Q250: Math and Logic Unit 1 Review Problems for 9/7/12
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1 2. i. p ( q p) p q q q p p ( q p) T T F F F T F T T T F T F F T F F T F T Q250: Math and Logic Unit 1 Review Problems for 9/7/12 ii. (p q) ( r q) p q r r p q r q (p q) ( r q) T T T F T T T T T F T T T T T F T F F F T T F F T F T F F T T F F T F F T F T F T F F F T F F F T F F F T F T F iii. (p r q) (q p) p q r p p r q q p (p r q) (q p) T T T F T F F T T F F T F F T F T F T T T T F F F T T T F T T T T T T F T F T T T T F F T T T F F F F F T F F T iv. (p r) ( p r) p r p r p r p r (p r) ( p r) T T F F T T T T F F T F T T F T T F F F F F F T T F T T v. ((p r) q) (p q) p q r q p r (p r) q p q ((p r) q) (p q) T T T F T T T T T T F F F F T F T F T T T T F F T F F T F T F F F T T F F F F T F T F F F F F T F F T T F T F F F F F T F T F F 1
2 4. Determine whether each of the following propositions is a tautology, a contradiction, or neither (a contingency). If it is a contingency, provide two interpretations that demonstrate this is the case. i. (p q) ( r q): Neither because entries 1-3 and 7 are true, but 4-6 and 8 are false. p q r r p q r q (p q) ( r q) T T T F T T T T T F T T T T T F T F F F T T F F T F T F F T T F F T F F T F T F T F F F T F F F T F F F T F T F ii. (p r) (p r): Tautology because all entries are true. p r r p r p r (p r) (p r) T T F T F T T F T F T T F T F F T T F F T F T T iii. ((p q) (r q)) (p q r): Neither because entries 3 and 4 are true, but 1, 2, and 5-8 are false. p q r q p q r q (p q) (r q) p q r ((p q) (r q)) (p q r) T T T F F T T F F T T F F F T T F F T F T T T T T T T T F F T T F F F T F T T F F T T F F F T F F F T T F F F F T T F T T F F F F F T F F T F F iv. ((a b) c) ((a b) c): Typo, so disregard, but here is how to do the problem as written anyway. Neither because entries 1, 3, 5, 7, and 8 are true, but 2, 4, and 6 are false. a b c c a b (a b) c ((a b) c) ((a b) c) ((a b) c) ((a b) c) T T T F T F F T T T F T T T T F T F T F T F F T T F F T T T T F F T T F T F F T F T F T T T T F F F T F F F F T F F F T F F F T 2
3 v. (q p) ( s q) (p s) q: Contradiction because all entries are false. p q s q s q p s q p s (p s) (q p) ( s q) (p s) q T T T F F T T T F F T T F F T T F F T F T F T T F T T T F F T F F T T T T F T F F T T F F F T T F F F T F F T F F T F F F F T T F T T T F F F F F T T T T T F F 6. For each of the following, prove that the conclusion follows logically from the premises. i. s: p q, r (p q), r p Prove: q r Conclusion does not follow. ii. s: ( q r) p, r (p q), (r p) q Conclusion does not follow. Prove: q iii. s: p q, q r 1. p q 2. q 3. q r 4. r 5. p q r iv. s: p q, p q, p Typo: Disregard Prove: p q r Prove: p Disjunction Elimination: 2,3 Disjunction Introduction: 4 v. s: p ( q r), r (q p), q Prove: r q 1. r (q p) 2. r 3. q p 4. p 5. p ( q r) 6. q r 7. q 8. q 9. r Modus Tollens: 1,2 Conjunction Elimination: 3 Disjunction Elimination: 4,5 Conjunction Elimination: 6 Contradiction: r q Conjunction Introduction: 8,9 vi. s: p (n s), (p r) (q m) 1. p (n s) 2. p 3. p r 4. (p r) (q m) 5. q m 6. q 7. p q Prove: p q Disjunction Introduction: 2 Modus Tollens: 3,4 Conjunction Elimination: 5 Conjunction Introduction: 2,6 3
