Chapter 1

LOGIC AND PROOF

To be able to understand mathematics and mathematical arguments, it is necessary

to have a solid understanding of logic and the way in which known facts can be

combined to prove new facts. In this chapter we take a careful look at the rules of

logic and the way in which mathematical arguments are constructed.

1.1 Logical statements

The study of logic is concerned with the truth of falsity of statements.

Definition 1 A statement∗ is a sentence which can be classified as true or false

without ambiguity. The truth or falsity of the statement is known as the truth value.

For a sentence to be a statement, it is not necessary that we actually know

whether it is true or false, but it must be clear that it is one or the other.

Example 2 Consider the following sentences:

1. “Every continuous function is differentiable” is a statement with truth value

“false”.

2. ”x < 2” is true for some x and false for some others. If we have a particular

context in mind, then it could be a statement. Otherwise it is not.

3. ”Every even number greater than 2 is the sum of two primes” is a statement,

whose truth value is not known...yet.

Exercise 3 Which of these sentences are statements?

1. If x is a real number, then x2 ≥ 0.

2. Seven is a prime number

3. This sentence is false

∗Some authors also call it proposition.

8 Logic and Proof

1.2 Logical Connectives

In studying mathematical logic we shall not be concerned with the truth value of

any particular simple statement. What will be important is how the truth value of

a compound statement is determined from the truth values of its simpler parts.

To obtain such compound statements it is necessary to introduce the concept of a

connective

Definition 4 A sentential connective is a logic symbol representing an operator

that combines statements into a new statement.

Statements with connectives are called compound statements. Statements

without connectives are known as atomic statements. The sentential connectives

are ”not”, ”and”, ”or”, ”if ...then”, and ”if and only if”. The respective operators

for these connectives are negation, conjunction, disjunction, implication and

equivalence respectively. We shall introduce each of them now

• Negation : Let p stand for a given statement. Then ˜p ( read not p ) representsthe logical opposite of p. When p is true, then ˜p is false and viceversa. This canbe summarized in a “truth table”, which gives the mapping from the truthvalue of the individual statements to the truth value of the resulting compoundstatement. Some authors also use � for the negation connective. In this case

p ˜p

T F

F T

.

where T stands for true and F for false. Note that the truth table of aconnective is an alternative way of defining a connective, since these are defined interms of the truth value of the resulting compound statement, given the truth valueof its components.

• Conjunction: If p and q are statements, then the statement p∧ q ( read p and

q) is true only when both p and q are true, and is false otherwise.

p q p ∧ q

T T T

T F F

F T F

F F F

.

Logical Connectives 9

For example, given p : ”2 > 0” and q : ”0 > 2” then the compound statementp ∧ q : ”2 > 0 and 0 > 2” is false.

• (Inclusive)Disjunction : If p and q are statements, then the statement p ∨ q

( read p or q) is true when at least one of the two statements is true, and isfalse when both are false.

p q p ∨ q

T T T

T F T

F T T

F F F

.

Note that the inclusive disjunction doesn’t complete the list of disjunctionsused in everyday life. In fact, we also have the exclusive disjunction, which is truewhen either p or q is true, but not when both are true. In logic the only use forthe connective or is for the inclusive meaning. For example, given p : ”2 > 0” andq : ”0 > 2” then the compound statement p ∨ q : ”2 > 0 or 0 > 2” is true.

• Implication : A statement of the form

If p, then q

is called an implication or a conditional statement. The if-statementp is called the antecedent and the then-statement q is called the consequent. Theconvention adopted for the truth value of the implication is that it will be calledfalse only when the antecedent is true and the consequent is false. If we denote theimplication ”if p, then q” by p⇒ q, we obtain the following truth table

p q p⇒ q

T T T

T F F

F T T

F F T

.

The last row of the table may seem counterintuitive. However, the usage of ’if... then” as a connective or as a mathematical language is quite different from thatof daily language. The reason for giving truth value T to the last case may be notunderstandable in the following mathematical expression: “3 + 1 = 7 ⇒ 6 − 1 = 2”

10 Logic and Proof

Nevertheless, we can easily prove the implication is indeed true: 3+1−3 = 7−3 = 4

so 1 = 4. But then 6 − 1 = 6 − 4 = 2.

