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Sections 1.5 & 3.1

Methods of Proof / Proof Strategy

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Definitions

• Theorem: a statement that can be shown to be true

• Proof: demonstration of truth of theorem– consists of series of arguments called axioms or

postulates– these are statements of underlying assumptions

about mathematical structures, hypotheses of theorem to be proved, and previously proved theorems

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More Definitions

• Lemma: simple theorem used in proof of other theorems

• Corollary: proposition that can be established directly from a proved theorem

• Conjecture: statement whose truth value is unknown

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Rules of Inference

• Means to draw conclusions from other assertions

• Rules of inference provide justification of steps used to show that a conclusion follows from a set of hypotheses

• The next several slides illustrate specific rules of inference

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Addition

A true hypothesis implies that the disjunctionof that hypothesis and another are true

p---------- p q

or p (p q)

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Simplification

If the conjunction of 2 propositions is true,then each proposition is true

p q---------- p

or (p q) p

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Conjunction

If p is true and q is true, then p q is true

p q---------- p q

or ((p) (q)) p q

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Modus Ponens

If a hypothesis and implication are both true,then the conclusion is true

p p q----------- q

or (p (p q)) q

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Modus Tollens

If a conclusion is false and its implicationis true, then the hypothesis must be false

q p q-----------p

or [q (p q)] p

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Hypothetical Syllogism

If an implication is true, and the implicationformed using its conclusion as the hypothesis isalso true, then the implication formed using theoriginal hypothesis and the new conclusion isalso true

p q q r----------- p r

or [(p q) (q r)] (p r)

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Disjunctive Syllogism

If a proposition is false, and the disjunction of itand another proposition is true, the second proposition is true

p qp--------- q

or, [(p q) p] q

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Using rules of inference

• We can use the rules of inference to form the basis for arguments

• A valid argument is an implication in which, when all hypotheses are true, the conclusion is true: (p1 p2 … pn) q

• When several premises are involved, several rules of inference my be needed to show that an argument is valid

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Example

Let p = “It is Monday” and p q = “If it is Monday, I have Discrete Math today”Since these statements are both true, then by Modus Ponens:

(p (p q)) qwe can conclude “I have Discrete Math today” (q)

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Another Example

Let q = “I don’t have Discrete Math today” andp q = “If it is Monday, I have Discrete Math today”If both of the above are true, then by Modus Tollens:

[q (p q)] pwe can conclude “It is not Monday” (p)

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Yet Another Example

Let p q = “If I am taking this class, I passed the test” andq r = “If I passed the test, I’m a happy camper”If both of the above are true, then by hypothetical syllogism:

[(p q) (q r)] (p r)we can conclude “If I am taking this class, I’m a happy camper” (p r)

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A More Complicated Example

• Randy works hard

• If Randy works hard, then he is a dull boy

• If Randy is a dull boy, then he will not get the job

Construct an argument using rules for inferenceto show that the hypotheses:

Imply the conclusion:Randy will not get the job

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Example Continued

Let p = Randy works hard, q = He is a dull boy, r = He will get the job

So we want to prove: (p (p q) (q r)) r

1.Applying modus ponens to the first part: p p q-----------q

2. We now have: (q (q r)) r

3. Applying modus ponensagain, substituting q for pand r for q:

q q r -----------r

4. We have a valid argument!

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Fallacies

• A fallacy is an argument based on contingencies rather than tautologies; some examples:– Fallacy of affirming the conclusion:

[(p q) q] p

This is not a tautology because it’s false when p is false and q is true

– Fallacy of denying the hypothesis:

[(p q) p] q

Like the previous fallacy, this is not a tautology because it is false when p is false and q is true

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Rules of Inference for Quantified Statements

• Universal instantiation:xP(x)

----------

P(c) if c U

• Universal generalization:P(c) for arbitrary c U

-----------------------------

xP(x) Note: c must be arbitrary

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Rules of Inference for Quantified Statements

• Existential instantiation:xP(x)---------- P(c) for some c UNote that value of c is not known; we only know it

exists• Existential generalization:

P(c) for some c U------------------------ xP(x)

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Example

Let P(x) = “A man is mortal”; then xP(x) = “All men are mortal”

