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V ALID A RGUMENTS ? If God does not exist, then it is not G (P A) - PowerPoint PPT Presentation

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V ALID A RGUMENTS ? If God does not exist, then it is not G (P A) true that if I pray, then my prayers P will be answered. I dont pray. G Therefore, there is a God. If it is true that if I pray then my (P A) G

  1. V ALID A RGUMENTS ? If God does not exist, then it is not ¬G → ¬(P → A) true that if I pray, then my prayers ¬P will be answered. I don’t pray. G Therefore, there is a God. If it is true that if I pray then my (P → A) → G prayers will be answered, then ¬P there is a God. But I don’t pray. G Therefore, there is a God. Friday, September 24, 2010

  2. P ROOFS W ITH C ONDITIONALS 3 Friday, 24 September Friday, September 24, 2010

  3. F ORMAL P ROOF R ULES ↔ Introduction: from a proof from P to Q and a proof from Q to P , we can infer P ↔ Q. 1. P … j. Q k. Q … m. P k. P ↔ Q ↔ Intro: 1-j, k-m Friday, September 24, 2010

  4. P ARADOXES OF M ATERIAL I MPLICATION We did this by showing that: Last time we showed: ¬P ∨ Q ¬P Q and P → Q P → Q P → Q 1. ¬P 1. Q 2. P for → Intro 2. P for → Intro 3. ⊥ ⊥ Intro 1,2 3. Q Reit 1 4. Q ⊥ Elim 3 5. P → Q → Intro 2-3 5. P → Q → Intro 2-4 Friday, September 24, 2010

  5. T HE O THER D IRECTION 1. P → Q Example: 2. ¬(¬P ∨ Q) for ¬Intro P → Q ¬P ∨ Q 3. ¬P for ¬Intro 4. ¬P ∨ Q vIntro 3 5. ⊥ ⊥ Intro 2,4 6. P ¬Intro 3-5 7. Q → Elim 1,6 8. ¬P ∨ Q vIntro 7 ⊥ ⊥ Intro ¬Intro 2- ¬P ∨ Q Friday, September 24, 2010

  6. T HE O THER D IRECTION 1. P → Q Example: 2. ¬(¬P ∨ Q) for ¬Intro P → Q ¬P ∨ Q 3. ¬P for ¬Intro 4. ¬P ∨ Q vIntro 3 5. ⊥ ⊥ Intro 2,4 6. P ¬Intro 3-5 7. Q → Elim 1,6 8. ¬P ∨ Q vIntro 7 9. ⊥ ⊥ Intro 2,8 ¬Intro 2-9 10. ¬P ∨ Q Friday, September 24, 2010

  7. P ROVING B ICONDITIONALS We have now proved: P → Q ¬P ∨ Q and P → Q ¬P ∨ Q Therefore we could prove: (P → Q) ↔ (¬P ∨ Q) Friday, September 24, 2010

  8. P ROVING B ICONDITIONALS 1. P → Q for ↔ Intro (P → Q) ↔ (¬P ∨ Q) ¬P ∨ Q ¬P ∨ Q for ↔ Intro P → Q (P → Q) ↔ (¬P ∨ Q) ↔ Intro Friday, September 24, 2010

  9. 1. P → Q ¬P ∨ Q ¬P ∨ Q P → Q (P → Q) ↔ (¬P ∨ Q) ↔ Intro 1-10, 11-18 Friday, September 24, 2010

  10. B ICONDITIONALS AND E QUIVALENCE We have now proved: (P → Q) ↔ (¬P ∨ Q) If a biconditional is a logical truth then the two parts are logically equivalent: (P → Q) ⇔ (¬P ∨ Q) Friday, September 24, 2010

  11. C HAINS OF E QUIVALENCE (P → Q) ⇔ (¬P ∨ Q) When a conditional is true By DeMorgan’s (¬P ∨ Q) ⇔ ¬(P ∧ ¬Q) Therefore (P → Q) ⇔ ¬(P ∧ ¬Q) and so ¬(P → Q) ⇔ (P ∧ ¬Q) When a conditional is false Friday, September 24, 2010

  12. N EGATED C ONDITIONALS 1. P → Q for ↔ Intro (P → Q) ↔ ¬(P ∧ ¬Q) ¬(P ∧ ¬Q) ¬(P ∧ ¬Q) for ↔ Intro P → Q (P → Q) ↔ ¬(P ∧ ¬Q) ↔ Intro Friday, September 24, 2010

