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Natural Deduction 2

2. . We shall learn the 10 rules of replacement.These rules express the logical equivalence between two statement forms such that one can replace another.We use the symbol :: to signify the relation of logical equivalence.. 3. 9. DeMorgan's Rule (DM). ~(p ? q) :: (~p v ~q)~(p v q) :: (~p ? ~q)E

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Natural Deduction 2

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    1. 1 Natural Deduction (2)

    2. 2 We shall learn the 10 rules of replacement. These rules express the logical equivalence between two statement forms such that one can replace another. We use the symbol :: to signify the relation of logical equivalence.

    3. 3 9. DeMorgans Rule (DM) ~(p ? q) :: (~p v ~q) ~(p v q) :: (~p ? ~q) Example: It is not the case that both Mary and Paul are late :: Mary is not late or Paul is not late. It is not the case that Mary or Paul is late :: Mary is not late and Paul is not late.

    4. 4 You can use truth tables to show that they are equivalent. DM simply requires you to change the connectives v and ? while shifting the negation symbol outside or inside the parentheses.

    5. 5 10. Commutativity (Com) (p v q) :: (q v p) (p ? q) :: (q ? p) This is obvious as v and ? behave similarly to addition and multiplication.

    6. 6 11. Associativity (Assoc) (p v (q v r)) :: ((p v q) v r) (p ? (q ? r)) :: ((p ? q) ? r) It is based on the fact that the meaning of a conjunction or a disjunction is unaffected by the placement of the parentheses if the same logical operator is used throughout.

    7. 7 12. Distribution (Dist) (p ? (q v r)) :: ((p ? q) v (p ? r)) (p v (q ? r)) :: ((p v q) ? (p v r)) This applies when a ? or a v appear in the same statement.

    8. 8 13. Double Negation (DN) P :: ~ ~p No explanation is needed.

    9. 9 Difference between implication and replacement Rules of implication apply to whole lines in a proof while rules of replacement apply to whole or any part of a line.

    10. 10 Rules of implication and rules of replacement cannot be applied in the same step. E.g.: Step 3 in the following is not allowed: 1. ~H 2. G v H 3. G 1, 2 Com, DS

    11. 11 Since we can replace part of a line with rules of replacement, we can first of all replace part of the conclusion line if that line is not already in the premises.

    12. 12 Example 1 K ? (F v B) 2. G ? K / (F ? G) v (B ? G) Working on the conclusion: (F ? G) v (B ? G) (G ? F) v (G ? B) Com, Com G ? (F v B) Dist

    13. 13 Now we have obtained (F v B), which is the consequent in the first premise. So we can start from the first step.

    14. 14 1. K ? (F v B) 2. G ? K / (F ? G) v (B ? G) 3. G 2, Simp 4. K ? G 2, Com 5. K 4, Simp 6. F v B 1, 5, MP 7. G ? (F v B) 3, 6, Conj 8. (G ? F) v (G ? B) 7, Dist 9. (F ? G) v (B ? G) 8, Com, Com3. G 2, Simp 4. K ? G 2, Com 5. K 4, Simp 6. F v B 1, 5, MP 7. G ? (F v B) 3, 6, Conj 8. (G ? F) v (G ? B) 7, Dist 9. (F ? G) v (B ? G) 8, Com, Com

    15. 15 Strategies 9. Conjunction can be used to set up DM. 1. ~A 2. ~B 3. ~A ? ~B 1, 2, Conj 4. ~(A v B) 3, DM

    16. 16 10. Constructive dilemma can be used to set up DM. 1. (A ? ~B) ? (C ? ~D) 2. A v C 3. ~B v ~D 1, 2, CD 4. ~(B ? D) 3, DM

    17. 17 11. Addition can be used to set up DM. 1. ~A 2. ~A v ~B 1, Add 3. ~(A ? B) 2, DM

    18. 18 12. Distribution can be used in two ways to set up DS. 1. (A v B) ? (A v C) 2. ~A 3. A v (B ? C) 1, Dist 4. B ? C 2, 3, DS 1 A ? (B v C) 2. ~(A ? B) 3. (A ? B) v (A ? C) 1, Dist 4. A ? C 2, 3, DS

    19. 19 13. Distribution can be used in two ways to set up Simp. 1. A v (B ? C) 2. (A v B) ? (A v C) 1, Dist 3. A v B 2, Simp 1. (A ? B) v (A ? C) 2. A ? (B v C) 1, Dist 3. A 2, Simp

    20. 20 14. Transposition (Trans) (p ? q) :: (~q ? ~p) This is obvious with our previous truth tables. Note that antecedent and consequent can only switch if they are both negated.

    21. 21 15. Material Implication (Impl) (p ? q) :: (~p v q) Example: If you get 100 marks, you will get the first place :: Either you dont get 100 marks or you get the first place.

    22. 22 16. Material Equivalence (Equiv) (p ? q) :: ((p ? q) ? (q ? p)) (p ? q) :: ((p ? q) v (~p ? ~q)) The first has been explained and the second simply reflects the fact of the truth table.

    23. 23 17. Exportation (Exp) ((p q) ? r) :: (p ? (q ? r)) Reason the above as: If we have both p and q, we have r :: If we have p, then if we have q we have r.

