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ProofsAdvanced Discrete Mathematics

Jim Skon

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Definition: A theorem is a valid logical assertion whichcan be proved using other theorems axioms (statements which are given to be true)

and rules of inference (logical rules which allow the

deduction of conclusions from premises).

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A lemma (not a “lemon”) is a 'pre-theorem' or a result which is needed to prove a theorem.

A corollary is a 'post-theorem' or a result which followsdirectly from a theorem.

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Valid reasoning in proofs

A mathematical proof is a sequence of statements, such that each statement:1. is an assumption, or

2. is a proposition already proved, or

3. Follow logically from one or more previous statements in proof.

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Valid Inference

Consider: H1 H2 ..... Hn C

whereHi are called the hypotheses

andC is the conclusion.

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Valid Inference: Argument

Argument - consists of a collection of statements, called premises of the argument, followed by a conclusion statement.

A1

A2

:

An

A

where means 'therefore' or 'it follows that.'}Premises

Conclusion

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Valid reasoning in proofs

Example: modus ponens P (P Q) Q modus ponens

P: The car is running

Q: The car has gas.

If we know that the car is running (P), we can prove that (Q) it has gas.

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Rules of Inference

Rules of Inference - used in proofs, or arguments, to move from what is known to what we want to prove.

modus ponens is a valid rule of inference.

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Valid Argument

An argument is said to be valid if whenever all the premises are true, the conclusion is also true.

If the premises are true, but the conclusion false, the argument is said to be invalid.

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Rules of Inference

Other famous rules of inference: PP Q Addition____________________________________P QPSimplification

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Rules of Inference

QP QPModus Tollens____________________________________ P QQRP RHypothetical syllogism

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Rules of Inference

P QPQDisjunctive syllogism ____________________________________ PQP QConjunction

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Rules of Inference

(P Q) (R S)P RQSConstructive dilemma

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Three Techniques Show true using logical inference

• Assume the hypotheses are true• Use the rules of inference and logical equivalences

to determine that the conclusion is true. Show true by showing that no way exists to make all

premises true but conclusion false Show false by finding a way to make premises true but

conclusion false.

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Formal Proofs

To prove an argument is valid or the conclusion followslogically from the hypotheses: Assume the hypotheses are true Use the rules of inference and logical

equivalences to determine that the conclusion is true.

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Proof Example

Consider the following logical argument:If horses fly or cows eat artichokes, then

the mosquito is the national bird. If the mosquito is the national bird then peanut butter takes good on hot dogs. But peanut butter tastes terrible on hot dogs. Therefore, cows don't eat artichokes.

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Proof Example

F Horses flyA Cows eat artichokesM The mosquito is the national birdP Peanut butter tastes good on hot

dogs

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Proof Example

Represent the formal argument using the variables1.(FA) M2.M P3.PA

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Proof Example

Use the hypotheses 1., 2., and 3. and the above rules of inference and any logical equivalencies to construct the proof.

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Proof Example

Assertion Reasons1.(FA) M Hypothesis 1.2.M P Hypothesis 2.3.(F A) P` steps 1 and 2 and hypothetical syll.4.P Hypothesis 3.5.(F A) steps 3 and 4 and modus tollens6.F A step 5 and DeMorgan7.A F step 6 and commutativity of 'and'8.A step 7 and simplificationQ. E. D.

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Proof Example

Consider:

p (q r) qp r

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Proof by contradiction

Consider:

r sp s r q p q

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Rules of Inference for Quantifiers

xP(x)P(c) Universal Instantiation (UI)_______________________________P(x)xP(x) Universal Generalization

(UG)

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Rules of Inference for Quantifiers

P(c)xP(x) Existential Generalization

(EG)_______________________________xP( x)P(c) Existential Instantiation (EI)

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Example

Every man has two legs. John Smith is a man.

Therefore, John Smith has two legs.

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Example

Define the predicates:M(x): x is a manL(x): x has two legsJ: John Smith, a member of the universe

The argument becomes1.x[M(x) L(x)]2.M( J )L( J)

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Example

The proof is1.x[M(x) L(x)] Hypothesis 12.M( J ) L(J ) step 1 and UI3.M(J) Hypothesis 24. L( J) steps 2 and 3 and modus ponensQ. E. D.

