info 2950

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Info 2950

Prof. Carla Gomesgomes@cs.cornell.edu

Module Logic (part 2 --- proof methods)

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Methods for Proving Theorems

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Theorems, proofs, and rules of inference

When is a mathematical argument (or “proof”) correct?What techniques can we use to construct a mathematical argument?

Theorem – statement that can be shown to be true.Axioms or postulates or premises – statements which are given and assumed to

be true.Proof – sequence of statements, a valid argument, to show that a theorem is true.Rules of Inference – rules used in a proof to draw conclusions from assertions

known to be true.

Note: Lemma is a “pre-theorem” or a result that needs to be proved to prove the theorem; A corollary is a “post-theorem”, a result which follows directly the theorem that has been proved. Conjecture is a statement believed to be true but for which there is not a proof yet. If the conjecture is proved true it becomes a thereom. Fermat’s theorem was a conjecture for a long time.

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Valid Arguments

Recall:

An argument is a sequence of propositions. The final proposition is called the conclusion of the argument while the other propositions are called the premises or hypotheses of the argument.

An argument is valid whenever the truth of all its premises implies the truth of its conclusion.

How to show that q logically follows from the hypotheses (p1 p2 …pn)?

Show that

(p1 p2 …pn) q is a tautology

One can use the rules of inference to show the validity of an argument.

Vacuous proof - if one of the premises is false then (p1 p2 …pn) q is vacuously True, since False implies anything.

(reminder)

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Note: Many theorems involve statements for universally quantified variables:

e.g.

“If x > y, where x and y are positive real numbers, then x2 > y2 ”

Quite often, when it is clear from the context, theorems are proved without explicitly using the laws of universal instantiation and universal generalization.

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Methods of Proof

1) Direct Proof

2) Proof by Contraposition

3) Proof by Contradiction

4) Proof of Equivalences

5) Proof by Cases

6) Existence Proofs

7) Counterexamples

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1) Direct Proof

Proof statement : p q

by:

Assume p

From p derive q.

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Direct proof --- Example 1

Here’s what you know:

Mary is a Math major or a CS major.If Mary does not like discrete math, she is not a CS major.If Mary likes discrete math, she is smart.Mary is not a math major.

Can you conclude Mary is smart?

M CD CD SM

((M C) (D C) (D S) (M)) S?

Let M - Mary is a Math majorC – Mary is a CS majorD – Mary likes discrete math S – Mary is smart

Informally, what’s the chain of reasoning?

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In general, to prove p q, assume p and show that q follows.

((M C) (D C) (D S) (M)) S?

Reminder: Propositional logic Rules of Inference or Method of Proof

Rule of Inference Tautology (Deduction Theorem) Name

P

P QP (P Q) Addition

P Q P

(P Q) P Simplification

P

Q

P Q

[(P) (Q)] (P Q) Conjunction

P

PQ

Q

[(P) (P Q)] P Modus Ponens

Q

P Q

P

[(Q) (P Q)] P Modus Tollens

P Q

Q R

P R

[(PQ) (Q R)] (PR) Hypothetical Syllogism

(“chaining”)

P Q P

Q

[(P Q) (P)] Q Disjunctive syllogism

P Q P R Q R

[(P Q) (P R)] (Q R) Resolution

See Table 1, p. 66, Rosen.

Subsumes MP

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1. M C Given (premise)2. D C Given3. D S Given4. M Given

DS (disjunctive syllogism; 1,4)

MT (modus tollens; 2,5)

5. C

6. D

7. S MP (modus ponens; 3,6)

Mary is smart!

Example 1 - direct proof

QED

QED or Q.E.D. --- quod erat demonstrandum

“which was to be demonstrated”or “I rest my case”

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Direct Proof --- Example 2

Theorem:

If n is odd integer, then n2 is odd.

Looks plausible, but…

How do we proceed? How do we prove this?

Start with

Definition: An integer is even if there exists an integer k such that n = 2k,

and n is odd if there exists an integer k such that n = 2k+1.

Properties: An integer is even or odd; and no integer is

both even and odd. (aside: would require proof.)

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Theorem:

(n) P(n) Q(n),

where P(n) is “n is an odd integer” and Q(n) is “n2 is odd.”

We will show P(n) Q(n)

Example 2: Direct Proof

Theorem:If n is odd integer, then n2 is odd.

Proof:Let P --- “n is odd integer” Q --- “n2 is odd” we want to show that P Q

• Assume P, i.e., n is odd.

• By definition n = 2k + 1, where k is some integer.

