Theory of Computer Science
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1 Theory of Computer Science A3. Proof Techniques Malte Helmert University of Basel February 22, 2017
2 Introduction
3 Mathematical Statements Mathematical Statement A mathematical statement consists of a set of preconditions and a set of conclusions. The statement is true if the conclusions are true whenever the preconditions are true. German: mathematische Aussage, Voraussetzung, Folgerung/Konklusion, wahr Notes: set of preconditions is sometimes empty often, assumptions is used instead of preconditions ; slightly unfortunate because assumption is also used with another meaning ( cf. indirect proofs)
4 Mathematical Statements Mathematical Statement A mathematical statement consists of a set of preconditions and a set of conclusions. The statement is true if the conclusions are true whenever the preconditions are true. German: mathematische Aussage, Voraussetzung, Folgerung/Konklusion, wahr Notes: set of preconditions is sometimes empty often, assumptions is used instead of preconditions ; slightly unfortunate because assumption is also used with another meaning ( cf. indirect proofs)
5 Examples of Mathematical Statements Examples (some true, some false): Let p N 0 be a prime number. Then p is odd. There exists an even prime number. Let p N 0 with p 3 be a prime number. Then p is odd. All prime numbers p 3 are odd. For all sets A, B, C: A (B C) = (A B) (A C) The equation a k + b k = c k has infinitely many solutions with a, b, c, k N 1 and k 2. The equation a k + b k = c k has no solutions with a, b, c, k N 1 and k 3. Which ones are true, which ones are false?
6 Proofs Proof A proof derives the correctness of a mathematical statement from a set of axioms and previously proven statements. It consists of a sequence of proof steps, each of which directly follows from the axioms, previously proven statements and the preconditions of the statement, ending with the conclusions of the theorem. German: Beweis, Axiom, Beweisschritt
7 Disproofs A disproof (refutation) analogously shows that a given mathematical statement is false by giving an example where the preconditions are true, but the conclusion is false. This requires deriving, in a sequence of proof steps, the opposite (negation) of the conclusion. German: Widerlegung Formally, disproofs are proofs of modified ( negated ) statements. Be careful about how to negate a statement!
8 Proof Strategies typical proof/disproof strategies: 1 All x S with the property P also have the property Q. For all x S: if x has property P, then x has property Q. To prove, assume you are given an arbitrary x S that has the property P. Give a sequence of proof steps showing that x must have the property Q. To disprove, find a counterexample, i. e., find an x S that has property P but not Q and prove this.
9 Proof Strategies typical proof/disproof strategies: 2 A is a subset of B. To prove, assume you have an arbitrary element x A and prove that x B. To disprove, find an element in x A \ B and prove that x A \ B.
10 Proof Strategies typical proof/disproof strategies: 3 For all x S: x has property P iff x has property Q. ( iff : if and only if ) To prove, separately prove if P then Q and if Q then P. To disprove, disprove if P then Q or disprove if Q then P. German: iff = gdw. ( genau dann, wenn )
11 Proof Strategies typical proof/disproof strategies: 4 A = B, where A and B are sets. To prove, separately prove A B and B A. To disprove, disprove A B or disprove B A.
12 Proof Techniques most common proof techniques: direct proof indirect proof (proof by contradiction) contraposition mathematical induction structural induction German: direkter Beweis, indirekter Beweis (Beweis durch Widerspruch), Kontraposition, vollständige Induktion, strukturelle Induktion
13 Direct Proof
14 Direct Proof Direct Proof Direct derivation of the statement by deducing or rewriting.
15 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C).
16 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof. We first show that x A (B C) implies x (A B) (A C) ( only-if part, part, part): German: Hin-Richtung...
17 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof. We first show that x A (B C) implies x (A B) (A C) ( only-if part, part, part): Let x A (B C). Then x A and x B C. German: Hin-Richtung...
18 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof. We first show that x A (B C) implies x (A B) (A C) ( only-if part, part, part): Let x A (B C). Then x A and x B C. If x B then, because x A is true, x A B must be true. German: Hin-Richtung...
