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Chapter One Notes The Foundations: Logic and Proofs Based on: Discrete Math & Its Applications - Kenneth Rosen CSC125 - Spring 2010 As MS document 1.1 - Propositional Logic Concepts:: Proposition Propositional variables Propositional logic (calculus) Logical connectives (operators) Compound propositions Logical operations Negation p not p Conjunction pq p and q Disjunction pq p or q (or both) Exclusive Or pq p xor q Conditional (implication) p q if p, then q (p implies q) Converse of p q is q p (p q) The converse does not necessarily follow. Example: If it rains then the streets are wet – true If the streets are wet then it rained – false Contrapositive of p q is q p The contrapositive does necessarily follow. If it rains then the streets are wet – true If the streets are not wet then it did not rain – true Inverse of p q is p q The inverse does not necessarily follow. Example: If it rains then the streets are wet – true If it does not rain then the streets are not wet – false 1 Biconditional p q if p, then q and if q, then p p implies q and q implies p p if and only if q p iff q Truth tables p p T F F T p q pq pq pq pq pq T T T T F T T T F F T T F F F T F T T T F F F F F F T T Bitwise And, Or, Xor And 0 0 0 0 1 0 1 0 0 1 1 1 Or 0 0 0 0 1 1 1 0 1 1 1 1 Xor 0 0 0 0 1 1 1 0 1 1 1 0 2 How many Boolean operations? x y f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f15 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Natural language, logic and ambiguity 1.2 - Propositional Equivalences Introduction:: From truth tables we can see that there are compound propositions that are equivalent to each other. In other words, whenever the first proposition is true, so is the second. And whenever the first is false, the second is also false. For example: p q p p q p q T T F T T T F F F F F T T T T F F T T T p q T T F F F T F T F F T T F T F T T F T F T F F T T T F T p q p q p q (p q) 3 Compare the above to Tables 3 and 4 on pp. 22 & 23 of Rosen. These are examples of important equivalences which we will study. Before we do, we need to learn some terminology. Tautology – a proposition that is always true Example: p p Contradiction (inconsistency) – a proposition that is always false Example: p p Contingency – a proposition that is neither a tautology nor a contradiction, i.e., it is sometimes true and sometimes false. Example: p q For verification of these examples, look at Table 1, p. 22 and Table 7, p. 10. Logical Equivalences:: p & q are logically equivalent if p ↔ q is a tautology In that case we can write: p ≡ q Truth tables can be used to establish whether or not p & q are logically equiv. DeMorgan's Laws – Table 2 {on quiz } Tables of important logical equivs. – Table 6 {on quiz} Table 7 – involving conditionals {on quiz} Table 8 – involving biconditionals {on quiz} Constructing New Logical Equivalences:: Given a small set of logical equivalences, we can derive new ones in a step by step process. Examples 6→8, pp. 26-27 illustrate how this is done. This process of derivation of new propositions from old is very useful and forms the basis of a lot of work in automated theorem proving. 1.3 - Predicates and Quantifiers Predicate logic predicate (propositional function) variable truth value of propositional function n-place predicate or n-ary predicate Quantifiers quantification - expresses extent to which a predicate is true over a range of 4 elements predicate calculus domain OR domain of discourse OR universe of discourse universal quantifier xP(x) counterexample – one single x for which P(x) is false existential quantifier xP(x) establishing example – one single x for which P(x) is true See Table 1, p. 34 uniqueness quantifier !xP(x) {or 1xP(x) } quantifiers with restricted domains precedence of quantifiers xP(x) Q(x) ≡ (xP(x)) Q(x) !