Assuming p = “Socrates is a man” is true, show thatq = “Socrates is mortal” is implied

This is an example of universal instantiation:P(Socrates) = “Socrates is mortal”;

Since xP(x) ---------

P(c)

Also, by modus ponens: (p (p q)) q

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Methods of Proof

• Many theorems are implications

• Recall that an implication (p q) is true when both p and q are true, or when p is false; it is only false if q is false

• To prove an implication, we need only prove that q is true if p is true (it is not common to prove q itself)

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Direct Proof

• Show that if p is true, q must also be true (so that the combination of p true, q false never occurs)– Assume p is true– Use rules of inference and theorems to show q

must also be true

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Example of Direct Proof

• Prove “if n is odd, n2 must be odd”– Let p = “n is odd”– Let q = “n2 is odd”

• Assume n is odd; then n = 2k + 1 for some integer k (by definition of an odd number)

• This means n2 = (2k + 1)2 = 2(2k2 + 2k) + 1

• Thus, by definition, n2 is odd

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Indirect Proof

• Uses the fact that an implication (p q) and its contrapositive q p have the same truth value

• Therefore proving the contrapositive proves the implication

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Indirect Proof Example

• Prove “if 3n + 2 is odd, then n is odd”– Let p = “3n + 2 is odd”– Let q = “n is odd”

• To prove q p , begin by assuming q is true– So n is even, and n = 2k for some integer k (by

definition of even numbers)– Then 3n + 2 = 3(2k) +2 = 6k + 2 = 2(3k + 1)

• Thus, 3n + 2 is even, q p and p q

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Vacuous Proof

• Suppose p is false - if so, then p q is true

• Thus, if p can be proven false, the implication is proven true

• This technique is often used to establish special cases of theorems that state an implication is true for all positive integers

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Vacuous Proof Example

• Show that P(0) is true where P(n) is:– “if n > 1, then n2 > n”– Let p = n>1 and q = n2 > n

• Since P(n) = P(0) and 0>1 is false, p is false• Since the premise is false, p q is true for

P(0)• Note that it doesn’t matter that the conclusion

(02 > 0 ) is false for P(0) - since the premise is false, the implication is true

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Trivial Proof

• If q can be proven true, then p q is true for all possible p’s, since:– T T and– F T are both true

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Example of Trivial Proof

• Let P(n) = “if a >= b then an >= bn” where a and b are positive integers; show that P(0) is true– so p = a >=b and– q = a0 >= b0

• Since a0 = b0, q is true for P(0)• Since q is true, p q is true• Note that this proof didn’t require examining

the hypothesis

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Proof by Contradiction

• Suppose q is false and p q is true

• This is possible only if p is true

• If q is a contradiction (e.g. r r), can prove p via p (r r)

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Example of proof by contradiction

• Prove 2 is irrational

• Suppose p is true - then 2 is rational

• If 2 is rational, then 2 = a/b for some numbers a and b with no common factors

• So (2 )2 = (a/b)2 or 2 = a2/b2

• If 2 = a2/b2 then 2b2 = a2

• So a2 must be even, and a must be even

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Example of proof by contradiction

• If a is even, then a = 2c and a2 = 4c2

• Thus 2b2 = 4c2 and b2 = 2c2 - which means b2 is even, and b must be even

• If a and b are both even, they have a common factor (2)

• This is a contradiction of the original premise, which states that a and b have no common factors

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Example of proof by contradiction

• So p (r r)– where p = 2 is rational, r = a & b have no

common factors, and r = a & b have a common factor

– r r is a contradiction– so p must be false– thus p is true and 2 is irrational

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Proof by contradiction and indirect proof

• Can write an indirect proof as a proof by contradiction

• Prove p q by proving q p

• Suppose p and q are both true

• Go through direct proof of q p to show p is also true

• Now we have a contradiction: p p is true

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Proof by Cases

• To prove (p1 p2 … pn) q, can use the tautology:((p1 p2 … pn) q) ((p1 q) (p2 q) … (pn q)) as a

rule for inference

• In other words, show that pi q for all values of i from 1 through n

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Proof by Cases

• To prove an equivalence (p q), can use the tautology:(p q) ((p q) (q p))

