Proof Review CS 202 Aaron Bloomfield Spring 2007.

Proof Review

CS 202Aaron Bloomfield

Spring 2007

Proof methods in this course• Logical equivalences

– via truth tables– via logical equivalences

• Set equivalences– via set identities– via membership tables– via mutual subset proof– via set builder notation and

logical equivalences• Rules of inference

– for propositions– for quantified statements

• Proofs without a category– Pigeonhole principle– Combinatorial proofs

• Ten proof methods in chapter 3:– Direct proofs– Indirect proofs– Vacuous proofs– Trivial proofs– Proof by contradiction– Proof by cases– Proofs of equivalence– Existence proofs

• Constructive• Non-constructive

– Uniqueness proofs– Counterexamples

• Induction– Weak mathematical induction– Strong mathematical induction– Structural induction

Logic proofsLogic proofs

Using Truth Tables

p q r p→r q →r (p→r)(q →r) pq (pq) →r

T T T T T T T TT T F F F F T F

T F T T T T F T

T F F F T T F T

F T T T T T F T

F T F T F T F T

F F T T T T F T

F F F T T T F T

rqprqrp )()()(

(pq) →rpq(p→r)(q →r)q →rp→rrqp

Idempotent Law

Associativity of Or

rqprqrp )()()( Definition of implication

Using Logical Equivalences

rqprqrp )()()(

rqprqp

rqprrqp

rqprqrp )()()(

rqprqrp

Re-arranging

Original statement

DeMorgan’s Law

qpqp

qpqp )(

rqrprqrp )()(

rrr

Set proofsSet proofs

Definition of difference


DeMorgan’s law

Complementation law

Distributive law

Complement law

Identity law

Commutative law

Proof by using basic set identities• Prove that A∩B=B-(B-A)

)AB-(BBA

)A(BB

)AB(B

A)B(B

A)(B)B(B

A)(BA)(BBA

Proof by membership tables

• The following membership table shows that A∩B=B-(B-A)

• Because the two indicated columns have the same values, the two expressions are identical

• This is similar to Boolean logic!

A B A ∩ B B-A B-(B-A)1 1 1 0 11 0 0 0 00 1 0 1 00 0 0 0 0

Proof by showing each set is a subset of the other 1

• Assume that xB-(B-A)– By definition of difference, we know that xB and xB-A

• Consider xB-A– If xB-A, then (by definition of difference) xB and xA– Since xB-A, then only one of the inverses has to be true

(DeMorgan’s law): xB or xA• So we have that xB and (xB or xA)

– It cannot be the case where xB and xB– Thus, xB and xA– This is the definition of intersection

• Thus, if xB-(B-A) then xA∩B

Proof by showing each set is a subset of the other 2

• Assume that xA∩B– By definition of intersection, xA and xB

• Thus, we know that xB-A– B-A includes all the elements in B that are also not in A not

include any of the elements of A (by definition of difference)• Consider B-(B-A)

– We know that xB-A– We also know that if xA∩B then xB (by definition of

intersection)– Thus, if xB and xB-A, we can restate that (using the definition

of difference) as xB-(B-A)• Thus, if xA∩B then xB-(B-A)

Proof by set builder notation and logical equivalences 2

Original statement


Negating “element of”


DeMorgan’s Law

Distributive Law

Negating “element of”

Negation Law

Identity Law

Definition of intersection

)}(|{ AxBxBxx

)( ABB )}(|{ ABxBxx

))}((|{ ABxBxx

}|{ AxBxx

)}(|{ AxBxBxx }|{ AxBxBxBxx })(|{ AxBxBxBxx

}|{ AxBxFx

BA

Rules of InferenceRules of Inference

Modus Ponens example• Assume you are given the following two

statements:– “you are in this class”– “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 will get a grade

q

qpp

Modus Tollens

• Assume that we know: ¬q and p → q– Recall that p → q = ¬q → ¬p

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

p

qpq

Generalization & Specialization

• Generalization: If you know that p is true, then p q will ALWAYS be true

• Specialization: If p q is true, then p will ALWAYS be true

qp

p

p

qp

More rules of inference• Conjunction: if p and q are

true separately, then pq is true

• Elimination: If pq is true, and p is false, then q must be true

• Transitivity: If p→q is true, and q→r is true, then p→r must be true

qp

qp

q

pqp

rp

rqqp

Even more rules of inference• Proof by division into cases:

if at least one of p or q is true, then r must be true

• Contradiction rule: If ¬p→c is true, we can conclude p (via the contra-positive)

• Resolution: If pq is true, and ¬pr is true, then qr must be true– Not in the textbook

p qp rq r

r

p c

p

rq

rpqp

Example of proof

• We have the hypotheses:– “It is not sunny this afternoon and it

is colder than yesterday”– “We will go swimming only if it is

sunny”– “If we do not go swimming, then we

will take a canoe trip”– “If we take a canoe trip, then we will

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

home by sunset”?