≡ x(P(x)) Q(x)) binding variables variable within scope of quantifier is bound variable not within scope of quantifier is free Logical equivalences x(P(x) Q(x)) ≡xP(x) xQ(x), if same domain is used throughout Negating quantified expressions (DeMorgan's Laws) x P(x) ≡ xP(x) xP(x) ≡ xP(x) Quantifiers related to and xP(x) operates like a super Let domain be {1,2,3} xP(x) ≡ P(1) P(2) P(3) xP(x) operates like a super Let domain be {1,2,3} xP(x) ≡ P(1) P(2) P(3) That is why the negation laws above are called DeMorgan's Laws x P(x) ≡ (P(1) P(2) P(3)) ≡ P(1) P(2) P(3) ≡ xP(x) 5 And xP(x) ≡ (P(1) P(2) P(3)) ≡ P(1) P(2) P(3) ≡ xP(x) See Table 2, p. 41 Translating to & from English Let T(x) represent "x is tall" Everyone is tall:: x T(x) It is not true that everyone is tall:: x T(x), is the same as Someone is not tall:: Someone is tall:: x T(x) xT(x) It is not true that someone is tall:: x T(x), is the same as No one is tall:: xT(x), which is the same as Everyone is not tall {although this is ambiguous; see below} Ambiguity and subtlety of natural languages It is important to realize that one of the difficulties in translating back and forth from English (or any other natural language) to symbolic logic comes from the fact that some expressions are used in more than one way, logically speaking. In addition it is difficult to disambiguate because closely related expressions are used to express subtle nuances of meaning. Everyone is not tall is a good example. It can be taken to mean no one is tall, as we took it to mean above. xT(x) Or it can be taken to mean not everyone is tall x T(x) or someone is not tall (which is logically equivalent) x T(x) That meaning comes out clearly in the following interchange. Person A: Everyone is tall. Person B: No, everyone is not tall. Attempts to disambiguate by changing our wording lead to other problems. We could try to express xT(x) as everyone is short, but that changes the meaning. If we have a basketball team 6 where the heights range from 5'9" to 6'6", we can say everyone is not tall; but we could not say everyone is short. We could make up new words and say everyone is untall, but that is not likely to be welcomed by the general population of English speakers. It soon becomes pretty clear that there is no cleancut solution to the translation problem. In fact, that is one of the reasons for using symbolic logic – to eliminate the ambiguity inherent and widespread in natural language. Read the examples below to see how prevalent ambiguity and subtlety are in our use of "the king's English." Everyone goes No, everyone does not Everyone stays away Everyone does not go Everyone is going Everyone is not going Everyone is so not going Someone goes No, someone does not go Some do not go Some stay away Everyone thinks Everyone is unthinking Everyone does not think Everyone is thoughtful Everyone is not thoughtful Everyone is thoughtless Someone thinks Someone does not think Some is thoughtless Someone is thoughtless Everyone is caring Everyone is uncaring Everyone is not caring Not everyone is caring Everyone is careless Everyone is careful Everyone is not careful 7 Not everyone is careful Not everyone is careless Quick Summary Realizing that there will still be ambiguous cases, here are some rules of thumb x D(x) – everyone does x D(x) – everyone does not; i.e., for everyone does not is true x D(x) – not everyone does; i.e., for some does not is true x D(x) – someone does x D(x) – someone does not; there is someone who does not x D(x) – no one does Equivalences x D(x) ≡ x D(x) No one does is the same as everyone does not x D(x) ≡ xD(x) Not everyone does is the same as someone does not 1.4 - Nested Quantifiers Think of quantification as loops Example 2 Order of Quantifiers Does not matter if all of same type Example 3 Very important when of mixed type Examples 4 & 5 Let domain be {1,2,3} x y P(x,y) ≡ yx P(x,y) ≡ P(1,1) P(1,2) P(1,3) P(2,1) P(2,2) P(2,3) P(3,1) P(3,2) P(3,3) x y P(x,y) ≡ yx P(x,y) ≡ P(1,1) P(1,2) P(1,3) P(2,1) P(2,2) P(2,3) P(3,1) P(3,2) P(3,3) 8 x y P(x,y) ≡ (P(1,1) P(1,2) P(1,3)) (P(2,1) P(2,2) P(2,3)) (P(3,1) P(3,2) P(3,3)) In other words, for x = 1, y P(x,y) for x = 2, y P(x,y) for x = 3, y P(x,y) x y P(x,y) ≡ (P(1,1) P(1,2) P(1,3)) (P(2,1) P(2,2) P(2,3)) (P(3,1) P(3,2) P(3,3)) In other words, either for x = 1, y P(x,y) for x = 2, y P(x,y) for x = 3, y P(x,y) Similarly, y x P(x,y) ≡ (P(1,1) P(2,1) P(3,1)) (P(1,2) P(2,2) P(3,2)) (P(1,3) P(2,3) P(3,3)) In other words, either for y = 1, x P(x,y) for y = 2, x P(x,y) for y = 3, x P(x,y) Therefore, x y P(x,y) ≠ y x P(x,y) ≠ x y P(x,y) See Table 1, p. 53 Translating mathematical statements Examples 6 & 7 Translating nested quantifiers into English Examples 9 & 10 Translating English into logical expressions Examples 11 & 13 Negating nested quantifiers Example 14 x y (xy = 1) ≡ x y (xy = 1) ≡ x y (xy = 1) ≡ x y (xy ≠ 1) 9 Example 15 w a f (P(w,f) Q(f,a)) w a f (P(w,f) Q(f,a)) w a f (P(w,f) Q(f,a)) w a f (P(w,f) Q(f,a)) w a f (P(w,f) Q(f,a)) Doubly quantified expression examples: Let W(x,y) be the statement “student x watched Eagles game y,” where the domain for x consists of all students in this class and the domain for y consists of all Philadelphia Eagles games last season. Express each proposition in English. 