In English there are several ways to express the same mathematical condition

p⇒ q. These are

”if p, then q” ”q provided that p”

”p implies q” ”q whenever p”

”p only if q” ”q is a necessary condition for p”

”q if p” ”p is a sufficient condition for q”

• Equivalence: The statement ”p if and only if q” is defined as the conjunctionof the two implications p⇒ q and q⇒ p. A statement of this form is called anequivalence and is denoted by p⇔ q. In written form, the abbreviation ”iff” isused instead of ”if and only if”. The truth table for equivalence can be obtainedby computing the truth table for the compund statement (p⇒ q) ∧ (q⇒ p)

p q p⇒ q q⇒ p p⇐⇒ q

T T T T T

T F F T F

F T T F F

F F T T T

Thus we see that p ⇔ q is true precisely when p and q have the same truthvalues. Since p ⇔ q is equivalent to (p⇒ q) ∧ (q⇒ p) we can use the terminologyseen above for the implication connective and say

”q if p” and ”q only if p””p is a sufficient condition for q” and ”p is a necessary condition for q”

Two compound statements p and q are said to be logically equivalent if oneis true if and only if the other is true. In other words, two compound propositionsare logically equivalent whenever they display the same truth table. In this case wewrite p ≡ q.

Example 5 (˜(˜p)) is logically equivalent to p

p ˜p ˜ (˜p)

T F T

F T F

Logical Connectives 11

Example 6 ˜ (p ∧ q) is logically equivalent to [(˜p) ∨ (˜q)]

p q ˜ (p ∧ q) ˜p ˜q [(˜p) ∨ (˜q)]

T T F F F F

T F T F T T

F T T T F T

F F T T T T

Exercise 7 (Implication) Show that the implication symbol is not a ”primitive”

connector. In other words, show that the implication p ⇒ q is logically equivalent

to [(˜p) ∨ q] . This is actually another convenient way to interpret the implication

connector.

When two compound statements p and q are logically equivalent, their equiv-

alence is a new statement with a truth value ”true” in all cases. Such a statement is

known as a tautology, being its definition

Definition 8 A compound proposition is said to be a tautology if it is always true

regardless of the truth value of the simple propositions from which it is constructed.

It is a contradiction if it is always false. Hence a contradiction and a tautology are

a negation of each other.

Example 9 p ∨ (˜p) is a tautology, while p ∧ (˜p) is a contradiction.

p (˜p) p ∨ (˜p) p ∧ (˜p)

T F T F

F T T F

Example 10 [p ∧ (p =⇒ q)] =⇒ q is a tautology

p q p⇒ q p ∧ (p⇒ q) [p ∧ (p⇒ q)]⇒ q

T T T T T

T F F F T

F T T F T

F F T F T

Lets now prove the equivalence of a compound statement that will be usefulwhen studying methods of proof

Example 11 (Contrapositive) p⇒ q is logically equivalent to [(˜q)⇒ (˜p)]

p q p⇒ q ˜p ˜q (˜q)⇒ (˜p) (p⇒ q)⇔ [(˜q)⇒ (˜p)]

T T T F F T T

T F F F T F T

F T T T F T T

F F T T T T T

12 Logic and Proof

Exercise 12 (De Morgan’s Law) Prove that ˜(p∨q) is logically equivalent to [(˜p) ∧ (˜q)]

Exercise 13 Establish the logical equivalence of these compound statements

1. ˜ (p⇒ q)⇔ [p ∧ ˜q]2. (p⇒ q) ⇔ (q ⇒ p)

Remark 14 As you have already proven p ⇒ q is not logically equivalent to its

converse q ⇒ p. Claiming the opposite is a very common mistake you should not

make in the future.

Exercise 15 Write the negation of each statement

a) If K is closed and bounded, then K is compact.b) If K is compact, then K is closed and bounded.c) A continuous function is differentiable.

