Methods of Proof Methods of Proof

(1)

Methods of Proof Methods of Proof

CS 202 CS 202

Rosen section 1.5

Aaron Bloomfield

(2)

In this slide set…

Rules of inference for propositions Rules of inference for propositions

Rules of inference for quantified Rules of inference for quantified

statements statements

Ten methods of proof

(3)

Proof methods in this slide set Proof methods in this slide set

Logical equivalences Logical equivalences

 via truth tablesvia truth tables

 via logical equivalencesvia logical equivalences

Set equivalences Set equivalences

 via membership tablesvia membership tables

 via set identitiesvia set identities

 via mutual subset proofvia mutual subset proof

 via set builder notation and via set builder notation and logical equivalences

logical equivalences

Rules of inference Rules of inference

 for propositionsfor propositions

 for quantified statementsfor quantified statements

Pigeonhole principle Pigeonhole principle Combinatorial proofs Combinatorial proofs

Ten proof methods in section 1.5:

 Direct proofsDirect proofs

 Indirect proofsIndirect proofs

 Vacuous proofsVacuous proofs

 Trivial proofsTrivial proofs

 Proof by contradictionProof by contradiction

 Proof by casesProof by cases

 Proofs of equivalenceProofs of equivalence

 Existence proofsExistence proofs

Constructive Constructive Non-constructive Non-constructive

 Uniqueness proofsUniqueness proofs

 CounterexamplesCounterexamples

Induction Induction

 Weak mathematical inductionWeak mathematical induction

 Strong mathematical inductionStrong mathematical induction

 Structural inductionStructural induction

(4)

Modus Ponens Modus Ponens

Consider (p

Consider (p   (p→q)) → q (p→q)) → q

p p q q p→q p→q p p   (p→q)) (p→q)) (p (p   (p→q)) → q (p→q)) → q

T T T T T T T T T T

T T F F F F F F T T

F F T T T T F F T T

F F F F T T F F T T

q

q p

p





(5)

Modus Ponens example Modus Ponens example

Assume you are given the following two Assume you are given the following two

statements:

 ““you are in this class”you are in this class”

 ““if you are in this class, you will get a grade”if you are in this class, you will get a grade”

Let p = “you are in this class”

Let q = “you will get a grade”

By Modus Ponens, you can conclude that you By Modus Ponens, you can conclude that you

will get a grade will get a grade

q

q p

p





(6)

Modus Tollens Modus Tollens

Assume that we know:

Assume that we know: ¬ ¬ q and p q and p → → q q



Recall that p Recall that p → → q = q = ¬ ¬ q q → → ¬ ¬ p p

Thus, we know

Thus, we know ¬ ¬ q and q and ¬ ¬ q q → → ¬ ¬ p p We can conclude

We can conclude ¬ ¬ p p

p q p

q









(7)

Modus Tollens example Modus Tollens example

Assume you are given the following two Assume you are given the following two

statements:

 ““you will not get a grade”you will not get a grade”

Let p = “you are in this class”

Let q = “you will get a grade”

By Modus Tollens, you can conclude that you By Modus Tollens, you can conclude that you

are not in this class are not in this class

p q p

q









(8)

Quick survey Quick survey



I feel I understand moduls ponens I feel I understand moduls ponens and modus tollens

and modus tollens

a)a)

Very well Very well

b)b)

With some review, I’ll be good With some review, I’ll be good

c)c)

Not really Not really

d)d)

Not at all Not at all

(9)

Addition & Simplification Addition & Simplification

Addition: If you know Addition: If you know

that p is true, then p

that p is true, then p   q q will ALWAYS be true

will ALWAYS be true

Simplification: If p

Simplification: If p   q is q is true, then p will true, then p will

ALWAYS be true ALWAYS be true

q p

p





p

q p





(10)

Example of proof Example of proof

Example 6 of Rosen, section 1.5 Example 6 of Rosen, section 1.5

We have the hypotheses:

 ““It is not sunny this afternoon and it It is not sunny this afternoon and it is colder than yesterday”

is colder than yesterday”

 ““We will go swimming only if it is We will go swimming only if it is sunny”

sunny”

 ““If we do not go swimming, then we If we do not go swimming, then we will take a canoe trip”

will take a canoe trip”

 ““If we take a canoe trip, then we will If we take a canoe trip, then we will be home by sunset”

be home by sunset”

Does this imply that “we will be Does this imply that “we will be

home by sunset”?

 ““It isIt is not not sunny this afternoonsunny this afternoon and it and it is colder than yesterday”

is colder than yesterday”

 ““We will go swimming only if We will go swimming only if it is it is sunny

sunny””

Does this imply that “we will be Does this imply that “we will be

home by sunset”?

 ““It isIt is not not sunny this afternoonsunny this afternoon and and it it is colder than yesterday

is colder than yesterday””

 ““We will go swimming only if We will go swimming only if it is it is sunny

sunny””

Does this imply that “we will be Does this imply that “we will be

home by sunset”?