19 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof. We first show that x A (B C) implies x (A B) (A C) ( only-if part, part, part): Let x A (B C). Then x A and x B C. If x B then, because x A is true, x A B must be true. Otherwise, because x B C we know that x C and thus with x A, that x A C. German: Hin-Richtung...
20 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof. We first show that x A (B C) implies x (A B) (A C) ( only-if part, part, part): Let x A (B C). Then x A and x B C. If x B then, because x A is true, x A B must be true. Otherwise, because x B C we know that x C and thus with x A, that x A C. In both cases x A B or x A C, and we conclude x (A B) (A C).... German: Hin-Richtung
21 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof (continued). if part, part, part: we must show that x (A B) (A C) implies x A (B C). Let x (A B) (A C). German: Rückrichtung...
22 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof (continued). if part, part, part: we must show that x (A B) (A C) implies x A (B C). Let x (A B) (A C). If x A B then x A and x B. The latter implies x B C and hence x A (B C). German: Rückrichtung...
23 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof (continued). if part, part, part: we must show that x (A B) (A C) implies x A (B C). Let x (A B) (A C). If x A B then x A and x B. The latter implies x B C and hence x A (B C). If x / A B we know x A C due to x (A B) (A C). This (analogously) implies x A and x C, and hence x B C and thus x A (B C). German: Rückrichtung...
24 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof (continued). if part, part, part: we must show that x (A B) (A C) implies x A (B C). Let x (A B) (A C). If x A B then x A and x B. The latter implies x B C and hence x A (B C). If x / A B we know x A C due to x (A B) (A C). This (analogously) implies x A and x C, and hence x B C and thus x A (B C). In both cases we conclude x A (B C).... German: Rückrichtung
25 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof (continued). We have shown that every element of A (B C) is an element of (A B) (A C) and vice versa. Thus, both sets are equal.
26 Direct Proof: Example Theorem (distributivity) For all sets A, B, C: A (B C) = (A B) (A C). Proof. Alternative: A (B C) = {x x A and x B C} = {x x A and (x B or x C)} = {x (x A and x B) or (x A and x C)} = {x x A B or x A C} = (A B) (A C)
27 Questions Questions?
28 Indirect Proof
29 Indirect Proof Indirect Proof (Proof by Contradiction) Make an assumption that the statement is false. Derive a contradiction from the assumption together with the preconditions of the statement. This shows that the assumption must be false given the preconditions of the statement, and hence the original statement must be true. German: Annahme, Widerspruch
30 Indirect Proof: Example Theorem There are infinitely many prime numbers.
31 Indirect Proof: Example Theorem There are infinitely many prime numbers. Proof. Assumption: There are only finitely many prime numbers.
32 Indirect Proof: Example Theorem There are infinitely many prime numbers. Proof. Assumption: There are only finitely many prime numbers. Let P = {p 1,..., p n } be the set of all prime numbers. Define m = p 1... p n + 1.
33 Indirect Proof: Example Theorem There are infinitely many prime numbers. Proof. Assumption: There are only finitely many prime numbers. Let P = {p 1,..., p n } be the set of all prime numbers. Define m = p 1... p n + 1. Since m 2, it must have a prime factor. Let p be such a prime factor.
34 Indirect Proof: Example Theorem There are infinitely many prime numbers. Proof. Assumption: There are only finitely many prime numbers. Let P = {p 1,..., p n } be the set of all prime numbers. Define m = p 1... p n + 1. Since m 2, it must have a prime factor. Let p be such a prime factor. Since p is a prime number, p has to be in P.
35 Indirect Proof: Example Theorem There are infinitely many prime numbers. Proof. Assumption: There are only finitely many prime numbers. Let P = {p 1,..., p n } be the set of all prime numbers. Define m = p 1... p n + 1. Since m 2, it must have a prime factor. Let p be such a prime factor. Since p is a prime number, p has to be in P. The number m is not divisible without remainder by any of the numbers in P. Hence p is no factor of m. Contradiction
36 Questions Questions?
37 Contraposition
38 Contraposition (Proof by) Contraposition Prove If A, then B by proving If not B, then not A. German: (Beweis durch) Kontraposition Examples: Prove For all n N 0 : if n 2 is odd, then n is odd by proving For all n N 0, if n is even, then n 2 is even. Prove For all n N 0 : if n is not a square number, then n is irrational by proving For all n N 0 : if n is rational, then n is a square number.