Exercise 16 Construct a truth table for each statement

a) [ p⇒ (q ∧ ˜q)]⇔ ˜pb) (q ⇒ ˜p) ⇔ (p ∧ q)c) (p ∧ q)⇔ (q ∧ p) (commutative property)d) (p ∨ q)⇔ (q ∨ p)e) [p ∧ (q ∧ r)]⇔ [(p ∧ q) ∧ r] (associative property)f) [p ∨ (q ∨ r)]⇔ [(p ∨ q) ∨ r]g) [p ∧ (q ∨ r)]⇔ [(p ∧ q) ∨ (p ∧ r)] (distributive property)h)[p ∨ (q ∧ r)]⇔ [(p ∨ q) ∧ (p ∨ r)]

1.3 Quantifiers

In Section 1 we saw that a mathematical sentence that involves a variable, for ex-

ample ”x > 3”, needs to be considered in a particular context in order to become a

statement. The quantifiers help create statements by limiting the role of variables in

a mathematical sentence.

Definition 17 A variable is a symbol that can assume several specifications. A

propositional function is a sentence which becomes a statement once we replace a

variable by one of its specifications.

Let p(x) denote a propositional function, p being the sentence and x being thevariable. Thus p (x) : x > 3 is a propositional function.

Let S denote a given set.† There are two quantifiers:

†You may think of a set S as a collection of objects, who are called the elements of S. When x

is an element of S we write x ∈ S and we read “x is an element of S”.

Quantifiers 13

• Universal quantifier (“for all”), denoted by ∀, with the following interpreta-tion:

[∀x ∈ S, p(x)] means that p(x) is true provided it is true for each x in S. Asan example, if S = {4,5}, then [∀x ∈ S, p(x) : x > 3] ≡ [(4 > 3) ∧ (5 > 3)].Sometimes the set S is implicit in the context, and can be omitted. The notationtherefore becomes more compact, leading to ∀x, p(x).

• Existential quantifier (“there exists”), denoted by ∃:

[∃x ∈ S p(x)] reads ” There exists an x belonging to S such that p (x) ”‡

meaning that p(x) is true provided there exists at least an x in S for whichp(x) is true. As an example, if S = {1, 2, 4}, then [∃x ∈ S � p(x) : x > 3] ≡[(1 > 3) ∨ (2 > 3) ∨ (4 > 3)].

Exercise 18 Let S be a finite set (i.e. {1, 2}). Use de De Morgan’s law to show that

˜[∃x ∈ S � p(x)] ⇔ [∀x ∈ S, ˜p(x)], and ˜ [∀x ∈ S , p(x)] ⇔ [∃x ∈ S, ˜p(x)] aretautologies.

Sometimes we use the symbol ∃! x to denote the case when a unique valueexists for the variable x that makes p(x) true. The universal and the existentialquantifiers are thus seen as extensions of the logical connectives ∧ and ∨, to deal withinfinitely many assertions, or assertions about infinitely many things, x. Moreover,one can combine the existential quantifier with negation: ˜∃ means “there exists no”(often written as /∃).

Example 19 /∃ x > 0 � (x+ 1) = 0 is logically equivalent to ∀x > 0, x+ 1 �= 0.

Exercise 20 Write the negation of the following statements

1. ∀x ∈ A, f(x) > 52. ∃ y > 0 � 0 ≤ g(y) ≤ 33. ∀ε > 0 ∃N � ∀n, if n > N, then ∀x ∈ S, | f

n(x) − f (x) | < ε (uniform

convergence)

As we saw in the last exercise, we can use both ∀ and ∃ in one statement. Itis important to clarify the following point about the order in which quantifiers areused. While [∀x,∀y, p(x, y)] ≡ [∀y,∀x, p(x, y)] ≡ [∀x, y, p(x, y)], the propositions[∃ y � ∀x p(x, y)] and [∀x,∃ y � p(x, y)] are not logically equivalent. In this case, infact, the order in which quantifiers appear affects the meaning — and the truth value— of the statement. The first statement says that for at least one y, p(x, y) is true forall x. In other words, the choice of y is independent of x. On the other hand, thesecond statement establishes that for all x there exist at least one y such that p(x, y)is true. This means that the choice of y is allowed to depend on x.