 ““We will go swimmingWe will go swimming only if only if it is it is sunny

sunny””

 ““If If we dowe do not not go swimming, then we go swimming, then we will take a canoe trip”

Does this imply that “we will be Does this imply that “we will be

home by sunset”?

sunny””

 ““If If we dowe do not not go swimming, then go swimming, then we we will take a canoe trip

will take a canoe trip””

 ““If If we take a canoe tripwe take a canoe trip, then we will , then we will be home by sunset”

Does this imply that “we will be Does this imply that “we will be

home by sunset”?

 ““It isIt is not not sunny this afternoon and sunny this afternoon and it it is colder than yesterday

sunny””

 ““If If we dowe do not not go swimminggo swimming, then , then we we will take a canoe trip

will take a canoe trip””

 ““If If we take a canoe tripwe take a canoe trip, then , then we will we will be home by sunset

be home by sunset””

Does this imply that “

Does this imply that “ we will be we will be home by sunset

home by sunset ”? ”?

p q r s

t

¬p  q r → p

¬r → s s → t t

(11)

Example of proof Example of proof

1.1. ¬p ¬p  q q 11^st^st hypothesis hypothesis

2.2. ¬p¬p Simplification using step 1Simplification using step 1 3.3. r → pr → p 22^nd^nd hypothesis hypothesis

4.4. ¬r¬r Modus tollens using steps 2 & 3Modus tollens using steps 2 & 3 5.5. ¬r → s¬r → s 33^rd^rd hypothesis hypothesis

6.6. ss Modus ponens using steps 4 & 5Modus ponens using steps 4 & 5 7.7. s → t s → t 44^th^th hypothesis hypothesis

8.8. tt Modus ponens using steps 6 & 7Modus ponens using steps 6 & 7 q

p 

p q

p



q p

q





(12)

So what did we show?

We showed that:

 [(¬p [(¬p  q)  q)  (r → p)  (r → p)  (¬r → s)  (¬r → s)  (s → t)] → t (s → t)] → t

 That when the 4 hypotheses are true, then the That when the 4 hypotheses are true, then the implication is true

implication is true

 In other words, we showed the above is a tautology!In other words, we showed the above is a tautology!

To show this, enter the following into the truth To show this, enter the following into the truth

table generator at table generator at

http://sciris.shu.edu/~borowski/Truth/:

((~P ^ Q) ^ (R => P) ^ (~R => S) ^ (S => T)) => T ((~P ^ Q) ^ (R => P) ^ (~R => S) ^ (S => T)) => T

(13)

More rules of inference More rules of inference

Conjunction: if p and q are true Conjunction: if p and q are true

separately, then p

separately, then pq is trueq is true Disjunctive syllogism: If p

Disjunctive syllogism: If pq is q is true, and p is false, then q must true, and p is false, then q must

be true be true

Resolution: If p

Resolution: If pq is true, and q is true, and

¬p¬pr is true, then qr is true, then qr must be truer must be true Hypothetical syllogism: If p→q is Hypothetical syllogism: If p→q is true, and q→r is true, then p→r true, and q→r is true, then p→r

q p





q p







r q

r p

q p











r q

q p







(14)

Example of proof Example of proof

Rosen, section 1.5, question 4 Rosen, section 1.5, question 4

Given the hypotheses:



“ “ If it does not rain or if it is not If it does not rain or if it is not foggy, then the sailing race will foggy, then the sailing race will be held and the lifesaving be held and the lifesaving

demonstration will go on”



“ “ If the sailing race is held, then If the sailing race is held, then the trophy will be awarded”

the trophy will be awarded”



“ “ The trophy was not awarded” The trophy was not awarded”

Can you conclude: “It rained”?

(¬r  ¬f) → (s  l)

s → t

¬t

r

(15)

Example of proof Example of proof

1.1. ¬t¬t 33^rd^rd hypothesis hypothesis 2.2. s → ts → t 22^nd^nd hypothesis hypothesis

3.3. ¬s¬s Modus tollens using steps 2 & 3Modus tollens using steps 2 & 3 4.4. (¬r(¬r¬f)→(s¬f)→(sl)l) 11^st^st hypothesis hypothesis

5.5. ¬(s¬(sl)→¬(¬rl)→¬(¬r¬f)¬f) Contrapositive of step 4Contrapositive of step 4

6.6. (¬s(¬s¬l)→(r¬l)→(rf)f) DeMorgan’s law and double negation lawDeMorgan’s law and double negation law 7.7. ¬s¬s¬l¬l Addition from step 3Addition from step 3

8.8. rrff Modus ponens using steps 6 & 7Modus ponens using steps 6 & 7 9.9. rr Simplification using step 8Simplification using step 8

q

p  q

p p



q p

q



 p

(16)