39 Contraposition (Proof by) Contraposition Prove If A, then B by proving If not B, then not A. German: (Beweis durch) Kontraposition Examples: Prove For all n N 0 : if n 2 is odd, then n is odd by proving For all n N 0, if n is even, then n 2 is even. Prove For all n N 0 : if n is not a square number, then n is irrational by proving For all n N 0 : if n is rational, then n is a square number.
40 Mathematical Induction
41 Mathematical Induction Mathematical Induction Proof of a statement for all natural numbers n with n m basis: proof of the statement for n = m induction hypothesis (IH): suppose that statement is true for all k with m k n inductive step: proof of the statement for n + 1 using the induction hypothesis German: vollständige Induktion, Induktionsanfang, Induktionsvoraussetzung, Induktionsschritt
42 Mathematical Induction: Example I Theorem For all n N 0 with n 1: n k=1 (2k 1) = n2
43 Mathematical Induction: Example I Theorem For all n N 0 with n 1: n k=1 (2k 1) = n2 Proof. Mathematical induction over n: basis n = 1: 1 k=1 (2k 1) = 2 1 = 1 = 12
44 Mathematical Induction: Example I Theorem For all n N 0 with n 1: n k=1 (2k 1) = n2 Proof. Mathematical induction over n: basis n = 1: 1 k=1 (2k 1) = 2 1 = 1 = 12 IH: m k=1 (2k 1) = m2 for all 1 m n
45 Mathematical Induction: Example I Theorem For all n N 0 with n 1: n k=1 (2k 1) = n2 Proof. Mathematical induction over n: basis n = 1: 1 k=1 (2k 1) = 2 1 = 1 = 12 IH: m k=1 (2k 1) = m2 for all 1 m n inductive step n n + 1: n+1 k=1 (2k 1) = ( n k=1 (2k 1)) + 2(n + 1) 1 IH = n 2 + 2(n + 1) 1 = n 2 + 2n + 1 = (n + 1) 2
46 Mathematical Induction: Example II Theorem Every natural number n 2 can be written as a product of prime numbers, i. e. n = p 1 p 2... p m with prime numbers p 1,..., p m.
47 Mathematical Induction: Example II Theorem Every natural number n 2 can be written as a product of prime numbers, i. e. n = p 1 p 2... p m with prime numbers p 1,..., p m. Proof. Mathematical Induction over n: basis n = 2: trivially satisfied, since 2 is prime...
48 Mathematical Induction: Example II Theorem Every natural number n 2 can be written as a product of prime numbers, i. e. n = p 1 p 2... p m with prime numbers p 1,..., p m. Proof. Mathematical Induction over n: basis n = 2: trivially satisfied, since 2 is prime IH: Every natural number k with 2 k n IH: can be written as a product of prime numbers....
49 Mathematical Induction: Example II Theorem Every natural number n 2 can be written as a product of prime numbers, i. e. n = p 1 p 2... p m with prime numbers p 1,..., p m. Proof (continued). inductive step n n + 1: Case 1: n + 1 is a prime number trivial Case 2: n + 1 is not a prime number. There are natural numbers 2 q, r n with n + 1 = q r. Using IH shows that there are prime numbers q 1,..., q s with q = q 1... q s and r 1,..., r t with r = r 1... r t. Together this means n + 1 = q 1... q s r 1... r t.
50 Mathematical Induction: Example II Theorem Every natural number n 2 can be written as a product of prime numbers, i. e. n = p 1 p 2... p m with prime numbers p 1,..., p m. Proof (continued). inductive step n n + 1: Case 1: n + 1 is a prime number trivial Case 2: n + 1 is not a prime number. There are natural numbers 2 q, r n with n + 1 = q r. Using IH shows that there are prime numbers q 1,..., q s with q = q 1... q s and r 1,..., r t with r = r 1... r t. Together this means n + 1 = q 1... q s r 1... r t.