• •

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

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

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

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

• Does this imply that “we will be home by sunset”?

• •






• •






• •






– “It is not sunny this afternoon and it

is colder than yesterday”– “We will go swimming only if it is

sunny”– “If we do not go swimming, then we

will take a canoe trip”– “If we take a canoe trip, then we will

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

home by sunset”?

pqrst

¬p q

r → p

¬r → s

s → t

t

Example of proof1. ¬p q 1st hypothesis2. ¬p Simplification using step 13. r → p 2nd hypothesis4. ¬r Modus tollens using steps 2 & 35. ¬r → s 3rd hypothesis6. s Modus ponens using steps 4 & 57. s → t 4th hypothesis8. t Modus ponens using steps 6 & 7

p

qp

q

qpp

p

qpq

Example of proof

• Given the hypotheses:– “If it does not rain or if it is not

foggy, then the sailing race will be held and the lifesaving demonstration will go on”

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

– “The trophy was not awarded”• Can you conclude: “It rained”?

(¬r ¬f) →(s l)

s → t

¬tr

Example of proof1. ¬t 3rd hypothesis2. s → t 2nd hypothesis3. ¬s Modus tollens using steps 2 & 34. (¬r¬f)→(sl) 1st hypothesis5. ¬(sl)→¬(¬r¬f) Contrapositive of step 46. (¬s¬l)→(rf) DeMorgan’s law and double negation law7. ¬s¬l Addition from step 38. rf Modus ponens using steps 6 & 79. r Simplification using step 8

p

qp

q

qpp

p

qpq

qp

p

Proofs without a categoryProofs without a category

Corollary 1: algebraic proofCorollary 1: algebraic proof

Let Let nn be a non-negative integer. Then be a non-negative integer. Then

Algebraic proofAlgebraic proof

n

k

n

kn

0

2

n

k

knk

kn

0

11

nn )11(2

n

k kn

0

n

j

jjnn yxjn

yx0

)(

Corollary 1: combinatorial proofCorollary 1: combinatorial proof

Let Let nn be a non-negative integer. Then be a non-negative integer. Then

Combinatorial proofCombinatorial proof A set with A set with nn elements has 2 elements has 2nn subsets subsets

By definition of power setBy definition of power set Each subset has either 0 or 1 or 2 or … or Each subset has either 0 or 1 or 2 or … or nn elements elements

There are subsets with 0 elements, subsets with 1 There are subsets with 0 elements, subsets with 1 element, … and subsets with element, … and subsets with nn elements elements

Thus, the total number of subsets isThus, the total number of subsets is Thus,Thus,

n

k

n

kn

0

2

nn

k kn

20

0n

n

k kn

0

1n

nn

Pigeonhole PrinciplePigeonhole PrincipleConsider 5 distinct points (Consider 5 distinct points (xxii, , yyii) with integer values, where ) with integer values, where ii = 1, = 1, 2, 3, 4, 52, 3, 4, 5Show that the midpoint of at least one pair of these five points Show that the midpoint of at least one pair of these five points also has integer coordinatesalso has integer coordinates

Thus, we are looking for the midpoint of a segment from (Thus, we are looking for the midpoint of a segment from (aa,,bb) to ) to ((cc,,dd))

The midpoint is ( (The midpoint is ( (aa++cc)/2, ()/2, (bb++dd)/2 ))/2 )Note that the midpoint will be integers if Note that the midpoint will be integers if aa and and cc have the same have the same parity: are either both even or both oddparity: are either both even or both odd

Same for Same for bb and and ddThere are four parity possibilitiesThere are four parity possibilities