Quick survey Quick survey



I feel I understand that proof… I feel I understand that proof…

a)a)

Very well Very well

b)b)

With some review, I’ll be good With some review, I’ll be good

c)c)

Not really Not really

d)d)

Not at all Not at all

(17)

Modus Badus Modus Badus

Consider the following:

Is this true?

p p q q p→q p→q q q   (p→q)) (p→q)) (q (q   (p→q)) → p (p→q)) → p

T T T T T T T T T T

T T F F F F F F T T

F F T T T T T T F F

F F F F T T F F T T

Not a valid

rule!

p

q p

q





p

p q

q









Fallacy of Fallacy of affirming the affirming the

conclusion

(18)

Modus Badus example Modus Badus example

Assume you are given the following two Assume you are given the following two

statements:

 ““you will get a grade”you will get a grade”

Let p = “you are in this class”

Let q = “you will get a grade”

You CANNOT conclude that you are in this class You CANNOT conclude that you are in this class

 You could be getting a grade for another classYou could be getting a grade for another class

p

q p

q





(19)

Modus Badus Modus Badus

Consider the following:

Is this true?

p p q q p→q p→q ¬p ¬p   (p→q)) (p→q)) (¬p (¬p   (p→q)) → ¬q (p→q)) → ¬q

T T T T T T F F T T

T T F F F F F F T T

F F T T T T T T F F

F F F F T T T T T T

Not a valid

rule!

q q p

p









Fallacy of

denying the

hypothesis

(20)

Modus Badus example Modus Badus example

Assume you are given the following two Assume you are given the following two

statements:

 ““you are not in this class”you are not in this class”

Let p = “you are in this class”

Let q = “you will get a grade”

You CANNOT conclude that you will not get a You CANNOT conclude that you will not get a

grade grade

 You could be getting a grade for another classYou could be getting a grade for another class

q q p

p









(21)

Quick survey Quick survey



I feel I understand rules of inference I feel I understand rules of inference for Boolean propositions…

for Boolean propositions…

a)a)

Very well Very well

b)b)

With some review, I’ll be good With some review, I’ll be good

c)c)

Not really Not really

d)d)

Not at all Not at all

(22)

Just in time for Valentine’s Day…

(23)

Bittersweets: Dejected sayings Bittersweets: Dejected sayings



I MISS MY EX I MISS MY EX



PEAKED AT 17 PEAKED AT 17



MAIL ORDER MAIL ORDER



TABLE FOR 1 TABLE FOR 1



I CRY ON Q I CRY ON Q



U C MY BLOG? U C MY BLOG?



REJECT PILE REJECT PILE



PILLOW HUGGIN PILLOW HUGGIN



ASYLUM BOUND ASYLUM BOUND



DIGNITY FREE DIGNITY FREE



PROG FAN PROG FAN



STATIC CLING STATIC CLING



WE HAD PLANS WE HAD PLANS



XANADU 2NITE XANADU 2NITE



SETTLE 4LESS SETTLE 4LESS



NOT AGAIN NOT AGAIN

(24)

Bittersweets: Dysfunctional sayings Bittersweets: Dysfunctional sayings



RUMORS TRUE RUMORS TRUE



PRENUP OKAY? PRENUP OKAY?



HE CAN LISTEN HE CAN LISTEN



GAME ON TV GAME ON TV



CALL A 900# CALL A 900#



P.S. I LUV ME P.S. I LUV ME



DO MY DISHES DO MY DISHES



UWATCH CMT UWATCH CMT



PAROLE IS UP! PAROLE IS UP!



BE MY YOKO BE MY YOKO



U+ME=GRIEF U+ME=GRIEF



I WANT HALF I WANT HALF



RETURN 2 PIT RETURN 2 PIT



NOT MY MOMMY NOT MY MOMMY



BE MY PRISON BE MY PRISON



C THAT DOOR? C THAT DOOR?

(25)

What we have shown What we have shown

Rules of inference for propositions Rules of inference for propositions

Next up: rules of inference for quantified Next up: rules of inference for quantified

statements

(26)

Rules of inference for the Rules of inference for the

universal quantifier universal quantifier

Assume that we know that

Assume that we know that   x P(x) is true x P(x) is true



Then we can conclude that P(c) is true Then we can conclude that P(c) is true

Here c stands for some specific constant Here c stands for some specific constant



This is called “universal instantiation” This is called “universal instantiation”

Assume that we know that P(c) is true for Assume that we know that P(c) is true for

any value of c any value of c



Then we can conclude that Then we can conclude that   x P(x) is true x P(x) is true



This is called “universal generalization” This is called “universal generalization”

(27)

Rules of inference for the Rules of inference for the

existential quantifier existential quantifier

Assume that we know that

Assume that we know that   x P(x) is true x P(x) is true



Then we can conclude that P(c) is true for Then we can conclude that P(c) is true for some value of c

some value of c



This is called “existential instantiation” This is called “existential instantiation”

Assume that we know that P(c) is true for Assume that we know that P(c) is true for

some value of c some value of c



Then we can conclude that Then we can conclude that   x P(x) is true x P(x) is true



This is called “existential generalization” This is called “existential generalization”

(28)

Example of proof Example of proof

Rosen, section 1.5, question 10a Rosen, section 1.5, question 10a

Given the hypotheses:

 ““Linda, a student in this class, owns Linda, a student in this class, owns a red convertible.”

a red convertible.”