• Therefore n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2 (2k2 + 2k ) + 1, which is by definition is an odd number (use k’ = (2k2 + 2k ) ).

QED

Proof strategy hint: Go back to definitions of conceptsand start by trying direct proof.

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2) Proof by Contraposition

Proof of a statement p q

Use the equivalence to the contrapositive:

¬q ¬ p

So, try: from ¬q derive ¬ p (i.e. via direct proof).

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Again, p q q p (the contrapositive)

So, we can prove the implication p q by first assuming q, and showing that p follows.

Example: Prove that if a and b are integers, and a + b ≥ 15, then a ≥ 8 or b ≥ 8.

(a + b ≥ 15) (a ≥ 8) v (b ≥ 8)

(Assume q) Suppose (a < 8) (b < 8).(Show p) Then (a ≤ 7) (b ≤ 7),

thus (a + b) ≤ 14,thus (a + b) < 15,

thus it’s not the case that (a+b >= 15).

Proof by Contraposition: Example 1

QED

Proof strategy:Note that negationof conclusion is

easier to start withhere. Why?

Proof by Contraposition: Example 2

TheoremFor n integer ,

if 3n + 2 is odd, then n is odd.

I.e. For n integer, 3n + 2 is odd n is odd

Proof by Contraposition: Let p --- “3n + 2” is odd; q --- “n is odd”; we want to show that p q • The contraposition of our theorem is ¬q ¬p n is even 3n + 2 is even • Now we can use a direct proof• Assume ¬q , i.e, n is even therefore n = 2 k for some k • Therefore 3 n + 2 = 3 (2k) + 2 = 6 k + 2 = 2 (3k + 1) which is even. QED

Again, negationof conclusion iseasy to start with.Try direct proof.

Intuitively, why?Start with intuitive understanding!

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3) Proof by Contradiction(arguably the most celebrated / used proof method)

A – We want to prove p.

We show that:

(1) ¬p F (i.e., ¬p leads to a False statement or contradiction,

say r ¬r)

(2) We conclude that ¬p is false since (1) is True,

and therefore p is True.

B – We want to show p q

(1) Assume the negation of the conclusion, i.e., ¬q

(2) Use show that (p ¬q) F (i.e., (p ¬q) leads to a False statement

or contradiction)

(3) Since ((p ¬q ) F) (p q) (why?) we are done

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Example:

Rainy days make gardens grow.Gardens don’t grow if it is not hot.When it is cold outside, it rains.

Prove that it’s hot.

Given: R GH GH R

Show: H

((R G) (H G) (H R)) H?

Example 1: Proof by Contradiction

Let R – Rainy dayG – Garden growsH – It is hot

Hmm. We will assume “not Hot” ≡ “Cold”

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Given: R GH GH R

Show: H

1. R G Given (“if it rains, plants grow”)2. H G Given (“if it’s cold, plants don’t

grow”)3. H R Given (“if it’s cold, it rains”)4. H assume the contrary

5. R MP (3,4)

6. G MP (1,5)

7. G MP (2,4)

8. G G contradiction! (so, our assumption must be wrong.

H

Aside: we assume it’s either Hot or it is not Hot.Called the “law of excluded middle”. In certain complexarguments, it’s not so obvious. (hmm…) This led to “constructive mathematics” and “intuitionistic mathematics”.

QED

Hmm… “it can’t be cold!”

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Thm. 2 is irrational.

Suppose 2 is rational. Then 2 = a/b for some integers a and b (assume relatively prime; no factor in common).

2 = a/b implies

2 = a2/b2

2b2 = a2

a2 is even, and so a is even (i.e., a = 2k for some k)

b2 = 2k2

2b2 = (2k)2 = 4k2

b2 is even, and so b is even (b = 2k’ for some k’)

But if a and b are both even, then they are not

relatively prime!Contradition. So,

2 cannot be rational. Q.E.D.

Proof by Contradiction: Example 2Our first interesting math proof!

It’s quite clever!!

Note: how we again first went to the definition of concepts (“rational”). Makes sense. Definitions provideimportant premises. In a sense, math is all about “What follows from thedefinitions!”

Aside: A number is rational if it can be written as the ratio of two integers.

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You’re going to let me get away with that?

a2 is even, and so a is even (i.e., a = 2k for some k)??

So, a really is even. QED

contradiction

Suppose to the contrary that a is odd.

Then a = 2k + 1 for some integer k

Then a2 = (2k + 1)(2k + 1) = 4k^2 + 4k + 1 = 2(2k^2+2k) + 1 = 2k’ + 1

and a2 is odd.

