CPSC 121: Models of Computation Unit 7: Proof Techniques (part 1) CPSC 121 – 2011W T2 1
Unit 7: Proof Techniques (part 1) th at Assignment #3 due Thursday February 16 17:00. Online quiz #8 very tentatively due Thursday st at 19:00 March 1 Epp, 4 th edition: 12.2, pages 791 to 795. Epp, 3 rd edition: 12.2, pages 745 to 747, 752 to 754 Rosen, 6 th edition: 12.2 pages 796 to 798, 12.3 Rosen, 7 th edition: 13.2 pages 858 to 861, 13.3 CPSC 121 – 2011W T2 2
Unit 7: Proof Techniques (part 1) By the start of class, you should be able to, for each proof strategy below: Identify the form of statement the strategy can prove. Sketch the structure of a proof that uses the strategy. Strategies: constructive/non-constructive proofs of existence generalizing from the generic particular direct proof (antecedent assumption) indirect proofs by contrapositive and contradiction proof by cases. CPSC 121 – 2011W T2 3
Unit 7: Proof Techniques (part 1) Quiz 7 feedback: Reasonably well done, although a bit hard to tell because of a board in Blackboard. We only have statistics for the graded attempts (44 out of 675 as of 23:38 last night) instead of all of them. We will do a lot more examples in class. CPSC 121 – 2011W T2 4
Unit 7: Proof Techniques (part 1) Open-ended question: when should you switch strategies? Monitor yourself When you are stuck. When the proof is going around in circles. When the proof is getting too messy. When it is taking too long. Through experience (how do you get that?) CPSC 121 – 2011W T2 5
Unit 7: Proof Techniques (part 1) CPSC 121: the BIG questions: ? ? ? How can we convince ourselves that an algorithm ? ? does what it's supposed to do? We need to prove its correctness. ? ? How do we determine whether or not one algorithm ? ? is better than another one? ? Sometimes, we need a proof to convince someone that ? the number of steps of our algorithm is what we claim it is. ? ? ? ? ? ? CPSC 121 – 2011W T2 6
Unit 7: Proof Techniques (part 1) By the end of this unit, you should be able to: Devise and attempt multiple different, appropriate proof strategie for a given theorem, including all those listed in the "pre-class" learning goals logical equivalences, rules of inference, universal modus ponens/tollens, For theorems requiring only simple insights beyond strategic choices or for which the insight is given/hinted, additionally prove the theorem. CPSC 121 – 2011W T2 7
Unit 7: Proof Techniques (part 1) Unit Summary Techniques for direct proofs. Existential quantifiers. More general term than in Epp. Universal quantifiers. Dealing with multiple quantifiers. Indirect proofs: contrapositive and contradiction Additional Examples CPSC 121 – 2011W T2 8
Unit 7: Proof Techniques (part 1) Direct Proofs (antecedent assumption) Assume the premises hold. Move one step at a time towards the conclusion. There are two general forms of statements: Those that start with an existential quantifier. Those that start with a universal quantifier. We use different techniques for them. CPSC 121 – 2011W T2 9
Unit 7: Proof Techniques (part 1) ∃ ∈ Form 1: x D, P(x) To prove this statement is true, we must Find a value of x (a “witness”) for which P(x) holds. So the proof will look like this: Choose x = <some value in D> Verify that the x we chose satisfies the predicate. Example: there is a prime number x such that 3 x +2 is not prime. CPSC 121 – 2011W T2 10
Unit 7: Proof Techniques (part 1) How do we translate There is a prime number x such that 3 x +2 is not prime into predicate logic? + , Prime(x) ~Prime(3 x +2) a) ∀ x ∈ Z ∧ b) ∃ x ∈ + , Prime(x) ~Prime(3 ∧ x +2) Z c) ∀ x ∈ + , Prime(x) → ~Prime(3 x +2) Z + , Prime(x) → ~Prime(3 x +2) d) ∃ x ∈ Z e) ∀ x ∈ x +2) where P is the set of all P, ~Prime(3 primes f) None of the above. CPSC 121 – 2011W T2 11
Unit 7: Proof Techniques (part 1) So the proof goes as follows: Proof: Choose x = It is prime because its only factors are 1 and x +2 = Now 3 and x +2 is not prime. Hence 3 QED. CPSC 121 – 2011W T2 12
Unit 7: Proof Techniques (part 1) Unit Summary Techniques for direct proofs. Existential quantifiers. Universal quantifiers. Dealing with multiple quantifiers. Indirect proofs: contrapositive and contradiction Additional Examples CPSC 121 – 2011W T2 13