 ““Everybody who owns a red Everybody who owns a red convertible has gotten at least one convertible has gotten at least one

speeding ticket”

Can you conclude: “Somebody in Can you conclude: “Somebody in this class has gotten a speeding this class has gotten a speeding

ticket”?

C(Linda) R(Linda)

x (R(x)→T(x))

x (C(x)T(x))

(29)

Example of proof Example of proof

1.1. x (R(x)→T(x))x (R(x)→T(x)) 33^rd^rd hypothesis hypothesis

2.2. R(Linda) → T(Linda)R(Linda) → T(Linda) Universal instantiation using step 1Universal instantiation using step 1 3.3. R(Linda)R(Linda) 22^nd^nd hypothesis hypothesis

4.4. T(Linda)T(Linda) Modes ponens using steps 2 & 3Modes ponens using steps 2 & 3 5.5. C(Linda)C(Linda) 11^st^st hypothesis hypothesis

6.6. C(Linda) C(Linda)  T(Linda) T(Linda) Conjunction using steps 4 & 5Conjunction using steps 4 & 5

7.7. x (C(x)x (C(x)T(x))T(x)) Existential generalization using Existential generalization using step 6

step 6

Thus, we have shown that “Somebody in this class has gotten a speeding ticket”

(30)

Example of proof Example of proof

Rosen, section 1.5, question Rosen, section 1.5, question 10d 10d

Given the hypotheses:



“ “ There is someone in this class There is someone in this class who has been to France”

who has been to France”



“ “ Everyone who goes to France Everyone who goes to France visits the Louvre”

visits the Louvre”

Can you conclude: “Someone Can you conclude: “Someone in this class has visited the in this class has visited the

Louvre”?

x (C(x)F(x))

x (F(x)→L(x))

x (C(x)L(x))

(31)

Example of proof Example of proof

1.1. x (C(x)x (C(x)F(x))F(x)) 11^st^st hypothesis hypothesis

2.2. C(y) C(y)  F(y) F(y) Existential instantiation using step 1Existential instantiation using step 1 3.3. F(y)F(y) Simplification using step 2Simplification using step 2

4.4. C(y)C(y) Simplification using step 2Simplification using step 2 5.5. x (F(x)→L(x))x (F(x)→L(x)) 22^nd^nd hypothesis hypothesis

6.6. F(y) → L(y)F(y) → L(y) Universal instantiation using step 5Universal instantiation using step 5 7.7. L(y)L(y) Modus ponens using steps 3 & 6Modus ponens using steps 3 & 6 8.8. C(y) C(y)  L(y) L(y) Conjunction using steps 4 & 7Conjunction using steps 4 & 7

9.9. x (C(x)x (C(x)L(x))L(x)) Existential generalization usingExistential generalization using step 8

step 8

Thus, we have shown that “Someone

(32)

How do you know which one to How do you know which one to

use? use?

Experience!

In general, use quantifiers with statements In general, use quantifiers with statements

like “for all” or “there exists”



Although the vacuous proof example on slide Although the vacuous proof example on slide 40 is a contradiction

40 is a contradiction

(33)

Quick survey Quick survey



I feel I understand rules of inference I feel I understand rules of inference for quantified statements…

for quantified statements…

a)a)

Very well Very well

b)b)

With some review, I’ll be good With some review, I’ll be good

c)c)

Not really Not really

d)d)

Not at all Not at all

(34)

Proof methods Proof methods

We will discuss ten proof methods:

1.

1. Direct proofsDirect proofs

2.2. Indirect proofsIndirect proofs

3.3. Vacuous proofsVacuous proofs

4.4. Trivial proofsTrivial proofs

5.

5. Proof by contradictionProof by contradiction

6.6. Proof by casesProof by cases

7.7. Proofs of equivalenceProofs of equivalence

8.8. Existence proofsExistence proofs

9.

