 5
bitwise, 113
logically implies, 47
loop invariant, see invariant
Lucas, Édouard, 80

Luyben, Karel, 10
Machine Learning, 1
main connective, 7
mapping, 117
mathematical generality, 5
mathematical induction, 70
mathematical structure, see formal structure
member (of a set), 91
modus ponens, 47, 52
modus tollens, 47, 52
NAND gate, 25
natural numbers, 62
necessary, 10
necessity, see necessary
negation, 6
of quantified statements, 42
non-Euclidean geometry, 56
NOR gate, 25
not (logical operator), 6
NOT gate, 25
null pointer, 82
Object-Oriented Programming, 9, 124
obviousness, 17, 57
octal number, 116
odd, 63
one-place predicate, 36
one-to-one, 122
one-to-one correspondence, 129
onto function, 121
open statement, 38
or (logical operator), 6, 102
inclusive vs. exclusive, 12
OR gate, 25
ordered n-tuple, 119
ordered pair, 119
ordered triple, 119
parameter, 125
partial function, 126
partial order, 141

161

partition, 141
Peirce, Charles, 37, 56
pigeonhole principle, 69
power set, 96
precedence rule, 7, 9, 108
predicate, 4, 35, 36, 92
predicate calculus, 35
predicate logic, 35
premise, 3, 46
primary key, 146
prime, 63
Principle of Explosion, 13, 24
principle of mathematical induction
see mathematical induction, 70
Probability Theory & Statistics, 1
product (of sets), 119
Prolog, 77, 87
Prometheus, 10
proof, 5, 17, 49, 55
by contradiction, 67
by contrapositive, 66
by division into cases, 66
by generalisation, 60
examples, 63
proper subset, 92
proposition, 4, 5
equivalent to one in DNF, 30
for a logic circuit, 26
propositional logic, 4, 5, 35
propositional variable, 17
prototype, 125
Python, 127
quantifier, 4
in English, 37
on a two-place predicate, 38
quantifiers, 37
query language, 145
query, database, 145
range, 121
rational numbers, 62

real numbers, 62
Reasoning & Logic, 1, 2, 4, 22, 24, 29, 32, 45,
56, 59, 62, 82, 111, 124, 135
recursion, 77
recursive definition, 85
recursive subroutine, 77
reflexive relation, 140
relation, 139, 146
relational database, 146
right subtree, 82
root, 82
Rubik’s Cube, 143
Russell’s Paradox, 56, 97
Russell, Bertrand, 56, 97
set, 89
of functions from A to B, 123
set difference, 94
set theory, 89
set-builder notation, 92
sets, 89
simple term, 29
Sinterklaas, 11, 14
situation, 7
Socrates, 3
sort tree, 84
SQL, see Structured Query Language
strong induction, see induction, second form
structural induction, 100
Structured Query Language, 146
subroutine, 77
subset, 91
substitution law, 18
sufficiency, see sufficient
sufficient, 10
summation notation, 73
surjective, 121
syllogism, 48
symmetric difference, 109
symmetric relation, 140
Tarski’s world, 40
162

Tarski, Alfred, 22, 41
tautology, 12, 18, 20, 44
ternary relation, 139
total function, 126
total order, 141
Towers of Hanoi, 79
transitive closure, 144
transitive relation, 140
tree, 81, 100
true (logical value), 15
truth table, 7, 16, 19
and logic circuits, 28
of a tautology, 13
TU Delft, 1, 10, 45
tuple, 42, 119
tuples, 42
Turing machine, 154
Turing, Alan, 154
uncountable set, 133
union, 93, 102
universal quantifier, 38
universal set, 104
valid argument, 47, 49
value (of a function), 118
variable, 38, 44
propositional, 5
Venn diagram, 90, 95
von Neumann, John, 154
Web & Database Technology, 146
wires in computers, 25
Zadeh, Lotfi, 22

163

Delftse Foundations of Computation
Stefan Hugtenburg and Neil Yorke-Smith
Delftse Foundations of Computation is a textbook for a one quarter
introductory course in theoretical computer science. It includes topics from
propositional and predicate logic, proof techniques, set theory and the theory
of computation, along with practical applications to computer science. It has
no prerequisites other than a general familiarity with computer programming.

Stefan Hugtenburg
TU Delft | Faculty of Electrical Engineering, Mathematics
and Computer Science: Software Technology, Distributed
Systems
Stefan Hugtenburg holds a MSc in Computer Science
from the Delft University of Technology, where he now
teaches in the undergraduate Computer Science and
Engineering programme. He is involved in all courses of
the Algorithmics track in the curriculum, starting with this
book and the course Reasoning & Logic, up until the final
year course Complexity Theory.

Neil Yorke-Smith
TU Delft | Faculty of Electrical Engineering, Mathematics
and Computer Science: Software Technology,
Algorithmics
Neil Yorke-Smith is an Associate Professor of
Algorithmics in the Faculty of Electrical Engineering,
Mathematics and Computer Science at the Delft
University of Technology. His research focuses on
intelligent decision making in complex socio-technical
situations, with a particular current interest in agentbased methodologies and behavioural factors in
automated planning and scheduling. He teaches
Reasoning & Logic and graduate courses in Artificial
Intelligence.

© 2018 TU Delft Open
ISBN 978-94-6366-083-9
DOI https://doi.org/10.5074/t.
isbn.9789463660839
textbooks.open.tudelft.nl
Cover image designed by
E. Walraven is licensed under CCBY TU Delft