Prime Factorization

Fundamental theorem of arithmetic

every positive integer has a unique prime factorization.

Many cryptographic protocols are based on the difficulty of

factoring large composite integers or a related problem, the

RSA problem. An algorithm which efficiently factors an

arbitrary integer would render RSA-based public-key

cryptography insecure.

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Theorem:There are infinitely many prime numbers.

Proof by contradictionLet P – “There are infinitely many primes”• Assume ¬P, i.e., “there is a finite number of primes” , call largest p_r.• Let’s define R the product of all the primes, i.e, R = p_1 x p_2 x … x p_r.• Consider R + 1• Now, R+1 is either prime or not:

– If it’s prime, we have prime larger than p_r. – If it’s not prime, let p* be a prime dividing (R+1). But p* cannot be

any of p_1, p_2, … p_r (remainder 1 after division); so, p* not among initial list and thus p* is larger than p_r.

• This contradicts our assumption that there is a finite set of primes, and therefore such an assumption has to be false which means that there are infinitely many primes. QED

Proof by Contradiction: Example 3

See e.g. http://odin.mdacc.tmc.edu/~krc/numbers/infitude.htmlhttp://primes.utm.edu/notes/proofs/infinite/euclids.html

Also, non-constructive flavor.

One of the most famousearly proofs. An earlyintellectual “tour the force”. (Euclid’s proof, ca 300 BC)

Clever “trick”. The key to the proof.

Obviously?

Can we use construction to generate new primes?

Answer:

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yes and no.

Construction leads to new primes not on list but onlyindirectly.

2 * 3 + 1 = 6 + 1 = 7 ok! (prime)2 * 3 *5 + 1 = 30 + 1 = 31 ok!2 * 3 * 5 * 7 + 1 = 210 + 1 = 211 ok!2 * 3 * 5 * 7 * 11 + 1 = 2310 + 1 = 2311 ok!

2 * 3 * 5 * 7 * 11 * 13 + 1 = 30,030 + 1 = 30,031 not prime! = 59 * 509 (both not on list!)How do we know? Requires factoring! Believed to be very hard…

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Theorem “If 3n+2 is odd, then n is odd”

Let p = “3n+2 is odd” and q = “n is odd”

1 – Assume p and ¬q i.e., 3n+2 is odd and n is not odd.2 – Because n is not odd, it is even.3 – If n is even, n = 2k for some k, and therefore 3n+2 = 3 (2k) + 2 = 2 (3k + 1), which is even.4 – So, we have a contradiction, 3n+2 is odd and 3n+2 is even.

Therefore, we conclude p q, i.e., “If 3n+2 is odd, then n is odd.”

Q.E.D.

Proof by Contradiction: Example 4

Compare with earlier proof by “contraposition”.Quite similar idea but slightly different proof strategy.

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Reminder: Methods of Proof

1) Direct Proof

2) Proof by Contraposition

3) Proof by Contradiction

4) Proof of Equivalences

5) Counterexamples

6) Proof by Cases

7) Existence Proofs

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4) Proof of Equivalences

To prove p q

show that p q

and q p.

The validity of this proof results from the fact that

(p q) [ (p q) (q p)] is a tautology

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5) Counterexamples

Show that (x) P(x) is false

we need only to find a counterexample.

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Counterexample: example

Show that the following statement is false:

“Every day of the week is a weekday”

Proof:

Saturday and Sunday are weekend days.

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6) Proof by Cases

To show

(p1 p2 … pn ) q

We use the tautology

[(p1 p2 … pn ) q ]

[(p1 q ) (p2 q) … (pn q )]

A particular case of a proof by cases is an exhaustive proof in which all the cases are considered

Note: equivalencevia re-writes or consider“when is expression False?”

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Theorem

“If n is an integer, then n2 ≥ n ”

Proof by casesCase 1 n=0 02 = 0Case 2 n > 0, i.e., n >= 1. We get n2 ≥ n since we can multiply both

sides of the inequality by n, which is positive.Case 3 n < 0 n2 ≥ 0 since we can multiply both sides of the

inequality by n, which is negative (changes the sign of the inequality); we have n2 ≥ 0 > n because n is negative. So, n2 ≥ n.

Proof by cases: Example

Aside: Hmm. Where is disjunction? (p1 p2 … pn ) q

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7) Existence ProofsExistence Proofs:

– Constructive existence proofs Example: “there is a positive integer that is the sum of cubes of

positive integers in two different ways” By “brute force”: 1729 = 103 + 93 = 123 + 13

– Non-constructive existence proofs Example: “n (integers), p so that p is prime, and p > n.”