(even, even), (even, odd), (odd, even), (odd, odd)(even, even), (even, odd), (odd, even), (odd, odd)Since we have 5 points, by the pigeonhole principle, there must Since we have 5 points, by the pigeonhole principle, there must be two points that have the same parity possibilitybe two points that have the same parity possibility

Thus, the midpoint of those two points will have integer coordinatesThus, the midpoint of those two points will have integer coordinates

Ten proof methods in Ten proof methods in chapter 3chapter 3

Direct proofs

• Consider an implication: p→q– If p is false, then the implication is always 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

Direct proof example

• Show that the square of an even number is an even number– Rephrased: if n is even, then n2 is even

• Assume n is even– Thus, n = 2k, for some k (definition of even

numbers)– n2 = (2k)2 = 4k2 = 2(2k2)– As n2 is 2 times an integer, n2 is thus even

Indirect proofs

• Consider an implication: p→q– It’s contrapositive is ¬q→¬p

• Is logically equivalent to the original implication!– If the antecedent (¬q) is false, then the

contrapositive is always true– Thus, show that if ¬q is true, then ¬p is true

• To perform an indirect proof, do a direct proof on the contrapositive

Indirect proof example

• If n2 is an odd integer then n is an odd integer

• Prove the contrapositive: If n is an even integer, then n2 is an even integer

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

• n2 = (2k)2 = 4k2 = 2(2k2)• Since n2 is 2 times an integer, it is even

Vacuous proofs

• Consider an implication: p→q

• If it can be shown that p is false, then the implication is always true– By definition of an implication

• Note that you are showing that the antecedent is false

Vacuous proof example

• Consider the statement:– All criminology majors in CS 202 are female– Rephrased: If you are a criminology major and

you are in CS 202, then you are female• Could also use quantifiers!

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

Trivial proofs

• Consider an implication: p→q

• If it can be shown that q is true, then the implication is always true– By definition of an implication

• Note that you are showing that the conclusion is true

Trivial proof example

• Consider the statement:– If you are tall and are in CS 202 then you are

a student

• Since all people in CS 202 are students, the implication is true regardless

Proof by contradiction

• Given a statement p, assume it is false– Assume ¬p

• Prove that ¬p cannot occur– A contradiction exists

• Given a statement of the form p→q– To assume it’s false, you only have to consider the

case where p is true and q is false

Proof by contradiction example• Theorem (by Euclid): There are infinitely many

prime numbers.

• Proof. Assume there are a finite number of primes• List them as follows: p1, p2 …, pn.• Consider the number q = p1p2 … pn + 1

– This number is not divisible by any of the listed primes• If we divided pi into q, there would result a remainder of 1

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

• This contradicts our assumption that all primes are in the list p1, p2 …, pn.

Proof by cases

• Show a statement is true by showing all possible cases are true

• Thus, you are showing a statement of the form:

is true by showing that:

qppp n ...21

qpqpqpqppp nn ...... 2121

Proof by cases example

• Prove that – Note that b ≠ 0

• Cases:– Case 1: a ≥ 0 and b > 0

• Then |a| = a, |b| = b, and– Case 2: a ≥ 0 and b < 0

• Then |a| = a, |b| = -b, and– Case 3: a < 0 and b > 0

• Then |a| = -a, |b| = b, and– Case 4: a < 0 and b < 0

• Then |a| = -a, |b| = -b, and

ba

ba

ba

ba

ba

ba

ba

ba

ba

ba

ba

ba

ba

ba

ba

ba

ba

Proofs of equivalences

• This is showing the definition of a bi-conditional

• Given a statement of the form “p if and only if q”– Show it is true by showing (p→q)(q→p) is

true

Proofs of equivalence example• Show that m2=n2 if and only if m=n or m=-n

– Rephrased: (m2=n2) ↔ [(m=n)(m=-n)]• Need to prove two parts:

– [(m=n)(m=-n)] → (m2=n2)• Proof by cases!• Case 1: (m=n) → (m2=n2)

– (m)2 = m2, and (n)2 = n2, so this case is proven• Case 2: (m=-n) → (m2=n2)

– (m)2 = m2, and (-n)2 = n2, so this case is proven– (m2=n2) → [(m=n)(m=-n)]

• Subtract n2 from both sides to get m2-n2=0• Factor to get (m+n)(m-n) = 0• Since that equals zero, one of the factors must be zero• Thus, either m+n=0 (which means m=n) or m-n=0 (which

means m=-n)