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Fallacies

Fallacies are incorrect inferences.Some common fallacies:

The Fallacy of Affirming the Consequent

The Fallacy of Denying the Antecedent

Begging the question or circular reasoning

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The Fallacy of Affirming the Consequent

If the butler did it he has blood on his hands.The butler had blood on his hands.Therefore, the butler did it.

This argument has the form:P QQ or [(P Q) Q] P P

which is not a tautology and therefore not a rule of inference!

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The Fallacy of Denying the Antecedent (or the hypothesis)

If the butler is nervous, he did it.The butler is really mellow.Therefore, the butler didn't do it.

This argument has the formP QP or [(P Q) P] Q Q

which is also not a tautology and hence not a rule of inference.

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Begging the question or circular reasoning

This occurs when we use the truth of statement being proved (or something equivalent) in the proof itself.

Example:Conjecture: if x2 is even then x is even.Proof: If x2 is even then x2 = 2k for some k. Then x = 2l for some l. Hence, x must be even.

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Proof Example

If the law is sufficient, then Christ died in vain

The law is sufficientTherefore Christ died in vain.

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Example

Babies are illogicalNobody is despised who can manage a

crocodileIllogical persons are despisedTherefore, babies cannot manage crocodiles

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Methods of Proof

We wish to establish the truth of the 'theorem’

P QP may be a conjunction of other

hypotheses.P Q is a conjecture until a proof is

produced.

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Types of proof:

Vacuous Proof of P Q The truth value of P Q is true if P is false. If P can be shown

false, then P Q holds. Thus prove P Q by showing P is false.

Example:If I am both rich and poor then hurricane Fran was a mild breeze.

This is of the form: (P P) Qand the hypotheses form a contradiction.

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Trivial Proof of P Q

If we know Q is true then P Q is true!

Example:If it's raining today then the empty set is a subset of every set.

The assertion is trivially true independent of the truth of P.

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Direct Proof of P Q

Prove Q, using P as an assumption.Thus prove P Q by showing Q is true whenever P is

true.assume the hypotheses are trueuse the rules of inference, axioms and any

logical equivalences to establish the truth of the conclusion.

Example: the Cows don’t eat artichokes proof above

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Direct Proof of P Q

Theorem: If 6x + 9y = 101, then x or y is not an integer.Proof: (Direct) Assume 6x + 9y = 101 is true.Then from the rules of algebra 3(2x + 3y) = 101.But 101/3 is not an integer so it must be the case that oneof 2x or 3y is not an integer (maybe both).Therefore, one of x or y must not be an integer.Q.E.D.

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Indirect Proof of P Q

Prove the contrapositive, e.g. Q P is true, using a direct proof methods.

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Indirect Proof of P QExample:

A perfect number is one which is the sum of all its divisors except itself. For example, 6 is perfect since 1 + 2 + 3 = 6. So is 28.

Theorem: A perfect number is not a prime. Proof: (Indirect). We assume the number p is a

prime and show it is not perfect. But the only divisors of a prime are 1 and itself. Hence the sum of the divisors less than p is 1 which

is not equal to p. Hence p cannot be perfect. Q. E. D..

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Proof by contradiction(reductio ad absurdum)

Assume the negation of the proposition is true, then derive a contradiction.

Thus to prove of P Q, assume P Q is true, then derive a contradiction.

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Proof by contradiction Theorem : 2 is irrational. Proof: Let P be the proposition “2 is irrational”Assume P or “2 is rational”Then 2 = a/b, where a and b are integers and have no common factors

(lowest terms). Then (2)2 = (a/b)2 is 2 = a2/b2. Thus 2b2 = a2. Thus a2 is even, implying a is even. Since a is even, a = 2c

for some integer c.Thus 2b2 = 4c2, so b2 = 2c2. Hence b is even.Contradiction! A and b are both even, so divisible by 2!

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Proof by cases of P Q

To prove P Q, find a set of propositions P1, P2, ..., Pn, n2, in which at least one Pj must be true for P to be true. P P1 P2 ... Pn

Then prove the n propositions P1 Q, P2 Q, ..., Pn Q.

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Proof by cases of P Q

Let be the operation 'max' on the set of integers:

if a b then ab = max{a, b} = a = ba.