9. Uniqueness proofsUniqueness proofs

10.10. CounterexamplesCounterexamples

(35)

Direct proofs Direct proofs

Consider an implication: p→q Consider an implication: p→q



If p is false, then the implication is always true If p is false, then the implication is always true



Thus, show that if p is true, then q is true Thus, show that if p is true, then q is true

To perform a direct proof, assume that p is

true, and show that q must therefore be

true true

(36)

Direct proof example Direct proof example

Rosen, section 1.5, question 20 Rosen, section 1.5, question 20



Show that the square of an even number is an Show that the square of an even number is an even number

even number



Rephrased: if n is even, then n Rephrased: if n is even, then n

²²

is even is even

Assume n is even Assume n is even



Thus, n = 2k, for some k (definition of even Thus, n = 2k, for some k (definition of even numbers)

numbers)



n n

²²

= (2k) = (2k)

²²

= 4k = 4k

²²

= 2(2k = 2(2k

²²

) )



As n As n

²²

is 2 times an integer, n is 2 times an integer, n

²²

is thus even is thus even

(37)

Quick survey Quick survey



These quick surveys are really These quick surveys are really getting on my nerves…

getting on my nerves…

a)a)

They’re great - keep ‘em coming! They’re great - keep ‘em coming!

b)b)

They’re fine They’re fine

c)c)

A bit tedious A bit tedious

d)d)

Enough already! Stop! Enough already! Stop!

(38)

Indirect proofs Indirect proofs

Consider an implication: p→q Consider an implication: p→q



It’s contrapositive is ¬q→¬p It’s contrapositive is ¬q→¬p

Is logically equivalent to the original implication!



If the antecedent (¬q) is false, then the If the antecedent (¬q) is false, then the contrapositive is always true

contrapositive is always true



Thus, show that if ¬q is true, then ¬p is true Thus, show that if ¬q is true, then ¬p is true

To perform an indirect proof, do a direct To perform an indirect proof, do a direct

proof on the contrapositive

(39)

Indirect proof example Indirect proof example

If n If n

²²

is an odd integer then n is an odd integer is an odd integer then n is an odd integer

Prove the contrapositive: If n is an even integer, Prove the contrapositive: If n is an even integer,

then n

²²

is an even integer is an even integer

Proof: n=2k for some integer k (definition of even Proof: n=2k for some integer k (definition of even

numbers) numbers)

n n

²²

= (2k) = (2k)

²²

= 4k = 4k

²²

= 2(2k = 2(2k

²²

) ) Since n

Since n

²²

is 2 times an integer, it is even is 2 times an integer, it is even

(40)

Which to use Which to use

When do you use a direct proof versus an When do you use a direct proof versus an

indirect proof?

If it’s not clear from the problem, try direct If it’s not clear from the problem, try direct

first, then indirect second first, then indirect second



If indirect fails, try the other proofs If indirect fails, try the other proofs

(41)

Example of which to use Example of which to use

Rosen, section 1.5, question 21 Rosen, section 1.5, question 21

 Prove that if n is an integer and nProve that if n is an integer and n³³+5 is odd, then n is +5 is odd, then n is eveneven

Via direct proof Via direct proof

 nn³³+5 = 2k+1 for some integer k (definition of odd +5 = 2k+1 for some integer k (definition of odd numbers)

numbers)

 nn³³ = 2k+6 = 2k+6



 Umm…Umm…

So direct proof didn’t work out. Next up: indirect So direct proof didn’t work out. Next up: indirect

3 2  6

 k n

(42)

Example of which to use Example of which to use

Rosen, section 1.5, question 21 (a) Rosen, section 1.5, question 21 (a)

 Prove that if n is an integer and nProve that if n is an integer and n³³+5 is odd, then n is +5 is odd, then n is eveneven

Via indirect proof Via indirect proof

 Contrapositive: If n is odd, then nContrapositive: If n is odd, then n³³+5 is even+5 is even

 Assume n is odd, and show that nAssume n is odd, and show that n³³+5 is even+5 is even

 n=2k+1 for some integer k (definition of odd numbers)n=2k+1 for some integer k (definition of odd numbers)

 nn³³+5 = (2k+1)+5 = (2k+1)³³+5 = 8k+5 = 8k³³+12k+12k²²+6k+6 = 2(4k+6k+6 = 2(4k³³+6k+6k²²+3k+3)+3k+3)

 As 2(4kAs 2(4k³³+6k+6k²²+3k+3) is 2 times an integer, it is even+3k+3) is 2 times an integer, it is even

(43)

Quick survey Quick survey



I feel I understand direct proofs and I feel I understand direct proofs and indirect proofs…

indirect proofs…

a)a)

Very well Very well

b)b)

With some review, I’ll be good With some review, I’ll be good

c)c)

Not really Not really

d)d)

Not at all Not at all

(44)

Proof by contradiction Proof by contradiction

Given a statement p, assume it is false Given a statement p, assume it is false

 Assume ¬pAssume ¬p

Prove that ¬p cannot occur Prove that ¬p cannot occur

 A contradiction existsA contradiction exists

Given a statement of the form p→q Given a statement of the form p→q

 To assume it’s false, you only have to consider the To assume it’s false, you only have to consider the case where p is true and q is false

case where p is true and q is false

(45)

Proof by contradiction example 1 Proof by contradiction example 1

Theorem (by Euclid): There are infinitely many Theorem (by Euclid): There are infinitely many

prime numbers.