  13. 1. P → Q for ↔ Intro 2. P ∧ ¬Q for ¬ Intro 3. P ∧ Elim2 4. ¬Q ∧ Elim2 5. Q → Elim1,3 6. ⊥ ⊥ Intro 4,5 7. ¬(P ∧ ¬Q) ¬Intro 2-6 ¬(P ∧ ¬Q) 8. ¬(P ∧ ¬Q) for ↔ Intro ¬(P ∧ ¬Q) for ↔ Intro 9. P for → Intro 10. ¬Q for ¬ Intro 11. P ∧ ¬Q ∧ Intro 9,10 12. ⊥ ⊥ Intro 8,11 Q ¬ Intro P → Q → Intro 9- (P → Q) ↔ ¬(P ∧ ¬Q) ↔ Intro Friday, September 24, 2010

  14. 1. P → Q for ↔ Intro 2. P ∧ ¬Q for ¬ Intro 3. P ∧ Elim2 4. ¬Q ∧ Elim2 5. Q → Elim1,3 6. ⊥ ⊥ Intro 4,5 7. ¬(P ∧ ¬Q) ¬Intro 2-6 8. ¬(P ∧ ¬Q) for ↔ Intro 9. P for → Intro 10. ¬Q for ¬ Intro 11. P ∧ ¬Q ∧ Intro 9,10 12. ⊥ ⊥ Intro 8,11 13. Q ¬ Intro 10-12 14. P → Q → Intro 9-13 15. (P → Q) ↔ ¬(P ∧ ¬Q) ↔ Intro 1-7, 8-14 Friday, September 24, 2010

  15. P USHING N EGATIONS I NSIDE DeMorgan’s Laws With repeated applications of ¬(P ∨ Q) ⇔ (¬P ∧ ¬Q) these rules, we can convert any sentence with main ¬(P ∧ Q) ⇔ (¬P ∨ ¬Q) connective ¬ into something Negated Conditional with a different main connective. ¬(P → Q) ⇔ (P ∧ ¬Q) Negated Biconditional Or get rid of any particular ¬(P ↔ Q) ⇔ (¬P ↔ Q) connectives that we don’t like Friday, September 24, 2010

  16. D OUBLE R EDUCTIOS The Law of the Excluded Middle P ∨ ¬P 1. ¬(P ∨ ¬P) for ¬Intro 2. P for ¬Intro 3. P ∨ ¬P ∨ Intro 2 4. ⊥ ⊥ Intro 1,3 5. ¬P ¬Intro 2-4 6. P ∨ ¬P ∨ Intro 5 ⊥ P ∨ ¬P ¬Intro Friday, September 24, 2010

  17. D OUBLE R EDUCTIOS The Law of the Excluded Middle P ∨ ¬P 1. ¬(P ∨ ¬P) for ¬Intro 2. P for ¬Intro 3. P ∨ ¬P ∨ Intro 2 4. ⊥ ⊥ Intro 1,3 5. ¬P ¬Intro 2-4 6. P ∨ ¬P ∨ Intro 5 7. ⊥ ⊥ Intro 1,6 P ∨ ¬P ¬Intro 1-7 Friday, September 24, 2010

  18. D OUBLE R EDUCTIOS 1. A → B A → B 2. ¬A → B ¬A → B 3. ¬B for ¬ Intro 4. A for ¬ Intro B 5. B → Elim 1,3 6. ⊥ ⊥ Intro 3,5 7. ¬A ¬ Intro 4-6 8. B → Elim 2,7 9. ⊥ ⊥ Intro 3,8 B ¬Intro 3-9 Friday, September 24, 2010

  19. U SING LEM 1. A → B 1. A → B 2. ¬A → B 2. ¬A → B 3. ¬B for ¬ Intro 3. A ∨ ¬A LEM 4. A for ¬ Intro 4. A for ∨ Elim 5. B → Elim 1,3 5. B → Elim 1,3 6. ⊥ ⊥ Intro 3,5 7. ¬A ¬ Intro 4-6 6. ¬A for ∨ Elim 8. B → Elim 2,7 7. B → Elim 2,6 9. ⊥ ⊥ Intro 3,8 10. B ¬Intro 3-9 8. B ∨ Elim 3,4-5,6-7 Friday, September 24, 2010

  20. H OW TO R EALLY DO P ROOFS 6.42 in LPL book 1. ¬(¬A ∨ ¬(¬B ∧ (¬A ∨ B))) for ¬I 2. A ∧ (¬B ∧ (¬A ∨ B)) DeMorgans ¬A ∨ ¬(¬B ∧ (¬A ∨ B)) 3. A ∧ Elim 4. ¬B ∧ Elim 5. ¬A ∨ B ∧ Elim 6. ⊥ from 3-5 ¬A ∨ ¬(¬B ∧ (¬A ∨ B)) Friday, September 24, 2010

  21. R EALLY H ARD P ROOFS P ↔ (Q ↔ R) This is by no means trivial! (P ↔ Q) ↔ R (like it is with ∧ and ∨ ) P ↔ (Q ↔ R) does NOT mean P ⇔ Q ⇔ R For example, P ⇔ P ↔ P P ↔ (P ↔ P) is NOT a tautology Friday, September 24, 2010

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