Theorem: The operation is associative.For all a, b, c(ab)c = a (bc).

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Proof by cases:The operation is

associative

Let a, b, c be arbitrary integers.Then one of the following 6 cases must hold

(are exhaustive):1. a b c2. a c b3. b a c4. b c a5. c a b6. c b a

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Proof by cases:The operation is

associative

Case 1: ab = a, ac = a, and bc = b.Hence (ab)c = a = a(bc).Therefore the equality holds for the first

case.The proofs of the remaining cases are

similar. Q. E. D.

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Vacuous Proof

Consider the proposition:If you your grandfather dies as a baby then you

will get an A in this class.

Proof of this statement: Your grandfather didn’t die, thus thus the

premise must be false. Thus P Q must be true.

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Trivial Proof

Consider the proposition: If 3n2 + 5n -2 2n2 + 7n - 16 then n = n2.

P(n).

Proof of P(0): 0 = 02, thus P(0) is trivially true. QED.

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Direct Proof

Consider: The sum of two even numbers is even.Restate as:

x:y: (x is even and y is even) x + y is evenProof:

1. Remember: x is even a:x = 2a (definition)2. Assume x is even and y is even (assume hypothesis)3. x + y = 2a + 2b (from 1 and 2)4. 2a + 2b = 2·(a+b)5. By 1, 2·(a+b) is even - QED.

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Direct Proof

Consider: Every multiple of 6 is also a multiple of 3.

Rewrite: x zy:(6·x = y 3·z = y)Proof:

1. Assume 6x = y (hypothesis)

2. 6x = y can be rewritten as 3 · 2x = y

3. Let z = 2x, then 3·z = y holds. QED.

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Indirect Proofs

Prove the contrapositive, e.g. Prove that:

Q P is true

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Indirect Proofs Prove: If x2 is even, then x is even.Rewrite: x : (EVEN(x2) EVEN(x))

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Indirect ProofsProve: If x2 is even, then x is even

1. x : (ODD(x) ODD(x2)) (contrapositive)2. Assume 1 ODD(n) true for some n (hypothesis)3. x is odd a:x = 2a + 1 (definition)4. n = 2a + 1 for some a (2 & 3)5. n2 = (2a + 1)2 (substitution)6. (2a + 1)2 = (2a + 1)(2a + 1)

= 4a 2 + 4a + 1= 2 (2a2 + 2a) + 1

7. 2 (2a2 + 2a) + 1 is odd (3 & 6) QED

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Proof by contradiction

To prove of P Q, assume P Q, derive a contradiction.

Recall that: P Q P Q

Then: P QP Q) P Q (Demorgan’s)

Thus to prove P Q we assume P Q and show a contradiction.

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Proof by contradiction

Consider Theorem: There is no largest prime number.

This can be stated as

"If x is a prime number, then there exists another prime y which is greater"

Formally: x y: (PRIME(x) PRIME(y) x < y)

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Proof by contradictionThere is no largest prime number

Assume largest prime number does exist. Call this number p. Restate implication as p is prime, and there does not exist a

prime which is greater.1. Form a product r = 2 · 3 · 5 · ... p) (e.g. r is the product of all primes)2. If we divide r+1 by any prime, it will have remainder 13. r+1 is prime, since any number not divisible by any prime which is

less must be prime.4. but r+1 > p , which contradicts that p is the greatest prime number.

QED.

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Proof by cases

To prove P Q, find a set of propositions P1, P2, ..., Pn, n2, in which at least one Pj must be true for P to be true. P P1 P2 ... Pn

Then prove the n propositions P1 Q, P2 Q, ..., Pn Q.

Thus:P(P1P2...Pn) and (P1Q)(P2Q)...(PnQ)(PQ)

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Proof by cases

Consider: For every nonzero integer x ,x2 > 0.

Let:P = "x is a nonzero integer”Q = x2 > 0

We want to prove P Q

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Proof by cases

If: P = "x is a nonzero integer” Q = x2 > 0

Prove P QP can be broken up into two cases:

P1 = x > 0

P2 = x < 0

Note that P (P1 P2).

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Proof by cases

For every nonzero integer x ,x2 > 0. Prove each case -

Prove P1 Q:

If x > 0, then x2 > 0, since the product of two positive numbers is always positive.