‡� stands for “such that”. Another ways of saying “such that” are given by the symbol “:” or

by writing ”s.t.”.

14 Logic and Proof

As an example consider p(x, y) being given by x + 1 = y. The statement[∃ y � ∀x p(x, y)] establishes that there exists an y such that for all x, x+1 = y. Thisis clearly a false statement. The second proposition, on the other hand, states thatfor all x there exists an y such that x + 1 = y. Unlike the former, this statement istrue.

Exercise 21 Find a p (x, y) such that [∃ y � ∀x p(x, y)] and [∀x,∃ y � p(x, y)] havethe same truth value.

Exercise 22 The following statements give properties of functions that we shall en-counter later in the course. You have to do two things. a) rewrite the definingconditions in logical symbolism and b) write the negation of part a) using the samesymbolism.

1. A function f is odd iff for every x, −f(x) = f(−x).

2. A function f is even iff for every x, f(x) = f(−x).

3. A function f is periodic iff there exists a k > 0 such that for every x, f (x+ k) =f(x).

4. A function f is increasing iff for every x and for every y, if x ≤ y, thenf (x) ≤ f (y) .

5. A function f is strictly decreasing iff for every x and every y, if x ≤ y, thenf (x) > f (y) .

6. A function f : A −→ B is injective iff for every x and y in A, if f (x) = f (y) ,then x = y.

7. A function f : A −→ B is surjective iff for every y in B there exists an x in

A such that f (x) = y.

8. A function f : D −→ R is continuous at c ∈ D iff for every ε > 0 there exists

a δ > 0 such that |f (x)− f (c)| < ε whenever |x− c| < δ and x ∈ D.

9. A function f is uniformly continuous on a set S iff for every ε > 0 there is

a δ > 0 such that |f (x)− f (y)| < ε whenever x and y are in S and |x− y| < δ.

10. The real number L is the limit of the function f : D −→ R at the point c iff

for each ε > 0 there is a δ > 0 such that |f (x)− L| < ε whenever x ∈ D and

0 < |x− c| < δ.

Techniques of Proof 15

1.4 Techniques of Proof

A proof is a method of establishing the truthfulness of an implication. Typically, one

has to prove proposition of the sort ifH1,H2,...,Hn, then C. PropositionsH1,...,Hn are

often referred to as the hypotheses of the proof, whereas proposition C is referred

to as the conclusion, or thesis. A formal proof of such a proposition consists of a

sequence of valid propositions ending with conclusion C. To be valid, a proposition in

the sequence must be either one of the hypotheses H1, ...Hn, or an axiom, a definition,a tautology or a previously proved proposition, or it must be derived from previouspropositions using either substitution or logical implications.

As an example, if you have to prove that given a set S, [∀x ∈ S, p(x)], thenyou need to prove that [x ∈ S =⇒ p(x)]. Similarly, in case you have to prove that[∃x ∈ S : p(x)], you “just” need to find an x in S such that p(x) is true.

There are several ways of proving a proposition, all of them being useful indifferent situations. These methods are enumerated below.

Direct To show that p =⇒ q is true, we first assume that p is true and conclude thatq is true.

Example 23 If x > 1, then x2 > x. Here p : x > 1 and q : x2 > x.The directproof goes as follows:

• Let p : x > 1 (Hypothesis);

• from the axiom ”If a > b and c > 0, then a c > b c” we have x > 1 ⇒

xx > 1 x;

• But by definition xx = x2 and 1 x = x;

• substituting in the above expression leads to xx > 1 x⇒ x2 > x

• which is the conclusion we wanted x > 1⇒ x2 > x.

Contrapositive Associated with the implication p⇒ q there is a logically equivalent statement˜q ⇒ ˜p, called the contrapositive (see example 11). Thus one way to provean implication is to give a direct proof of its contrapositive. In other words,assume that ˜q is true and conclude that ˜p is true. Then we can conclude that(p⇒ q) is true.