1. xyW(x,y) A student from this class watched an Eagles game last season. 2. xyW(x,y) No student from this class watched an Eagles game last season. Here are equivalent forms, by DeMorgan's Law ≡ xyW(x,y) ≡ xyW(x,y) Note: Equivalent to #18 {For all students and all Eagles games, it not true that any student watched any game} 3. x yW(x,y) A student from this class watched every Eagles game last season. 4. x yW(x,y) No student from this class watched every Eagles game last season. Here are equivalent forms, by DeMorgan's Law ≡ xyW(x,y) ≡ xyW(x,y) Note: Equivalent to #16 5. yx W(x,y) Every Eagles game last season was watched by a student in this class. 6. yx W(x,y) 10 Not every Eagles game last season was watched by a student in this class. Here are equivalent forms, by DeMorgan's Law ≡ yxW(x,y) ≡ yxW(x,y) Note: Equivalent to #17 7. xyW(x,y) Every student from this class watched an Eagles game last season. 8. xyW(x,y) Not every student from this class watched an Eagles game last season. Here are equivalent forms, by DeMorgan's Law ≡ xyW(x,y) ≡ xyW(x,y) Note: Equivalent to #14 {A student from this class did not watch any Eagles games last season; OR: It is not true that every student watched an Eagles game last season} 9. yxW(x,y) There is an Eagles game last season that every student watched. 10. yxW(x,y) None of the Eagles games last season were watched by every student. Here are equivalent forms, by DeMorgan's Law ≡ yxW(x,y) ≡ yxW(x,y) Note: Equivalent to #15 11. x yW(x,y) Every student from this class watched every Eagles game last season. 12. x yW(x,y) Not every student from this class watched every Eagles game last season. Here are equivalent forms, by DeMorgan's Law ≡ xyW(x,y) ≡ xyW(x,y) Note: Equivalent to #13 11 13. xyW(x,y) A student from this class who missed an Eagles game last season. Here are equivalent forms, by DeMorgan's Law ≡ xyW(x,y) ≡ xyW(x,y) Note: Equivalent to #12 {There is a student from this class who did not watch every Eagles game last season; OR: Not every student from this class watched every Eagles game last season.} 14. x yW(x,y) A student from this class did not watch any Eagles games last season. Here are equivalent forms, by DeMorgan's Law ≡ xyW(x,y) ≡ xyW(x,y) Note: Equivalent to #8 {Not every student from this class watched an Eagles game last season; OR: It is not true that every student watched an Eagles game last season} 15. y x W(x,y) No Eagles game was watched by every student in this class. Here are equivalent forms, by DeMorgan's Law ≡ yxW(x,y) ≡ yxW(x,y) Note: Equivalent to #10 16. xyW(x,y) Every student from this class missed an Eagles game last season. Here are equivalent forms, by DeMorgan's Law ≡ xyW(x,y) ≡ xyW(x,y) Note: Equivalent to #4 {No student watched every Eagles game last season; OR: For every student, there is an Eagles game they did not watch} 17. yxW(x,y) There is an Eagles game that no one watched. 12 Here are equivalent forms, by DeMorgan's Law ≡ yxW(x,y) ≡ yxW(x,y) Note: Equivalent to #6 18. xyW(x,y) No student watched any Eagles game last season. Here are equivalent forms, by DeMorgan's Law ≡ xyW(x,y) ≡ xyW(x,y) Note: Equivalent to #2 Every student in this class missed watching every Eagles game last season. 