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1 Sections 1.5 & 3.1 Methods of Proof / Proof Strategy.

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1 1 Sections 1.5 & 3.1 Methods of Proof / Proof Strategy

2 2 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

3 3 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

4 4 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

5 5 Addition A true hypothesis implies that the disjunction of that hypothesis and another are true p ----------  p  q or p  (p  q)

6 6 Simplification If the conjunction of 2 propositions is true, then each proposition is true p  q ----------  p or (p  q)  p

7 7 Conjunction If p is true and q is true, then p  q is true p q ----------  p  q or ((p)  (q))  p  q

8 8 Modus Ponens If a hypothesis and implication are both true, then the conclusion is true p p  q -----------  q or (p  (p  q))  q

9 9 Modus Tollens If a conclusion is false and its implication is true, then the hypothesis must be false  q p  q -----------  p or [  q  (p  q)]   p

10 10 Hypothetical Syllogism If an implication is true, and the implication formed using its conclusion as the hypothesis is also true, then the implication formed using the original hypothesis and the new conclusion is also true p  q q  r -----------  p  r or [(p  q)  (q  r)]  (p  r)

11 11 Disjunctive Syllogism If a proposition is false, and the disjunction of it and another proposition is true, the second proposition is true p  q  p ---------  q or, [(p  q)   p]  q

12 12 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: (p 1  p 2  …  p n )  q When several premises are involved, several rules of inference my be needed to show that an argument is valid

13 13 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))  q we can conclude “I have Discrete Math today” (q)

14 14 Another Example Let  q = “I don’t have Discrete Math today” and p  q = “If it is Monday, I have Discrete Math today” If both of the above are true, then by Modus Tollens: [  q  (p  q)]   p we can conclude “It is not Monday” (  p)

15 15 Yet Another Example Let p  q = “If I am taking this class, I passed the test” and q  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)

16 16 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 inference to show that the hypotheses: Imply the conclusion: Randy will not get the job

17 17 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 ponens again, substituting q for p and  r for q: q q   r -----------  r 4. We have a valid argument!

18 18 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

19 19 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

20 20 Rules of Inference for Quantified Statements Existential instantiation:  xP(x) ----------  P(c) for some c  U Note that value of c is not known; we only know it exists Existential generalization: P(c) for some c  U ------------------------   xP(x)

21 21 Example Let P(x) = “A man is mortal”; then  xP(x) = “All men are mortal” Assuming p = “Socrates is a man” is true, show that q = “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

22 22 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)

23 23 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

24 24 Example of Direct Proof Prove “if n is odd, n 2 must be odd” –Let p = “n is odd” –Let q = “n 2 is odd” Assume n is odd; then n = 2k + 1 for some integer k (by definition of an odd number) This means n 2 = (2k + 1) 2 = 2(2k 2 + 2k) + 1 Thus, by definition, n 2 is odd

25 25 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

26 26 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

27 27 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

28 28 Vacuous Proof Example Show that P(0) is true where P(n) is: –“if n > 1, then n 2 > n” –Let p = n>1 and q = n 2 > 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 (0 2 > 0 ) is false for P(0) - since the premise is false, the implication is true

29 29 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

30 30 Example of Trivial Proof Let P(n) = “if a >= b then a n >= b n ” where a and b are positive integers; show that P(0) is true –so p = a >=b and –q = a 0 >= b 0 Since a 0 = b 0, q is true for P(0) Since q is true, p  q is true Note that this proof didn’t require examining the hypothesis

31 31 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)

32 32 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 = a 2 /b 2 If 2 = a 2 /b 2 then 2b 2 = a 2 So a 2 must be even, and a must be even

33 33 Example of proof by contradiction If a is even, then a = 2c and a 2 = 4c 2 Thus 2b 2 = 4c 2 and b 2 = 2c 2 - which means b 2 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

34 34 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

35 35 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

36 36 Proof by Cases To prove (p 1  p 2  …  p n )  q, can use the tautology: ((p 1  p 2  …  p n )  q)  ((p 1  q)  (p 2  q)  …  (p n  q)) as a rule for inference In other words, show that p i  q for all values of i from 1 through n

37 37 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 (p 1  p 2  …  p n ), can use the tautology: (p 1  p 2  …  p n )  ((p 1  p 2 )  (p 2  p 3 )  …  (p n  p 1 ))

38 38 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)

39 39 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

40 40 Constructive Existence Proof Example For every positive integer n, there is an integer divisible by >n primes Stated formally, this is:  n  x(x:x is divisible by >n primes) Assume we know the prime numbers and can list them: p 1, p 2, … If so, the number p 1 * p 2 * … * p n+1 is divisible by >n primes

41 41 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)

42 42 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  x  P(x) is true –Many examples exist - 3, 6 and 7 are all candidates

43 43 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

44 44 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

45 45 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

46 46 Backward reasoning - example Prove 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 n 2 = (2i + 1) 2 = 4i 2 + 4i + 1 We need to show that n 2 has the form 8k + 1 Reasoning backwards, this follows if 4i 2 + 4i can be written as 8k 2.But 4i 2 + 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, 4i 2 + 4i = 4i(i + 1) = 4(2k) = 8k; it follows that, since n 2 = 4i 2 + 4i + 1 and 4i 2 + 4i = 8k, that n 2 = 8k + 1

47 47 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

48 48 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)

49 49 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

50 50 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

51 51 Section 3.1 Methods of Proof - ends -

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