Proof. Assume there are a finite number of primes Proof. Assume there are a finite number of primes

List them as follows: p

₁₁

, p , p

₂₂

…, p …, p

_n_n

. . Consider the number q = p

Consider the number q = p

₁₁

p p

₂₂

… p … p

_n_n

+ 1 + 1

 This number is not divisible by any of the listed primesThis number is not divisible by any of the listed primes

If we divided p

If we divided p_i_i into q, there would result a remainder of 1 into q, there would result a remainder of 1

 We must conclude that q is a prime number, not among We must conclude that q is a prime number, not among the primes listed above

the primes listed above

This contradicts our assumption that all primes are in the list This contradicts our assumption that all primes are in the list

(46)

Proof by contradiction example 2 Proof by contradiction example 2

Rosen, section 1.5, question 21 (b) Rosen, section 1.5, question 21 (b)

 Prove that if n is an integer and nProve that if n is an integer and n³³+5 is odd, then n is even+5 is odd, then n is even

 Rephrased: If nRephrased: If n³³+5 is odd, then n is even+5 is odd, then n is even

Assume p is true and q is false Assume p is true and q is false

 Assume that nAssume that n³³+5 is odd, and n is odd+5 is odd, and n is odd

n=2k+1 for some integer k (definition of odd numbers) n=2k+1 for some integer k (definition of odd numbers) nn³³+5 = (2k+1)+5 = (2k+1)³³+5 = 8k+5 = 8k³³+12k+12k²²+6k+6 = 2(4k+6k+6 = 2(4k³³+6k+6k²²+3k+3)+3k+3)

As 2(4k

As 2(4k³³+6k+6k²²+3k+3) is 2 times an integer, it must be +3k+3) is 2 times an integer, it must be eveneven

Contradiction!

(47)

A note on that problem…

Rosen, section 1.5, question 21 Rosen, section 1.5, question 21

 Prove that if n is an integer and nProve that if n is an integer and n³³+5 is odd, then n is even+5 is odd, then n is even

 Here, our implication is: If nHere, our implication is: If n³³+5 is odd, then n is even+5 is odd, then n is even

The indirect proof proved the contrapositive: ¬q → ¬p The indirect proof proved the contrapositive: ¬q → ¬p

 I.e., If n is odd, then nI.e., If n is odd, then n³³+5 is even+5 is even

The proof by contradiction assumed that the implication The proof by contradiction assumed that the implication

was false, and showed a contradiction was false, and showed a contradiction

 If we assume p and ¬q, we can show that implies qIf we assume p and ¬q, we can show that implies q

 The contradiction is q and ¬qThe contradiction is q and ¬q

Note that both used similar steps, but are different Note that both used similar steps, but are different

(48)

How the book explains How the book explains

proof by contradiction proof by contradiction

A very poor explanation, IMHO A very poor explanation, IMHO

Suppose q is a contradiction (i.e. is always false) Suppose q is a contradiction (i.e. is always false)

Show that ¬p→q is true Show that ¬p→q is true

 Since the consequence is false, the antecedent must be Since the consequence is false, the antecedent must be false

false

 Thus, p must be trueThus, p must be true

Find a contradiction, such as (r

Find a contradiction, such as (r   ¬r), to represent q ¬r), to represent q Thus, you are showing that ¬p→(r

Thus, you are showing that ¬p→(r   ¬r) ¬r)

 Or that assuming p is false leads to a contradictionOr that assuming p is false leads to a contradiction

(49)

A note on proofs by contradiction A note on proofs by contradiction

You can DISPROVE something by using a proof You can DISPROVE something by using a proof

by contradiction by contradiction

 You are finding an example to show that something is You are finding an example to show that something is not true

not true

You cannot PROVE something by example You cannot PROVE something by example

Example: prove or disprove that all numbers are Example: prove or disprove that all numbers are even even

 Proof by contradiction: 1 is not evenProof by contradiction: 1 is not even

 (Invalid) proof by example: 2 is even(Invalid) proof by example: 2 is even

(50)

Quick survey Quick survey



I feel I understand proof by I feel I understand proof by contradiction…

contradiction…

a)a)

Very well Very well

b)b)

With some review, I’ll be good With some review, I’ll be good

c)c)

Not really Not really

d)d)

Not at all Not at all

(51)

Vacuous proofs Vacuous proofs

Consider an implication: p→q Consider an implication: p→q

If it can be shown that p is false, then the If it can be shown that p is false, then the

implication is always true implication is always true



By definition of an implication By definition of an implication

Note that you are showing that the Note that you are showing that the

antecedent is false

(52)

Vacuous proof example Vacuous proof example

Consider the statement:



All criminology majors in CS 202 are female All criminology majors in CS 202 are female



Rephrased: If you are a criminology major and Rephrased: If you are a criminology major and you are in CS 202, then you are female

you are in CS 202, then you are female

Could also use quantifiers!