Quite subtle in general… We showed it implicitly before. (a corollary from “infinitely many primes”)

Uniqueness proofsInvolves:

Existence proofShow uniqueness

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Existence Proofs: Example 1

n (integers), p so that p is prime, and p > n.

Proof: Let n be an arbitrary integer, and consider n! + 1. If (n! + 1) is prime, we are done since (n! + 1) > n. But what if

(n! + 1) is composite?

If (n! + 1) is composite then it has a prime factorization, p1p2…pn = (n! + 1)

NON-CONSTRUCTIVE

Consider the smallest pi, and call it p. How small can it be?

Can it be 2?Can it be 3?Can it be 4?Can it be n?

So, p > n, and we are done. QED

But we don’t know what p is!!

Thm. There exist irrational numbers x and y such that xy is rational.

Proof:2 is irrational (see earlier proof).

Consider: z = 22

We have two possible cases:1) z is rational. Then, we’re done (take x = 2 and y = 2 ). or1) z is irrational. Now, let x = z and y = 2. And consider:

xy = (22 )2

= 2(2 2) = 22 = 2 , which is rational. So, we’re done.

Since, either 1) or 2) must be true, it follows that there does exist irrational x

and y such that xy is rational. Q.E.D.

Existence proof: Example 2

Non-constructive!

So… what is it: is 22 rational or not?? guess?

It’s irrational but requires very different proof…

“Start with something you know

about rational / irrational numbers.”

Existence proof: Example 3

From game theory.

Consider the game “Chomp”.

Two players. Players take turn eating

at least one of the remaining cookies.

At each turn, the player also eats all cookies to the left and below the cookie he or she selects.

The player who is “forced” to eat the poisened cookie loses.

Is there a winning strategy for either player?

Poisonous

m x n cookies

Winning strategy for a player:“A way of making moves” that is guaranteedto lead to a win, no matter what theopponent does. (How big to write down?)

Claim: First player has a winning strategy!

Proof. (non-constructive)

First, note that the game cannot end in a “draw”. After at most m x n moves, someone must have eaten the last cookie. Consider the following strategy for the first player:

--- Start by eating the cookie in the bottom right corner.

Three possible moves

--- Now, two possibilities:

1) This is part of a winning strategy for 1st player (and thus player has winning strategy). OR

2) 2nd player can now make a move that is part of the winning strategy for the

2nd player.But, if 2) is the case, then 1st player can followa winning strategy by on the first move makingthe move of the second player and following hisor her winning strategy! So, again, 1st player haswinning strategy. Q.E.D.

This is called a “strategy stealing” argument. Think through carefully to convince yourself!(Actual strategy not known for general boards!)

Is first choice of the bottom right cookie essential? If so, why?

Corner is “null move”

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Fallacies

Fallacies are incorrect inferences. Some common fallacies:

1. The Fallacy of Affirming the Consequent

2. The Fallacy of Denying the Antecedent

3. 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 formPQQ P

or ((PQ) Q) P which is not a tautology and therefore not a valid rule of inference (try: Q --- True and P --- False)

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The Fallacy of Denying the Antecedent

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 Q¬P

¬Q

or ((PQ) ¬P) ¬Q which is not a tautology and therefore not a valid rule of inference.(Consider P --- False, and Q --- True.)

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Begging the question or circular reasoning

This occurs when we use the truth of the statement being proved (or

something equivalent) in the proof itself.

Example:

Conjecture: if n2 is even then n is even.

Proof: If n2 is even then n2 = 2k for some k. Let n = 2m for some m. Hence, n must be even.

Note that the statement n = 2m is introduced without any argument showing it.

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Final example:Tiling

Standard checkerboard. 8x8 = 64 squares

A dominoCan you use 32 dominos tocover the board? Easily!

(many ways!)What about the mutilatedcheckerboard? Hmm…Use counting? No! Why?

Proof uses clever coloringand counting argument.Note: also valid for boardand dominos without b&w pattern!(use proof by contradiction)

What is the proof based upon?

Notoriously hard problemautomated theorem prover --- requires “true cleverness”

X

X

62 squares: 32 black 30 white31 doms.: 31 black 31 white squares!

Do we need the “colors”?

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Additional Proof Methods Covered in Info2950

• Induction Proofs

• Combinatorial proofs

But first we have to cover some basic notions on sets, functions, and counting.