51 Structural Induction
52 Inductively Defined Sets: Examples Example (Natural Numbers) The set N 0 of natural numbers is inductively defined as follows: 0 is a natural number. If n is a natural number, then n + 1 is a natural number.
53 Inductively Defined Sets: Examples Example (Natural Numbers) The set N 0 of natural numbers is inductively defined as follows: 0 is a natural number. If n is a natural number, then n + 1 is a natural number. Example (Binary Tree) The set B of binary trees is inductively defined as follows: is a binary tree (a leaf) If L and R are binary trees, then L,, R is a binary tree (with inner node ). German: Binärbaum, Blatt, innerer Knoten
54 Inductively Defined Sets: Examples Example (Natural Numbers) The set N 0 of natural numbers is inductively defined as follows: 0 is a natural number. If n is a natural number, then n + 1 is a natural number. Example (Binary Tree) The set B of binary trees is inductively defined as follows: is a binary tree (a leaf) If L and R are binary trees, then L,, R is a binary tree (with inner node ). German: Binärbaum, Blatt, innerer Knoten Implicit statement: all elements of the set can be constructed Implicit statement: by finite application of these rules
55 Inductive Definition of a Set Inductive Definition A set M can be defined inductively by specifying basic elements that are contained in M construction rules of the form Given some elements of M, another element of M can be constructed like this. German: induktive Definition, Basiselemente, Konstruktionsregeln
56 Structural Induction Structural Induction Proof of statement for all elements of an inductively defined set basis: proof of the statement for the basic elements induction hypothesis (IH): suppose that statement is true for some elements M inductive step: proof of the statement for elements constructed by applying a construction rule to M (one inductive step for each construction rule) German: strukturelle Induktion, Induktionsanfang, Induktionsvoraussetzung, Induktionsschritt
57 Structural Induction: Example (1) Definition (Leaves of a Binary Tree) The number of leaves of a binary tree B, written leaves(b), is defined as follows: leaves( ) = 1 leaves( L,, R ) = leaves(l) + leaves(r) Definition (Inner Nodes of a Binary Tree) The number of inner nodes of a binary tree B, written inner(b), is defined as follows: inner( ) = 0 inner( L,, R ) = inner(l) + inner(r) + 1
58 Structural Induction: Example (2) Theorem For all binary trees B: inner(b) = leaves(b) 1.
59 Structural Induction: Example (2) Theorem For all binary trees B: inner(b) = leaves(b) 1. Proof. induction basis: inner( ) = 0 = 1 1 = leaves( ) 1 statement is true for base case...
60 Structural Induction: Example (3) Proof (continued). induction hypothesis: to prove that the statement is true for a composite tree L,, R, we may use that it is true for the subtrees L and R.
61 Structural Induction: Example (3) Proof (continued). induction hypothesis: to prove that the statement is true for a composite tree L,, R, we may use that it is true for the subtrees L and R. inductive step for B = L,, R : inner(b) = inner(l) + inner(r) + 1 IH = (leaves(l) 1) + (leaves(r) 1) + 1 = leaves(l) + leaves(r) 1 = leaves(b) 1
62 Questions Questions?
63 Summary
64 Summary A proof is based on axioms and previously proven statements. Individual proof steps must be obvious derivations. direct proof: sequence of derivations or rewriting indirect proof: refute the negated statement contraposition: prove A B as not B not A mathematical induction: prove statement for a starting point and show that it always carries over to next number structural induction: generalization of mathematical induction to arbitrary recursive structures
65 Preparation for the Next Lecture What s the secret of your long life? Simplify this advice! I am on a strict diet: If I don t drink beer to a meal, then I always eat fish. Whenever I have fish and beer with the same meal, I abstain from ice cream. When I eat ice cream or don t drink beer, then I never touch fish. Exercise from U. Schöning: Logik für Informatiker Picture courtesy of graur razvan ionut/freedigitalphotos.net
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