• If a theorem states that several propositions are equivalent (p1 p2 … pn), can use the tautology:(p1 p2 … pn) ((p1 p2) (p2 p3) … (pn p1))

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Theorems & Quantifiers

• Existence proof: proof of a theorem asserting that objects of a particular type exist, aka propositions of the form xP(x)

• Proof by counter-example: proof of a theorem of the form xP(x)

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Types of Existence Proofs

• Constructive: find an element a such that P(a) is true

• Non-constructive: prove xP(x) without finding a specific element - often uses proof by contradiction to show xP(x) implies a contradiction

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Constructive Existence Proof Example

• For every positive integer n, there is an integer divisible by >n primes

• Stated formally, this is: nx(x:x is divisible by >n primes)

• Assume we know the prime numbers and can list them: p1, p2, …

• If so, the number p1 * p2 * … * pn+1 is divisible by >n primes

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Non-constructive Existence Proof Example

• Show that for every positive integer n there is a prime greater than n

• This is xQ(x) where Q(x) is the proposition x is prime and x > n– Let n be a positive integer; to show there is a prime > n,

consider n! + 1– Every integer has a prime factor, so n! + 1 has at least one

prime factor– When n! + 1 is divided by an integer <= n, remainder is 1– Thus, any prime factor of this integer must be > n

• Proof is non-constructive because we never have to actually produce a prime (or n)

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Proof by Counter-example

• To prove xP(x) is false, need find only one element e such that P(e) is false

• Example: Prove or disprove that every positive integer can be written as the sum of 2 squares– We need to show xP(x) is true– Many examples exist - 3, 6 and 7 are all

candidates

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Choosing a method of proof

• When confronted with a statement to prove:– Replace terms by their definitions– Analyze what hypotheses & conclusion mean

• If statement is an implication, try direct proof;– If that fails, try indirect proof– If neither of the above works, try proof by

contradiction

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Forward reasoning

• Start with the hypothesis• Together with axioms and known theorems,

construct a proof using a sequence of steps that leads to the conclusion

• With indirect reasoning, can start with negation of conclusion, work through a similar sequence to reach negation of hypothesis

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Backward reasoning

• To reason backward to prove a statement q, we find a statement p that we can prove with the property p q

• The next slide provides an example of this type of reasoning

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Backward reasoning - exampleProve that the square of every odd integer has the form 8k + 1 for some integer k:

1. Begin with some odd integer n, which by definition has the form n = 2i + 1 for some integer i. • Then n2 = (2i + 1)2 = 4i2 + 4i + 1• We need to show that n2 has the form 8k + 1• Reasoning backwards, this follows if 4i2 + 4i can be written as 8k

2. But 4i2 + 4i = 4i(i + 1)• i(i+ 1) is the product of 2 consecutive integers• Since every other integer is even, either i or i+1 is even• This means their product is even, and can be written 2k for some integer k

3. Therefore, 4i2 + 4i = 4i(i + 1) = 4(2k) = 8k; it follows that, since n2 = 4i2 + 4i + 1 and 4i2 + 4i = 8k, that n2 = 8k + 1

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The Halting Problem

• The halting problem is a famous theorem in computer science:– Is there a procedure that can take as input a computer

program and its input and determine whether the program will stop with the given input?

– The short answer is no. You can’t just run a program and observe to determine if it will terminate:

• you’d know if it does halt, but not if it doesn’t

• some algorithms would outlast your ability to observe them

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Proof

• Suppose such a procedure (H) exists, and can be called with H(P,I) where P is a program and I is the input to P

• H prints “halt” if the P halts and “loops forever” if not

• Since a program’s source code can be used as input to itself, we could call H with H(P,P)

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Proof

• To show that no H exists, construct procedure K(P), which takes the result of H(P,P) as input– if output of H(P,P) is “loops forever”, K(P)

halts– if output of H(P,P) is “halts”, K(P) loops

forever

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Proof

• Suppose we provide K as input to K:– then if H(K,K) produces “loops forever”, K halts– and if H(K,K) produces “halts”, K loops forever

• This is a violation of what H says - so we have a contradiction

• H can’t always provide correct answers - so there is no procedure that solves the halting problem

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Section 3.1

Methods of Proof

- ends -