Example 24 Consider the statement for m ∈ N , ”If 7m is an odd number,

then m is an odd number”. The contrapositive of the statement is ”If m is not

an odd number, then 7m is not an odd number”, or equivalently ”If m is an

even number, then 7m is an even number”. This statement is much easier to

prove.

16 Logic and Proof

Proof. Let ∼ q : ”m is an even number”. Then, by definition m = 2k for some

k ∈ N. But then substituting into m we have 7m = 7 (2k) = 2 (7k) = 2 k′ which

is also even.

Exercise 25 Try the direct proof of the above statement.

Decomposition Suppose we want to prove that p ⇒ q, and that p can be decomposed into

two disjoint propositions p1, p2 such that p1 ∧ p2 is a contradiction. Then p ≡

(p1 ∨ p2) ∧ ˜(p1 ∧ p2) ≡ (p1 ∨ p2) .

Given this choice of p1 and p2 we have

(p⇒ q) ⇔ (˜p ∨ q)⇔ [˜ (p1 ∨ p2) ∨ q]

⇔ [(˜p1 ∧ ˜p2) ∨ q]⇔ [(˜p1 ∨ q) ∧ (˜p2 ∨ q)]

⇔ [(p1 ⇒ q) ∧ (p2 ⇒ q)]

meaning that you only need to show that p1 ⇒ q and p2 ⇒ q. Note that thismethod works also if you decompose p into a number of propositions biggerthan 2 as far as these propositions are mutually exclusive (which means thatevery pair of them is a contradiction).

Example 26 Take the contrapositive of the example given for the direct proof

method. The contrapositive is ”x2 ≤ x⇒ x ≤ 1”.

Proof. By definition x2 ≥ 0. Then p : 0 ≤ x2 ≤ x and q : x ≤ 1. Decompose

p into p1 : x > 0 and p2 : x = 0. Then p1 ∧ p2 is always false and p ≡

(p1 ∨ p2) ∧ ˜(p1 ∧ p2). To show that p2 ⇒ q is trivial since 0 ≤ 1. To prove

p1 ⇒ q, we only need to use the implication ”∀ c > 0, x ≤ y ⇒ x

c≤

y

c”. Given

that x > 0 by assumption, x2 ≤ x⇒ x2

x≤ x

x⇒ x ≤ 1

Exercise 27 Show the following statement ”Let x ∈ R. If |x| ≥ 1 ⇒ x2 ≥ x”.

Construction This approach is used when the statement includes an existential quantifier. i.e.,the conclusion is of the form: ∃x, p(x). To prove this, simply find (“construct”)a value of x such that p(x) is true when H is true. For example, if A ={1, 2, 3, 4,5} and the proposition is: ∃x ∈ A � x > 2, observe that for x = 3 wehave x > 2 and x ∈ A.

Exercise 28 Show that f (x) = x is a continuous function at every x0 ∈ R . You

have to prove that ”∀ε > 0 ,∃ δ (x0, ε) > 0 � |x− x0| < δ (x0, ε)⇒ |f (x)− f (x0)| <ε”.

Techniques of Proof 17

Contradiction Suppose we want to prove p ⇒ q. This method of proof — also known as ”re-ductio ad absurdum”— consists of assuming that [p ∧ (˜q)] is true and derivinga contradiction c. To show that both statements are logically equivalent, wecompare their truth tables

p q p⇒ q ˜q (p ∧ ˜q) c (p ∧ ˜q)⇒ c

T T T F F F T

T F F T T F F

F T T F F F T

F F T T F F T

.

An equivalent way of making a proof by contradiction is given by the follow-ing tautology: (p =⇒ q)⇔ [(p ∧ (˜q)) =⇒ (˜p)]. This is a particular case of thegeneral definition, since [(p ∧ (˜q)) =⇒ (˜p)] is equivalent to [(p ∧ (˜q)) =⇒ ((˜p) ∧ p)](since p was already an assumption), and given that ((˜p) ∧ p) ⇔ c then[(p ∧ (˜q)) =⇒ c] .The proof of their equivalence is left as an exercise.