Existence proofs

• Given a statement: x P(x)• We only have to show that a P(c) exists for

some value of c

• Two types:– Constructive: Find a specific value of c for

which P(c) exists– Nonconstructive: Show that such a c exists,

but don’t actually find it• Assume it does not exist, and show a contradiction

Constructive existence proof example

• Show that a square exists that is the sum of two other squares– Proof: 32 + 42 = 52

• Show that a cube exists that is the sum of three other cubes– Proof: 33 + 43 + 53 = 63

Non-constructive existence proof example

• Prove that either 2*10500+15 or 2*10500+16 is not a perfect square– A perfect square is a square of an integer– Rephrased: Show that a non-perfect square exists in

the set {2*10500+15, 2*10500+16}

• Proof: The only two perfect squares that differ by 1 are 0 and 1– Thus, any other numbers that differ by 1 cannot both

be perfect squares– Thus, a non-perfect square must exist in any set that

contains two numbers that differ by 1– Note that we didn’t specify which one it was!

Uniqueness proofs

• A theorem may state that only one such value exists

• To prove this, you need to show:– Existence: that such a value does indeed

exist• Either via a constructive or non-constructive

existence proof– Uniqueness: that there is only one such value

Uniqueness proof example• If the real number equation 5x+3=a has a

solution then it is unique

• Existence– We can manipulate 5x+3=a to yield x=(a-3)/5– Is this constructive or non-constructive?

• Uniqueness– If there are two such numbers, then they would fulfill

the following: a = 5x+3 = 5y+3– We can manipulate this to yield that x = y

• Thus, the one solution is unique!

Counterexamples• Given a universally quantified statement, find a single

example which it is not true

• Note that this is DISPROVING a UNIVERSAL statement by a counterexample

x ¬R(x), where R(x) means “x has red hair”– Find one person (in the domain) who has red hair

• Every positive integer is the square of another integer– The square root of 5 is 2.236, which is not an integer

A note on counterexamples• You can DISPROVE something by showing a

single counter example– You are finding an example to show that something is

not true

• You cannot PROVE something by example

• Example: prove or disprove that all numbers are even– Proof by contradiction: 1 is not even– (Invalid) proof by example: 2 is even

InductionInduction

Mathematical induction example

• Inductive step: show 6

)1)1(2)(1)1)((1(1

1

2

kkkik

i

6)32)(2)(1()1(

1

22

kkkikk

i

6139261392 2323 kkkkkk

)32)(2)(1()12)(1()1(6 2 kkkkkkk

6)32)(2)(1(

6)12)(1()1( 2

kkkkkkk

6)1)1(2)(1)1)((1(1

1

2

kkkik

i

6)12)(1(

1

2

kkkik

i

Strong induction example

• Show that any number > 1 can be written as the product of one or more primes

• Base case: P(2)– 2 is the product of 2 (remember that 1 is not

prime!)• Inductive hypothesis: assume P(2), P(3),

…, P(k) are all true• Inductive step: Show that P(k+1) is true

Strong induction example 1

• Inductive step: Show that P(k+1) is true• There are two cases:

– k+1 is prime• It can then be written as the product of k+1

– k+1 is composite• It can be written as the product of two composites,

a and b, where 2 ≤ a ≤ b < k+1• By the inductive hypothesis, both P(a) and P(b) are

true

Chess and induction7

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7

Can the knight reach any square in a finite number of moves?

Show that the knight can reach any square (i, j) for which i+j=k where k > 1.

Base case: k = 2

Inductive hypothesis: assume the knight can reach any square (i, j) for which i+j=k where k > 1.

Inductive step: show the knight can reach any square (i, j) for which i+j=k+1 where k > 1.

Chess and induction

• Inductive step: show the knight can reach any square (i, j) for which i+j=k+1 where k > 1.– Note that k+1 ≥ 3, and one of i or j is ≥ 2– If i ≥ 2, the knight could have moved from (i-2, j+1)

• Since i+j = k+1, i-2 + j+1 = k, which is assumed true

– If j ≥ 2, the knight could have moved from (i+1, j-2)• Since i+j = k+1, i+1 + j-2 = k, which is assumed true

Proof Review CS 202 Aaron Bloomfield Spring 2007.

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Transcript of Proof Review CS 202 Aaron Bloomfield Spring 2007.