Download presentation

Presentation is loading. Please wait.

1
Lecture 11 1.5, 3.1 Methods of Proof

2
Last time in 1.5 To prove theorems we use rules of inference such as: p, p q, therefore, q NOT q, p q, therefore NOT p. p AND q, therefore p FORALL x P(x), therefore for arbitrary c, P(c) EXISTS x P(x), therefore for some c, P(c) It is easy to make mistakes, make sure that: 1) All premises pi are true when you prove (p1 AND p2 AND...pn) q 2) Every rule of inference you use is correct. Some proof strategies: To proof p q 1) direct proof: assume p is true, use rules to prove that q is true. 2) indirect proof, assume q is NOT true, use rules to prove p is NOT true. To prove p is true: 3) By contradiction: assume p is NOT true, use rules to show that NOT p F i.e. it leads to a contradiction.

3
Vacuous –Trivial Proofs Lets say we want to prove p q but the premise p can be shown to be false! Then p q is always true because (F T) = T and (F F) = T. This is a vacuous prove. Old example: prove that for any set S: Proof: The following must be shown to be true: However: the empty set does not contain any elements and the premise is always false. Therefore the implication must always be true! Trivial Proof: We want to prove p q, and we can show that q is true. Then, because (T T) = T and (F T) = T we have proven the implication. Example: P(n): a>=b a^n >= b^n for positive integers. Is P(0) true? P(0): a^0 >= b^0 is equivalent to 1>=1. Therefore, q is true and thus p q is true.

4
Example Indirect Proof Prove that: if n is an integer and n^2 is odd, then n is odd. Direct prove is hard in this case. Indirect proof: Assume NOT q : n is even. n = 2k n^2 = 4k^2 = 2(2k^2) is even, is not odd. Thus NOT q NOT p, p q

5
Example of Proof by Contradiction def: rational number is a number that can be written as a/b for integers a,b, where b should not be 0. Reals that are not rational are irrational. sqrt(2) is irrational. (note: we are not proving an implication now, although we could have written it as: if x=sqrt(2) x = irrational.) Assume sqrt(2) is not irrational. sqrt(2) = a/b where there is no common divisor (otherwise divide by this number). 2 = a^2 / b^2 2b^2 = a^2 a^2 = even a = even a = 2c 2b^2 = 4c^2 b^2 = 2c^2 b^2 = even b=even both a and b can be divided by 2 (contradiction).

6
1.5 An indirect proof is in fact also a prove by contradiction: Assume p = T and (NOT q) = T indirect prove: (NOT q) (NOT p) Therefore: p AND (NOT p) = T, contradiction Proof by cases: (p1 OR p2 OR p3... OR pn) q (if at least one of the premises hold, q follows) This is equivalent to (p1 q) AND (p2 q) AND... AND (pn q) Example: Prove x^2 >= 0 p1: x =0 p2: x=0 x^2 >=0 p3: x>0 x^2 >=0

7
Equivalence Proofs These are bidirectional statements of the form p q. Equivalent to proving 2 cases: p q AND q p. Example: Prove integer n is odd if and only if (iff) n^2 is odd. Lecture 10: if n is odd n^2 is odd Lecture 11: if n^2 is odd n is odd.

8
Lists of Equivalences p1 p2 p3 ..... pn This means they are all equivalent. There are C(n,2) pairs to prove! There is however a smarter way: Design one path through all pi’s that can bring you from any pi to any other then you are done: p1 p2 p3 p4 p5 p6

9
Theorems with Quantifiers Existence proofs: Proofs of the form: There exists an element x such that..... these proofs may be constructive (construct some x) or non-constructive. Uniqueness: Proofs of the form: there exists a unique element x such that... Example: Prove that proof: for arbitrary x, y=-x makes the proposition true. Is it unique. Assume it is not true and show contradiction: Let say there is a r such that x+r=0 but r is not –x. Then it follows that x+r=x+y r=y which contradicts our assumption.

10
3.1 self reading

11
The Halting Problem Is it possible to design an algorithm that always predicts for a given program P and a given input to that program I, if it will stop or run forever? Proposition: there isn’t Proof. Assume there is such a predictor H(P,I). A program can be represented as a bit-string, and therefore as input to another program. Imagine a program P that can take itself as input. H(P,P) should predict if it stops: H = 1 if it stops, H=0 if it doesn’t stop. Define a new program K(P) as follows: that is K(P) loops forever if H(P,P) = 1 K(P) = stops if H(P,P) = 0 Can H(K,K) predict whether this program stops? K(K) loops forever if H(K,K) = 1 K(K) stops if H(K,K) = 0 So H(K,K) does not predict correctly which is in contradiction with the assumption!

12
Sequences & Summations definition: A sequence is a function from the set of integers to a set S: a1,a2,a3,.... (function from 1,2... a1,a2,...) an is a term in the sequence which is sometimes denoted with {an} (not a set!) Example: {an} with an= 1/n over n=1,.... 1,1/2,1/3,... Geometric progression: Sequence of the form: a,ar,ar^2,...,ar^n,... a = initial term, r = common ratio are real numbers 3,6,12,24,... (a=3,r=2): ratios are constant Arithmetic progression Sequence of the form a,a+d,a+2d,...,a+nd,... a = initial term and d = common difference are real numbers 3,5,7,9,11,...(a=3,r=2): differences are constant A string: finite sequence a1,...,an.

13
Sums Notation notation: single sum double sum

14
Summations Theorem proof trivial

15
Summations Theorem: sketch of proof: Split into 2 cases: even and odd n. In both cases argue as follows:

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google