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Work these exercises to see how well you understand this material.

Solutions

  1. 3.5.4.1. Answer.


  2. 3. Answer:
    1. Direct proof:
      1. d → (a ∨ c)
      2. d
      3. a ∨ c 
      4. a → b 
      5. ¬a ∨ b
      6. c → b 
      7. ¬c ∨ b
      8. (¬a ∨ b) ∧ (¬c ∨ b)
      9. (¬a ∧ ¬c) ∨ b
      10. ¬(a ∨ c) ∨ b 
      11. b □
    2. Indirect proof:
      1. ¬b Negated conclusion
      2. a → b Premise
      3. ¬a Indirect Reasoning (1), (2)
      4. c → b Premise
      5. ¬c Indirect Reasoning (1), (4)
      6. (¬a ∧ ¬c) Conjunctive (3), (5)
      7. ¬(a ∨ c) DeMorgan's law (6)
      8. d → (a ∨ c) Premise
      9. ¬d Indirect Reasoning (7), (8)
      10. d Premise
      11. ⊬ (9), (10) □
    3. Direct proof:
      1. (p → q) ∧ (r → s)
      2. p → q
      3. (p → t) ∧ (s → u)
      4. q → t 
      5. p → t 
      6. r → s 
      7. s → u 
      8. r → u 
      9. p → r
      10. p → u
      11. p → (t ∧ u) Use (x → y) ∧ (x → z) ⇔ x → (y ∧ z)
      12. ¬(t ∧ u) → ¬p
      13. ¬(t ∧ u)
      14. ¬p □
    4. Indirect proof:
      1. p
      2. p → q
      3. q
      4. q → t
      5. t
      6. ¬(t ∧ u)
      7. ¬t ∨ ¬u 
      8. ¬u
      9. s → u 
      10. ¬s
      11. r → s 
      12. ¬r 
      13. p → r 
      14. r
      15. 0 □
    5. Direct proof:
      1. ¬s ∨ p Premise
      2. s Added premise (conditional conclusion)
      3. ¬(¬s) Involution (2)
      4. p Disjunctive simplification (1), (3)
      5. p → (q → r) Premise
      6. q → r Detachment (4), (5)
      7. q Premise
      8. r Detachment (6), (7) □
    6. Indirect proof:
      1. ¬(s → r) Negated conclusion
      2. ¬(¬s ∨ r) Conditional equivalence (I)
      3. s ∧ ¬r DeMorgan (2)
      4. s Conjunctive simplification (3)
      5. ¬s ∨ p Premise
      6. s → p Conditional equivalence (5)
      7. p Detachment (4), (6)
      8. p → (q → r) Premise
      9. q → r Detachment (7), (8)
      10. q Premise
      11. r Detachment (9), (10)
      12. ¬r Conjunctive simplification (3)
      13. 0 Conjunction (11), (12) □
    7. Direct proof:
      1. p → q
      2. q → r
      3. p → r
      4. p ∨ r
      5. ¬p ∨ r
      6. (p ∨ r) ∧ (¬p ∨ r)
      7. (p ∧ ¬p) ∨ r
      8. 0 ∨ r
      9. r □
    8. Indirect proof:
      1. ¬r Negated conclusion
      2. p ∨ r Premise
      3. p (1), (2)
      4. p → q Premise
      5. q Detachment (3), (4)
      6. q → r Premise
      7. r Detachment (5), (6)
      8. 0 (1), (7) □

  3. Answer:
    1. Let W stand for "Wages will increase", I stand for "there will be in- flation", and C stand for "cost of living will increase". Therefore the argument is: W → I , ¬I → ¬C, W ⇒ C . The argument is invalid. The easiest way to see this is through a truth table, which has one case, the seventh, that this false. Let x be the conjunction of all premises.

    2. Let r stand for "the races are fixed", c stand for "casinos are crooked", t stand for "the tourist trade will decline", and p stand for "the police will be happy". Therefore, the argument is:

      (r ∨ c) → t, t → p, ¬p → ¬r.

      The argument is valid. Proof:
      1. t → p Premise
      2. ¬p Premise
      3. ¬t Indirect Reasoning (1), (2)
      4. (r ∨ c) → t Premise
      5. ¬(r ∨ c) Indirect Reasoning (3), (4)
      6. (¬r) ∧ (¬c) DeMorgan (5)
      7. ¬r Conjunction simplification (6) □

  4. Answer: p1→ pkand pk→ pk+1 implies p1→ pk+1. It takes two steps to get to p1→ pk+1 from p1→ pkThis means it takes 2(100 − 1) steps to get to p1→ p100 (subtract 1 because p1→ p2 is stated as a premise). A final step is needed to apply detachment to imply p100