Exercise 29 Show that (p =⇒ q)⇔ [(p ∧ (˜q)) =⇒ (˜p)] .

Example 30 We will show again that ”If 7m is odd, then m is odd”.

Proof. Here the statements are p :” 7m is odd ” and q : ”m is odd ”.To prove the result by contradiction we will use the equivalence (p =⇒ q) ⇔[(p ∧ (˜q)) =⇒ (˜p)] , so we have to assume first that (p ∧ (˜q)) , that is ” 7mis odd and m is even ”. Since m is even by definition we have m = 2k for somek. Then 7m= 7(2k) = 2 (7k) = 2 j for j=7k. But this means that 7m is even(˜p) , contradicting the assumption.

Remark 31 Be very careful when writing proofs by contradiction ! A very strongword of caution can be found in Royden§ ”All students are enjoined in the strongestpossible terms to eschew proofs by contradiction! There are two reasons for the pro-hibition: First such proofs are very often fallacious, the contradiction on the finalpage arising from an erroneous deduction on an earlier page, rather than from theincompatibility of p with ˜q. Second, even when correct, such a proof gives little in-sight into the connection between p and q whereas both the direct proof and the proofby contraposition construct a chain of argument connecting p and q.One reason whymistakes are so much more likely in proofs by contradiction than in direct proofs isthat in a direct proof ( assuming the hypothesis is not always false) all deductionfrom the hypothesis are true in those cases where the hypothesis holds. One is dealingwith true statements, and ones’s intuition and knowledge about what it is true help tokeep one from making erroneous statements. In proofs by contradiction, however, youare (assuming the theorem is true) in the unreal world where any statement can bederived, and so the falsity of a statement is no indication of an erroneous deduction.

§H.L. Royden ”Real Analysis” pg3.

18 Logic and Proof

Exercise 32 Let x be a real number. Prove that if x>0 then 1/x>0.

Induction It is used for statements of the form p(n), ∀n ∈ {m, m + 1, m + 2, ...}. Theproof consists of two steps:

Basis of Induction Prove that for the first element in the set of interest — in this case m —,p(m) is true,

Inductive step Show p(k) ⇒ p(k+ 1), for k � m. In other words, assume p(k) is true andprove that p(k + 1) is also true.

Exercise 33 The following tautologies are widely used in the methods of proof. Someof them have already been seen before. a) Verify that they are indeed tautologies, b)Interpret them and associate them with the different methods of proof

1. (p⇔ q)⇔ [(p⇒ q) ∧ (q ⇒ p)]2. (p⇔ q)⇔ [(p⇒ q) ∧ (˜p⇒ ˜q)]3. (p⇒ q)⇔ (˜q ⇒ ˜p)4. p ∨ ˜p5. (p ∧ ˜p)⇔ c

6. (˜p⇒ c)⇔ p

7. (p⇒ q)⇔ [(p ∧ ˜ q)⇒ c]8. [ p ∧ (p⇒ q)]⇒ q

9. [˜q ∧ (p⇒ q)]⇒ ˜p10. [˜p ∧ (p ∨ q)]⇒ q

11. (p ∧ q)⇒ p

12. [(p⇒ q) ∧ (q ⇒ r)]⇒ (p⇒ r)13. [(p1 ⇒ p2) ∧ (p2 ⇒ p3) ∧ ... ∧ (pn−1 ⇒ pn)]⇒ (p1 ⇒ pn)14. [(p ∧ q)⇒ r]⇔ [p⇒ (q ⇒ r)]15. [(p⇒ q) ∧ (r ⇒ s) ∧ (p ∨ r)]⇒ (q ∨ s)16. [p⇒ (q ∨ r)]⇔ [(p ∧ ˜q)⇒ r]17. [(p ∨ q)⇒ r]⇔ [(p⇒ r) ∧ (q⇒ r)]

1.5 References

S.R Lay , Analysis with an Introduction to Proof . Chapter 1. Third Edition. Pren-tice Hall.

A. Matozzi, Lecture Notes Econ 897 University of Pennsylvania Summer

2001.

H.L. Royden, Real Analysis . Third Edition. Prentice Hall.