1.5 Rules of Inference Valid arguments in propositional logic p→q p . q Definitions: argument - sequence of propositions premises - all the propositions in the sequence except the final one conclusion - the final proposition in the sequence argument form - sequence involving propositional variables valid - an argument form is valid if conclusion follows from premises, no matter what values the variables have. conclusion follows from premises - conclusion is true if all premises are true Rules of inference for propositional logic Study Table 1 Examples 1, 3, 4, 5 Using rules of inference to build arguments Example 6 Hypotheses: H1: p q H2: r p H3: r s H4: s t Argument: 13 1: 2: 3: 4: 5: 6: 7: 8: p q p rp r r s s st t Example 7 Hypotheses: H1: p q H2: p r H3: r s Argument: 1: p q 2: q p 3: p r 4: q r 5: r s 6: q s H1 Simp (1) H2 MT (2,3) H3 MP (4,5) H4 MP(6,7) H1 Contrapos (1) H2 HypSyl (2,3) H3 HypSyl (4,5) Resolution ((p q) (~p r)) → (q r) Clause - disjunction of variables or their negations Resolvent - conclusion of application of resolution rule Fallacies Affirming the conclusion Denying the hypothesis Rules of inference for quantified statements Study Table 2 Examples 12 & 13 Example 12 Premises: P1: x (D(x) C(x)) P2: D(Marla) Argument: 14 1: 2: 3: 4: x (D(x) C(x)) D(Marla) C(Marla) D(Marla) C(Marla) P1 UI (1) P2 MP (2,3) Example 13 Hypotheses: P1: x (C(x) B(x)) P2: x (C(x) P(x)) Argument: 1: x (C(x) B(x)) 2: C(a) B(a) 3: C(a) 4: x (C(x) P(x)) 5: C(a) P(a) 6: P(a) 7: B(a) 8: P(a) B(a) 9: x (P(x) B(x)) P1 EI (1) Simp (2) P2 UI (4) MP (3,5) Simp (2) Conj (6,7) EG (8) Combining rules of inference for propositions and quantified statements Universal modus ponens Universal modus tollens Summary of Valid Rules of Inference (Valid Argument Forms) Modus Ponens p p→q q Modus Tollens q p→q p Hypothetical Syllogism p→q q→r p→r 15 Disjunctive Syllogism pq p . q Addition p . pq Simplification pq p Conjunction p q pq Resolution pq p r qr Universal Instantiation x P(x) P(c) for any element c Universal Generalization P(c) for an arbitrary c x P(x) Existential Instantiation x P(x) P(c) for some element c Existential Generalization P(c) for some element c x P(x) Universal Modus Ponens x (P(x) → Q(x)) 16 P(a), where a is any element in the domain Q(a) Universal Modus Tollens x (P(x) → Q(x)) Q(a), where a is any element in the domain P(a) Summary of Fallacies (Invalid Argument Forms) Affirming the conclusion q p→q p Denying the hypothesis p p→q q Unwarranted addition p . pq Unwarranted simplification pq p Unwarranted Universal Generalization P(c) for a particular c x P(x) Existential Instantiation x P(x) P(c) for a particular element c Universal Affirming the conclusion x (P(x) → Q(x)) Q(a), where a is any element in the domain P(a) Universal Denying the Hypothesis 17 x (P(x) → Q(x)) P(a), where a is any element in the domain Q(a) 1.6 Introduction to Proofs Terminology Theorem – statement that can be proven to be true. Propositions, facts, results – less important true statements. Proof – valid argument that establishes truth of a theorem. Axiom, postulate – a statement taken to be true, though not proven (e.g., Appendix 1) Lemma – intermediate result proven, then used in proof of a theorem. Corollary – a theorem that can be established directly as the result of another theorem Conjecture – statement thought to be true, but not proven to be, E.g., Collatz Conjecture, twin primes conjecture, Goldbach's conjecture Definitions Even integer n is even if integer k, s.t., n = 2k. Odd integer n is odd if integer k, s.t., n = 2k+1. Rational number real number r is rational if integers p,q, q ≠ 0, s.t., r = p/q. Irrational number real number r is irrational if it is not rational. FOREWORD: QUANTIFIERS, UI, UG Quantifiers – yes or no? :: p. 76 – use of UI followed by UG requires choice of arbitrary element. – this is not the same as choosing any specific element. Fallacious proof – To Prove: All integers are even. – Proof: UI: Choose any element of the domain, say 44. Proof step: 44 = 2*22; therefore 44 is even. 18 UG: Since true of element chosen, it is true for all elements; therefore all integers are even! – Analysis: The UI/UG combination requires choice of arbitrary element; choosing a specific element, even if chosen at random, does not satisfy that requirement. A correct proof using UI/UG: Prove: The sum of two odd integers is even. Proof: UI: Let: a = 2k + 1, where k is any integer b = 2 l + 1, where l is any integer Comment: Since we are using the definition of odd integer and k & l are any arbitrary integers, this sets the stage allowing us to use UG later in the proof. Then: Then: a + b = 2k + 1 + 2 Let: m=k+ l + 1 = 2(k + l +1) l +1 Then: a + b = 2m, which is even by definition. UG: For all odd integers, a & b, a + b is even. Comment: We can apply UG here because a and b were chosen to be arbitrary odd integers. By using a representation which adheres strictly to the definition of odd integer, our representation fits all odd integers. And therefore, anything we derive as true of a and b is also true of all odd integers. Mathematicians shorthand: Because the above proof process is used so often, mathematicians usually omit mentioning all of the things that are obvious by virtue of their frequent use. Thus, the above proof can be stated much more succinctly, as below. 1. Prove: The sum of two odd integers is even. 19 Proof: Let: a = 2k + 1 b = 2l + 1 Then: a + b = 2k + 1 + 2 l + 1 = 2(k + l +1) EVEN PROOF METHODS Direct Proof Direct proofs start with a proposition (mathematical fact) known to be true and proceed, step by step, to derive the conclusion directly. The above proof is a good example of direct proof. It starts with the definition of odd integer and then uses laws of arithmetic to derive other facts, leading eventually to the desired conclusion. Indirect Proof – Proof by Contraposition Proof by contraposition can be used when the proposition to be proved is of the form: p q. Recall that the Law of Contraposition states that p q is equivalent to q p. Sometimes the latter is easier to prove than the former. A proof by contraposition proceeds by showing that if q is true, then p must also be true. Here is Example #3 from Rosen, p. 78. Prove: For integer n: 3n + 2 is odd n is odd. Prove contrapositive: For integer n: n is even 3n + 2 is even. Proof: Let: n = 2k (definition of even integer) Then: 3n+2 = 6k+2 = 2(3k+1) EVEN Indirect Proof - Proof by Contradiction Proof by contradiction is based upon the axiom that every proposition must be either True or False (Rosen, p. 1). Therefore, if we show that a proposition cannot be false, then it must be true. Here is Example #10 from Rosen, p. 80. Prove: 2 is irrational. Proof by contradiction: 20 Assume: 2 is rational. Then: There exist integers p & q, q ≠ 0, s.t., 2 = p/q. WOLG, we assume p & q have no common factors since, any common factors can be eliminated by expressing p/q in lowest terms. By the laws of arithmetic: (2)2 = (p/q)2 2 = p2/q2 2q2 = p2 Thus: p2 is even By Exercise 16 (to be done later): p is even By definition: p = 2k, for some integer k. Therefore: 2q2 = (2k)2= 4k2 q2 = 2k2 And so: q2 is even; and q is even. q = 2l for some l. But we now have a contradiction: Since p = 2k and q = 2l, they have a common factor, namely 2. Therefore: our assumption that 2 is rational is false; and we have proven that: 2 is irrational. This proof can also be written in an abbreviated form, as below. Prove: 2 is irrational. Proof by contradiction: Assume: 2 is rational. Then: There exist integers p & q, q ≠ 0, s.t., 2 = p/q, with no common factors for p & q. By the laws of arithmetic: (2)2 = (p/q)2 2 = p2/q2 2q2 = p2 Then: p is even, i.e., p = 2k, for some integer k. 21 2q2 = (2k)2= 4k2 q2 = 2k2 Therefore: q is even, i.e., q = 2 l, for some integer l. p and q have a common factor, namely 2. Therefore: 2 is irrational. Mistakes in proofs: Division by zero Example #15, p. 83 Logical fallacies Affirming the conclusion, Example #16, pp. 83-84 Denying the hypothesis, Example #17, p. 84 Begging the question (circular reasoning), Example #18, p. 84 1.7 Proof Methods and Strategies PROOF METHODS Exhaustive Proof Examine all of a relatively small number of cases; show true to all. Prove: The only consecutive perfect powers less than 100 are 8 & 9. {Example 2, p. 87} Exhaustive proof: List all powers < 100: N N2 N3 N4 1 1 1 1 2 4 8 16 3 9 27 81 4 16 64 5 25 6 36 7 49 8 64 9 81 N5 1 32 N6 1 64 N7 1 Examine this list. The only consecutive integers in that list are 8 & 9. Proof by Cases: 22 Use when no obvious way to prove for all values, but can individual cases readily provable. Caveat: Make sure all cases are considered. Prove: n2 ≥ n, for all integers. {Example 3, p. 88} Proof, by cases: Integers naturally fall into 3 categories: negative, zero, and positive. Case 1: n < 0. Then n = -k, for some k ≥ 1. And n2 = (-k)2 = k2 ≥ 1 > 0 > n. So n2 ≥ n. Case 2: n = 0. Then n2 = 02 = 0 = n. So n2 ≥ n. Case 3: n > 0. Then n ≥ 1. And n*n ≥ 1*n. So n2 ≥ n. Exhaustive Proof Errors 1. Draw a conclusion for some, but not all instances. No matter how many instances are considered, no conclusion is warranted unless all instances are considered. Mathematicians sometimes call this the "Law of Small Numbers". Example 8, p. 90. 