Unit 7: Proof Techniques (part 1) ∀ ∈ Form 2: x D, P(x) To prove this statement is true, we must Show that P(x) holds no matter how we choose x. So the proof will look like this: Consider an unspecified element x of D Verify that the predicate P holds for this x. Note: the only assumption we can make about x is the fact that it belongs to D. So we can only use properties common to all elements of D. CPSC 121 – 2011W T2 14
Unit 7: Proof Techniques (part 1) Example: every Racket function is at least 12 characters long. The proof goes as follows: Proof: Consider an unspecified Racket function f This function Therefore f is at least 12 characters long. CPSC 121 – 2011W T2 15
Unit 7: Proof Techniques (part 1) Terminology: the following statements all mean the same thing: Consider an unspecified element x of D Without loss of generality consider a valid element x of D. Suppose x is a particular but arbitrarily chosen element of D. CPSC 121 – 2011W T2 16
Unit 7: Proof Techniques (part 1) Another example: Prove that if a, b are positive integers, then gcd(a,b) = gcd(b, a mod b) gcd(x,y) is the greatest common divisor of x and y. x mod y is the remainder after you divide x by y. For instance 17 mod 5 = 2. Why is this theorem useful? Racket supports rational numbers. It simplifies x/y by dividing x and y by their gcd. CPSC 121 – 2011W T2 17
Unit 7: Proof Techniques (part 1) ∀ ∈ Form 2*: x D, P(x) → Q(x) This is a special case of form 2 The textbook calls this (and only this) a direct proof. The proof looks like this: Proof: Consider an unspecified element x of D. Assume that P(x) is true. Use this and properties of the element of D to verify that the predicate Q holds for this x. CPSC 121 – 2011W T2 18
Unit 7: Proof Techniques (part 1) Why is the line Assume that P(x) is true valid? a) Because these are the only cases where Q(x) matters. b) Because P(x) is preceded by a universal quantifier. c) Because we know that P(x) is true. d) Both (a) and (c) e) Both (b) and (c) CPSC 121 – 2011W T2 19
Unit 7: Proof Techniques (part 1) Example: prove that ∀ n ∈ N, n ≥ 1024 → 10n ≤ nlog 2 n Proof: Consider an unspecified natural number n. Assume that n ≥ 1024. Then ... CPSC 121 – 2011W T2 20
Unit 7: Proof Techniques (part 1) Other interesting techniques for direct proofs ☺ Proof by intimidation Proof by lack of space (Fermat's favorite!) Proof by authority Proof by never-ending revision For the full list, see: http://school.maths.uwa.edu.au/~berwin/humour/invali d.proofs.html CPSC 121 – 2011W T2 21
Unit 7: Proof Techniques (part 1) Unit Summary Techniques for direct proofs. Existential quantifiers. Universal quantifiers. Dealing with multiple quantifiers. Indirect proofs: contrapositive and contradiction Additional Examples CPSC 121 – 2011W T2 22
Unit 7: Proof Techniques (part 1) How do we deal with theorems that involve multiple quantifiers? Start the proof from the outermost quantifier. Work our way inwards. Example: for every positive integer n, there is a prime p that is larger than n. Written using predicate logic: CPSC 121 – 2011W T2 23
Unit 7: Proof Techniques (part 1) The proof goes as follows: Proof: Consider an unspecified positive integer n Choose p as follows: + + ∀ n ∀ n ∈ ∈ Z Z ∃ p ∃ p ∈ ∈ + + Z Z Now prove that p > n and that p is prime. CPSC 121 – 2011W T2 24
Unit 7: Proof Techniques (part 1) Details (part 1) How do we choose p? First we compute x = n! + 1 (where n! = 1∙2∙3∙ ∙∙∙ ∙(n-1)∙n). By the fundamental theorem of arithmetic, x can be written as a product of primes: x = p 1 ∙p 2 ∙ ∙∙∙ p t We use any one of these as p (say p 1 ). The integer p is a prime by definition. CPSC 121 – 2011W T2 25
Unit 7: Proof Techniques (part 1) Details (part 2). Now we need to prove that p > n. Which of the following should we prove? + , i ≤ n → i divides n! a) ∀ i ∈ Z b) ∃ i ∈ + , i ≤ n i does not divide x ∧ Z c) ∀ i ∈ + , i ≤ p → i does not divide x Z d) ∀ i ∈ + , i ≤ n → i does not divide x Z e) None of the above. CPSC 121 – 2011W T2 26
Unit 7: Proof Techniques (part 1) Details (part 3). Now the proof: Pick an unspecified integer 2 ≤ i ≤ n. Observe that i = n! 1 x = n! i 1 i = 1 ⋅ 2 ⋯ i − 1 ⋅ i 1 ⋯ n 1 i i 1 ⋅ 2 ⋯ i − 1 ⋅ i 1 ⋯ n Since is an integer, but 1/i is not an integer, this means that x/i is not an integer. Hence i does not divide x. CPSC 121 – 2011W T2 27
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