    24. 24 18. Tautology (Taut) p :: (p v p) p :: (p p) Repetition is still correct.

    25. 25 More strategies 14. Material implications can be used to set up HS. 1. ~A v B 2. ~B v C 3. A ? B 1, Impl 4. B ? C 2, Impl 5. A ? C 3, 4, HS

    26. 26 15. Exportation can be used to set up MP. 1. (A B) ? C 2. A 3. A ? (B ? C) 1, Exp 4. B ? C 2, 3, MP

    27. 27 16. Exportation can be used to set up MT. 1. A ? (B ? C) 2. ~C 3. (A B) ? C 1, Exp 4. ~(A B) 2, 3, MT

    28. 28 17. Addition can be used to set up material implication. 1. A 2. A v ~B 1, Add 3. ~B v A 2, Com 4. B ? A 3, Impl

    29. 29 18. Transposition can be used to set up HS. 1. A ? B 2. ~C ? ~B 3. B ? C 2, Trans 4. A ? C 1, 3, HS

    30. 30 19. Transposition can be used to set up CD. 1. (A ? B) (C ? D) 2. ~B v ~D 3. (~B ? ~A) (~D ? ~C) 1, Trans, Trans 4. ~A v ~C 2, 3, CD

    31. 31 20. Constructive dilemma can be used to set up tautology. 1. (A ? C) (B ? C) 2. A v B 3. C v C 1, 2, CD 4. C 3, Taut

    32. 32 21. Material implication can be used to set up tautology. 1. A ? ~A 2. ~A v ~A 1, Impl 3. ~A 2, Taut

    33. 33 22. Material implication can be used to set up distribution. 1. A ? (B C) 2. ~A v (B C) 1, Impl 3. (~A v B) (~A v C) 2, Dist

    34. 34 Conditional Proof It is a shorter method used for deriving conditional propositions. First, we assume the antecedent of the conditional proposition we want to get. Then we try to derive the consequent from the assumption and other premises. Finally we use ?to connect the antecedent we have assumed and the consequent. By doing this, we also discharge the assumption we have made.

    35. 35 Example If the conclusion is a conditional statement: 1. A ? (B C) 2. (B v D) ? E / A ? E

    36. 36 1. A ? (B C) 2. (B v D) ? E / A ? E 3. A ACP (assumption for conditional proof) 4. B C 1, 3, MP 5. B 4, Simp 6. B v D 5, Add 7. E 2, 6, MP 8. A ? E 37, CP (assumption discharged)

    37. 37 1. G ? (H I) 2. J ? (K L) 3. G v J / H v K

    38. 38 If a ND contains two conditional proof, no steps in the first conditional proof can be used again when the first conditional proof has been completed.

    39. 39 Indirect Proof Usually we use this method when no other methods seem to be promising. First, we assume the opposite of what is to be proved. Second, we derive a contradiction from the assumption and other given premises. This shows that our assumption is false. Consequently, we can conclude what is to be proved. Actually it is a special case of conditional proof.

    40. 40 - - - / q ~q - - - r ~r

    41. 41 After the contradiction r ~r, we have: r Simp ~r r Com ~r Simp r v q Add q DS ~q ? q CP ~~q v q Impl q v q DN q Taut

    42. 42 Since the previous part applies to all indirect proofs, we can skip it to have: - - - / q ~q AIP - - r ~r q IP

    43. 43 Example of Indirect Proof 1. By "God" I shall mean "the being than which none greater is possible." 2. God is a possible being. 3. Supposition: God does not exist in reality but exists only in the understanding. 4. If something exists only in the understanding, but is a possible being, then it might have been greater than it is. 5. Therefore, God might have been greater than he is. 6. Statement 5 contradicts with statement 1. 7. Therefore, our supposition (3) is false; God exists in reality as well as in the understanding. Premise 4 is false. The number 2 is greater than 1 whether 2 things exist or not. Premise 4 is false. The number 2 is greater than 1 whether 2 things exist or not.

    44. 44 Examples 1. (S v T) ? ~S / ~S

    45. 45 1. (K ? K) ? R 2. (R v M) ? N / N

    46. 46 Proving Logical Truth (Tautology) This is based on the method of conditional proof to show that certain propositions are logical truths. In the topic of truth table, we use a complete truth table to prove that a compound proposition is a tautology.

    47. 47 If the tautology is a conditional proposition p ? q, then the argument p // q is a valid argument. To show that p ? q is a tautology, we can use CP. If the tautology is not a conditional proposition, we can use IP to show that it is a tautology.

    48. 48 p ACP - - - - q p ? q CP

    49. 49 ~p AIP - - - q ~q p IP

    50. 50 Example /P ? ((P ? Q) ? Q) 1. P ACP 2. P ? Q ACP 3. Q 1, 2, MP 4. (P ? Q) ? Q 23, CP 5. P ? ((P ? Q) ? Q) 14, CP

    51. 51 Please note that at the end of the proof, we obtain the tautology from nothing. All the assumptions we made earlier have been discharged. So we may also say that a tautology/logical truth is the conclusion of an argument with no premises. Hence, a tautology is necessarily true.

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