4 vii. s: s t, p q, (p r) s 1. p q 2. p 3. p r 4. (p r) s 5. s 6. s t 7. t Prove: t Disjunction Introduction: 2 Modus Ponens: 3,4 Modus Ponens: 5,6 viii. s: (m n) (q m), m q, (m q) (m n) Prove: n m 1. m q 2. (m q) (m n) 3. m n 4. (m n) (q m) 5. q m 6. m 7. q 8. q 9. m Modus Ponens: 1,2 Modus Ponens: 3,4 Modus Tollens: 5,6 Disjunction Elimination: 1,6 Contradiction: n m Disjunction Introduction: 9 ix. s: ( w y) (z a), w x, x, a 1. x 2. w x 3. w 4. w y 5. ( w y) (z a) 6. z a 7. a 8. z Prove: z Modus Tollens: 1,2 Disjunction Introduction: 3 Modus Ponens: 4,5 Disjunction Elimination: 6,7 x. s: D E, E F Prove: (D A) (F E) 1. D E 2. E F 3. D A 4. D 5. E 6. F 7. F E 8. (D A) (F E) Conjunction Elimination: 3 Modus Ponens: 1,4 Modus Ponens: 2,5 Conjunction Introduction: 5,6 Implication Introduction: Transform each informal argument into predicate logic. Then give a formal proof. i. All monkeys who play chess drive well. Therefore, if there is a monkey that plays chess, then there is a monkey that can drive well. Let Chess(x) mean that x plays chess, and let Drives(x) mean that x drives well, where the domain of discourse of x is monkeys. 1. x Chess(x) Drives(x) 2. Chess(c) Drives(c) 3. x Chess(x) x Drives(x) Existential Introduction: 2 4
5 ii. All superheroes who wear a cape get stuck in sliding doors. Therefore, if all superheroes wear a cape, then at least some superheroes get stuck in sliding doors. Let Cape(x) mean that x wears a cape, and let Stuck(x) mean that x gets stuck in sliding doors, where the domain of discourse of x is superheroes. 1. x Cape(x) Stuck(x) 2. Cape(c) Stuck(c) 3. x Cape(x) x Stuck(x) Universal Intro and Existential Intro: 2 iii. Everyone who is sane can do logic. No lunatics are fit to serve on a jury. None of your friends can do logic. Therefore, if a person is fit to serve on a jury, then he/she is not your friend. Let Sane(x) mean that x is sane, let Logic(x) mean that x can do logic, let Friend(x) mean that x is your friend, and let Jury(x) mean that x is fit to serve on a jury, where the domain of discourse of x is people. 1. x Sane(x) Logic(x) 2. x Sane(x) Jury(x) 3. x Friend(x) Logic(x) 4. Sane(c) Logic(c) 5. Sane(c) Jury(c) 6. Friend(c) Logic(c) 7. Jury(c) 8. Sane(c) 9. Logic(c) 10. Friend(c) 11. Jury(c) Friend(c) 12. x Jury(x) Friend(x) Universal Elimination: 2 Universal Elimination: 3 Modus Tollens: 5,7 Modus Ponens: 4,8 Modus Tollens: 6,9 Implication Introduction: 7-10 Universal Introduction: 7-11 iv. Babies are illogical. Nobody is despised who can manage a crocodile. Illogical persons are despised. Therefore babies cannot manage crocodiles. Let Baby(x) mean that x is a baby, let Logical(x) mean that x is Logical, let Despised(x) mean that x is despised, and let Croc(x) mean that x can manage crocodiles, where the domain of discourse of x is people. 1. x Baby(x) Logical(x) 2. x Croc(x) Despised(x) 3. x Logical(x) Despised(x) 4. Baby(c) Logical(c) 5. Croc(c) Despised(c) Universal Elimination: 2 6. Logical(c) Despised(c) Universal Elimination: 3 7. Baby(c) 8. Logical(c) Modus Ponens: 4,7 9. Despised(c) Modus Ponens: 6,8 10. Croc(c) Modus Tollens: 5,9 11. Baby(c) Croc(c) Implication Introduction: x Baby(x) Croc(x) Universal Introduction:
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