2. Attempt proof by cases, but omit some cases. Example 9, p. 90. Existence Proofs - Constructive Provide an element a such that P(a) is true. Show: There is a positive integer that can be written as the sum of cubes of positive integers in two different ways. {Example 10, p. 91} Proof: 1729 = 103 + 93 = 123 + 13 23 Existence Proofs - Nonconstructive Prove that xP(x) is true. Often this is done via proof by contradiction. I.e., derive a contradiction from xP(x). Show: There exist irrational numbers x and y such that xy is rational. {Example 11, p. 91} Proof: We know 2 is irrational. Consider 22. Either it is rational or irrational. Case 1: 22 is rational. Then let x = 2 and y = 2. Case 2: 22 is irrational. Then let x = 22 and y = 2. So xy = (22)2= 22*2= 22= 2. So either in case 1 or in case 2 we have xy rational, though x and y are both irrational. Although we do not know which case meets the specification, we know that one must. Exercise: Use the calculator to determine which of the two cases meets the specification. Uniqueness Proofs There are two parts to uniqueness proofs: Existence: Show element with desired property exists. Uniqueness: Show x ≠ y, x does not have desired property; or, show if x and y both have desired property, then x = y. Show: For odd integers a & b, a ≠ b, ! integer c, s.t. |a – c| =|b – c|. {Exercise 15, p. 103} Proof: Existence: We show existence constructively. WOLG, assume a > b. If |a – c| = |b – c|, then either a – c = b – c or a – c = c – b. But a – c = b – c a = b, so a – c = c – b. Reworking the equation: a–c=c–b a + b = 2c c = (a + b)/2 24 Finally, we know c is an integer since a+b, the sum of two odd integers, is even. Uniqueness: Solutions of linear equations are unique. Forward and Backward Reasoning: Start with what you want to prove and working backward derive the premises. If that succeeds, we construct the proof by reversing the reasoning steps. Caveat: This only works if at each step we produce statements that are equivalent, and therefore reversible. Examples 14 & 15, p. 95 illustrate this process. Adapting existing proofs: Often a known proof can be adapted to prove a similar (or parallel) conclusion. This is illustrated in Example 16, p. 96. Also, in 1.6 homework answers, the proof of Exercise 1 was adapted in the proofs of Exercises 2 and 6. Proof strategy in action On p. 97 Rosen presents many strategies for generating hypotheses and conjectures and for seeking proofs or counterexamples. Among others, this includes "trying it on for size," the method applied to Exercise 7 of 1.6. In general, before we attempt a proof we should look at enough examples to convince us that what we are attempting to prove is indeed true. Sometime this may surprise us (see immediately below). Look for counterexamples: Sometimes, in the process of looking at examples, we will be surprised to find an example for which the proposed proposition does not hold. This will, of course, be a counterexample and will enable us to disprove by counterexample. Exercise 38 of 1.6 illustrates this. Tilings: Tiling problems can be fiendishly difficult, but sometimes a single observation can lead to a simple and easy solution. Examples 18-22, pp. 98-100 illustrate. Role of Open Problems 25 There are some fascinating stories in the history of mathematics involving attempts to solve open problems. Among these are: Fermat's Last Theorem and the Collatz Conjecture. One of keen interest to computer scientists is this question: is P = NP, where P is the set of problems that can be solved in polynomial time and NP is the set of problems which can be solved in non-deterministic polynomial time. Most computer scientists and mathematicians think that P ≠ NP, and that problems in the class NP require exponential (or worse) time. We will look at some exponential and factorial algorithms later in this course. 26