Prove P2 Q:If x < 0, then x2 > 0, since the product of two negative numbers is always positive. QED.

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Existence Proofs

We wish to establish the truth ofxP( x).

Constructive existence proof:Establish P(c) is true for some c in

the universe.Then xP( x) is true by Existential

Generalization (EG).

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Constructive Existence Proofs

Theorem: There exists an integer solution to the equation

x2 + y2 = z2 .

Proof: Choose x = 3, y = 4, z = 5.

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Constructive Existence Proofs

Theorem: There exists a bijection from A= [0,1] to B= [0, 2].Proof: We build two injections and conclude there must be a bijection without

ever exhibiting the bijection.Let f be the identity map from A to B.Then f is an injection (and we conclude that | A | | B | ).Define the function g from B to A as g(x) = x/4.Then g is an injection.Therefore, | B | | A |.We now apply a previous theorem which states thatif | A | | B | and | B | | A | then | A | = | B |.Hence, there must be a bijection from A to B.(Note that we could have chosen g(x) = x/2 and obtained abijection directly).Q. E. D.

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Nonconstructive existence proof

Assume no c exists which makes P(c) true and derive a contradiction.Example:

Theorem: There exists an irrational number.Proof:Assume there doesn’t exist an irrational number.Then all numbers must be rational.Then the set of all numbers must be countable.Then the real numbers in the interval [0, 1] is a countable set. But we have already shown this set is not countable.Hence, we have a contradiction (The set [0,1] is countableand not countable).Therefore, there must exist an irrational number.Q. E. D.Note: we have not produced such a number!

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Disproof by Counterexample:

Recall that xP(x) xP(x ).

To establish that xP(x ) is true (or xP(x) is false) construct a c such that P(c) is true or P(c) is false.

In this case c is called a counterexample to the assertion xP(x)

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Nonexistence Proofs

We wish to establish the truth of

xP( x) (which is equivalent to xP(x) ).

Use a proof by contradiction by assuming there is a c which makes P(c) true.

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The (infamous) Halting Problem

We wish to establish the nonexistence of a universal debugging program.Theorem: There does not exist a program which will always determine if an arbitrary program P halts.We say the Halting Problem is undecidable.

UDPP1

P2

P3

Yes (Halts)

No (Infinite Loop)

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Halting ProblemProof: Suppose there is such a program called HALT

which will determine if any input-free program P halts.HALT(P) prints 'yes' and halts if P halts,

otherwise,HALT(P) prints 'no' and halts.We now construct another procedure as follows:procedure ABSURD;

if HALT(ABSURD) = 'yes' thenwhile true do print 'ha'(Note that ABSURD is input-free.)

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Halting Problem If ABSURD halts then we execute the loop which

prints unending gales of laughter and thus the procedure does not halt.

If ABSURD does not halt then we will exit the program and halt.Hence, ABSURD halts if it doesn'tand doesn't halt if it does

which is an obvious contradiction. Hence such a program does not exist.Q. E. D.

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Universally Quantified Assertions

We wish to establish the truth of

xP(x)We assume that x is an arbitrary

member of the universe and show P(x) must be true. Using UG it follows that xP(x) .

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Universally Quantified Assertions

Example:Theorem: For the universe of

integers, x is even iff x2 is even.Proof: The quantified assertion isx[x is even x2 is even]

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Universally Quantified Assertions

Proof:We assume x is arbitrary.Recall that P Q is equivalent to (P Q) (QP).Case 1. We show if x is even then x2 is even using a

directproof (the only if part or necessity).If x is even then x = 2k for some integer k.Hence, x2 = 4k2 = 2(2k2 ) which is even since it is aninteger which is divisible by 2.This completes the proof of case 1.

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Universally Quantified Assertions

Case 2. We show that if x 2 is even then x must be even(the if part or sufficiency) .We use an indirect proof:Assume x is not even and show x2 is not even.If x is not even then it must be odd.So, x = 2k + 1 for some k.Thenx2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1which is odd and hence not even.This completes the proof of the second case.Therefore we have shown x is even iff x2 is even.Since x was arbitrary, the result follows by UG.Q.E.D.

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