Since there are no criminology majors in Since there are no criminology majors in this class, the antecedent is false, and the this class, the antecedent is false, and the

implication is true

(53)

Trivial proofs Trivial proofs

Consider an implication: p→q Consider an implication: p→q

If it can be shown that q is true, then the If it can be shown that q is true, then the

implication is always true implication is always true



By definition of an implication By definition of an implication

Note that you are showing that the Note that you are showing that the

conclusion is true

(54)

Trivial proof example Trivial proof example

Consider the statement:



If you are tall and are in CS 202 then you are If you are tall and are in CS 202 then you are a student

a student

Since all people in CS 202 are students, Since all people in CS 202 are students,

the implication is true regardless

(55)

Proof by cases Proof by cases

Show a statement is true by showing all Show a statement is true by showing all

possible cases are true possible cases are true

Thus, you are showing a statement of the Thus, you are showing a statement of the

form:

is true by showing that:



^p₁ ^ ^p₂ ^^...^ ^p_n



^ ^q

 



^p₁ ^ ^p₂ ^^...^ ^p_n ^ ^q



^

 

^p₁ ^ ^q

 

^ ^p₂ ^ ^q



^^...^



^p_n ^ ^q

 

(56)

Proof by cases example Proof by cases example

Prove that Prove that



Note that b ≠ 0 Note that b ≠ 0

Cases:



Case 1: a ≥ 0 and b > 0 Case 1: a ≥ 0 and b > 0

Then |a| = a, |b| = b, and Then |a| = a, |b| = b, and



Case 2: a ≥ 0 and b < 0 Case 2: a ≥ 0 and b < 0

Then |a| = a, |b| = -b, and Then |a| = a, |b| = -b, and



Case 3: a < 0 and b > 0 Case 3: a < 0 and b > 0

Then |a| = -a, |b| = b, and Then |a| = -a, |b| = b, and



Case 4: a < 0 and b < 0 Case 4: a < 0 and b < 0

Then |a| = -a, |b| = -b, and Then |a| = -a, |b| = -b, and

b a b

a 

b a b a b

a  

b a b a b

a b

a 

 





b a b

a b

a     

a a a

a    

(57)

The think about proof by cases The think about proof by cases

Make sure you get ALL the cases Make sure you get ALL the cases



The biggest mistake is to leave out some of The biggest mistake is to leave out some of the cases

the cases

(58)

Quick survey Quick survey



I feel I understand trivial and vacuous I feel I understand trivial and vacuous proofs and proof by cases…

proofs and proof by cases…

a)a)

Very well Very well

b)b)

With some review, I’ll be good With some review, I’ll be good

c)c)

Not really Not really

d)d)

Not at all Not at all

(59)

End of prepared slides

(60)

Proofs of equivalences Proofs of equivalences

This is showing the definition of a bi- This is showing the definition of a bi-

conditional conditional

Given a statement of the form “p if and Given a statement of the form “p if and

only if q”



Show it is true by showing (p→q) Show it is true by showing (p→q)   (q→p) is (q→p) is

true true

(61)

Proofs of equivalence example Proofs of equivalence example

 Show that mShow that m²²=n=n²² if and only if m=n or m=-n if and only if m=n or m=-n

 Rephrased: (mRephrased: (m²²=n=n²²) ↔ [(m=n)) ↔ [(m=n)(m=-n)](m=-n)]

Need to prove two parts:

 [(m=n)[(m=n)(m=-n)] → (m(m=-n)] → (m²²=n=n²²))

Proof by cases!

Case 1: (m=n) → (m

Case 1: (m=n) → (m²²=n=n²²))

 (m)(m)²² = m = m²², and (n), and (n)²² = n = n²², so this case is proven, so this case is proven

Case 2: (m=-n) → (m

Case 2: (m=-n) → (m²²=n=n²²))

 (m)(m)²² = m = m²², and (-n), and (-n)²² = n = n²², so this case is proven, so this case is proven

 (m(m²²=n=n²²) → [(m=n)) → [(m=n)(m=-n)](m=-n)]

Subtract n

Subtract n²² from both sides to get m from both sides to get m²²-n-n²²=0=0 Factor to get (m+n)(m-n) = 0

Factor to get (m+n)(m-n) = 0

Since that equals zero, one of the factors must be zero Since that equals zero, one of the factors must be zero

(62)

Existence proofs Existence proofs

Given a statement:

Given a statement:   x P(x) x P(x)

We only have to show that a P(c) exists for We only have to show that a P(c) exists for

some value of c some value of c

Two types:



Constructive: Find a specific value of c for Constructive: Find a specific value of c for which P(c) exists

which P(c) exists



Nonconstructive: Show that such a c exists, Nonconstructive: Show that such a c exists, but don’t actually find it

but don’t actually find it

Assume it does not exist, and show a contradiction Assume it does not exist, and show a contradiction

(63)

Constructive existence proof Constructive existence proof

example example

Show that a square exists that is the sum Show that a square exists that is the sum

of two other squares of two other squares



Proof: 3 Proof: 3

²²

+ 4 + 4

²²

= 5 = 5

²²

Show that a cube exists that is the sum of Show that a cube exists that is the sum of

three other cubes three other cubes



Proof: 3 Proof: 3

³³

+ 4 + 4

³³

+ 5 + 5

³³

= 6 = 6

³³

(64)

Non-constructive existence proof Non-constructive existence proof

example example

Prove that either 2*10

Prove that either 2*10⁵⁰⁰⁵⁰⁰+15 or 2*10+15 or 2*10⁵⁰⁰⁵⁰⁰+16 is not a +16 is not a perfect square

perfect square

 A perfect square is a square of an integerA perfect square is a square of an integer

 Rephrased: Show that a non-perfect square exists in the set Rephrased: Show that a non-perfect square exists in the set {2*10

{2*10⁵⁰⁰⁵⁰⁰+15, 2*10+15, 2*10⁵⁰⁰⁵⁰⁰+16}+16}

Proof: The only two perfect squares that differ by 1 are 0 Proof: The only two perfect squares that differ by 1 are 0

and 1 and 1

 Thus, any other numbers that differ by 1 cannot both be perfect Thus, any other numbers that differ by 1 cannot both be perfect squares

squares

 Thus, a non-perfect square must exist in any set that contains Thus, a non-perfect square must exist in any set that contains two numbers that differ by 1

two numbers that differ by 1

 Note that we didn’t specify which one it was!Note that we didn’t specify which one it was!

(65)

Uniqueness proofs Uniqueness proofs

A theorem may state that only one such A theorem may state that only one such

value exists value exists

To prove this, you need to show:



Existence: that such a value does indeed Existence: that such a value does indeed exist

exist

Either via a constructive or non-constructive Either via a constructive or non-constructive existence proof

existence proof



Uniqueness: that there is only one such value Uniqueness: that there is only one such value

(66)

Uniqueness proof example Uniqueness proof example

If the real number equation 5x+3=a has a If the real number equation 5x+3=a has a

solution then it is unique solution then it is unique

Existence Existence

 We can manipulate 5x+3=a to yield x=(a-3)/5We can manipulate 5x+3=a to yield x=(a-3)/5

 Is this constructive or non-constructive?Is this constructive or non-constructive?

Uniqueness Uniqueness

 If there are two such numbers, then they would fulfill If there are two such numbers, then they would fulfill the following: a = 5x+3 = 5y+3

the following: a = 5x+3 = 5y+3

 We can manipulate this to yield that x = yWe can manipulate this to yield that x = y

Thus, the one solution is unique!

(67)

Counterexamples Counterexamples

Given a universally quantified statement, find a single Given a universally quantified statement, find a single

example which it is not true example which it is not true

Note that this is DISPROVING a UNIVERSAL statement Note that this is DISPROVING a UNIVERSAL statement

by a counterexample by a counterexample

x ¬R(x), where R(x) means “x has red hair”x ¬R(x), where R(x) means “x has red hair”

 Find one person (in the domain) who has red hairFind one person (in the domain) who has red hair

Every positive integer is the square of another integer Every positive integer is the square of another integer

 The square root of 5 is 2.236, which is not an integerThe square root of 5 is 2.236, which is not an integer

(68)

Mistakes in proofs Mistakes in proofs

Modus Badus Modus Badus



Fallacy of denying the hypothesis Fallacy of denying the hypothesis



Fallacy of affirming the conclusion Fallacy of affirming the conclusion

Proving a universal by example Proving a universal by example



You can only prove an existential by example! You can only prove an existential by example!

(69)

Quick survey Quick survey



I felt I understood the material in this I felt I understood the material in this slide set…

slide set…

a)a)

Very well Very well

b)b)

With some review, I’ll be good With some review, I’ll be good

c)c)

Not really Not really

d)d)

Not at all Not at all

(70)

Quick survey Quick survey



The pace of the lecture for this The pace of the lecture for this slide set was…

slide set was…

a)a)

Fast Fast

b)b)

About right About right

c)c)

A little slow A little slow

d)d)

Too slow Too slow

(71)

Quick survey Quick survey



How interesting was the material in How interesting was the material in this slide set? Be honest!

this slide set? Be honest!

a)a)

Wow! That was SOOOOOO cool! Wow! That was SOOOOOO cool!

b)b)

Somewhat interesting Somewhat interesting

c)c)

Rather borting Rather borting

d)d)