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Introductory Logic PHI 120

Presentation: " n ->I (m) and m,n RAA (k) “. Introductory Logic PHI 120. Homework. Get Proofs handout (online) Identify and Solve first two ->I problems on handout. Solve S14* : ~P -> Q, ~Q ⊢ P Read pp.28-9 "double turnstile“ Study this presentation at home esp. S14

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Introductory Logic PHI 120

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  1. Presentation: "n ->I(m) and m,n RAA(k)“ Introductory LogicPHI 120

  2. Homework • Get Proofshandout (online) • Identify and Solve first two ->I problems on handout. • Solve S14* : ~P -> Q, ~Q ⊢ P • Read pp.28-9 "double turnstile“ • Study this presentation at home • esp. S14 • All 10 rules committed to memory!!! TAs may collect this assignment

  3. The 10 Primitive Rules • You should have the following in hand: • “The Rules” Handout • See bottom of handout

  4. Two Rules of Importance • Arrow – Introduction: ->I n ->I(m) • Reductio ad absurdum: RAA m, n RAA(k) Discharging assumption n ->I(m) m, n RAA(k) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . One premise rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Two premise rule

  5. Two Rules of Importance • Arrow – Introduction: ->I • Reductio ad absurdum: RAA Discharging assumption n ->I(m) m, n RAA(k) • Strategy

  6. Arrow - Introduction n ->I(m)

  7. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A (3) ???

  8. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A (3) ??? “P -> R” is not in the premises. Hence, we have to make it. Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  9. n ->I(m) ⊢ P -> R ⊢ P -> R ⊢ P -> R ⊢ P -> R Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  10. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A (3) ??? Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  11. n ->I(m) S16: P -> Q, Q -> R ⊢P -> R • (1) P -> Q A • (2) Q -> R A (3) ??? possible premise of an ->E possible premise of an ->E Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  12. n ->I(m) • Apply ->I Strategy S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A 3 (3) P A(step 1 in strategy of ->I) Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  13. n ->I(m) S16: P -> Q, Q -> R ⊢P-> R • (1) P -> Q A • (2) Q -> R A 3 (3) P A (step 1 in strategy of ->I) Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  14. n ->I(m) Read the problem properly! S16: P -> Q, Q -> R ⊢P-> R • (1) P -> Q A • (2) Q -> R A 3 (3) P A (4) What kind of statement is “R” (the consequent)? Where is it located in premises? Step 2(often more than one line) Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  15. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A 3 (3) P A antecedent of (1) (4) Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  16. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A (4) 1,3 ->E Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  17. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A (4) Q1,3 ->E Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  18. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1(1) P -> Q A • (2) Q -> R A 3(3) P A 1,3 (4) Q 1,3 ->E Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  19. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A 1,3 (4) Q 1,3 ->Eantecedent of (2) (5) Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  20. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A 1,3 (4) Q 1,3 ->E (5) 2,4 ->E Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  21. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A 1,3 (4) Q 1,3 ->E (5) R2,4 ->E Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  22. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1(1) P -> Q A 2(2) Q -> R A 3(3) P A 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  23. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E (6) Strategy to make an ->I: 1) Assume the antecedent of conclusion 2) Solve for the consequent (i.e., as a conclusion) 3) Apply ->I rule to generate the conditional

  24. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E (6) P -> R n ->I(m) Step 3

  25. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E (6) P -> R5 ->I(3)

  26. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) P A 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E (6) P -> R5 ->I(3)

  27. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R • (1) P -> Q A • (2) Q -> R A • (3) PA 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E (6) P -> R 5 ->I(3) This must be an assumption

  28. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1(1) P -> Q A 2(2) Q -> R A 3 (3) P A 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E (6) P -> R 5 ->I(3)

  29. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1(1) P -> Q A 2(2) Q -> R A 3 (3) P A 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E 1,2 (6) P -> R 5 ->I(3)

  30. n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1(1) P -> Q A 2(2) Q -> R A 3 (3) P A 1,3 (4) Q 1,3 ->E 1,2,3 (5) R 2,4 ->E 1,2 (6) P -> R 5 ->I(3) The Two Questions • Is (6) the conclusion of the sequent? • Are the assumptions correct?

  31. must be an assumption n ->I(m) Any kind of wff (will be the consequent) Any kind of wff (will be the antecedent)

  32. Reductio ad absurdum m,n RAA(k)

  33. The Key to RAA A B Denials • If the proof contains incompatible premises, you are allowed to deny any assumption within the proof. m, n RAA(k) Premises: denials of one another Conclusion: will be the denial of some assumption (k)

  34. P & Q, ~P ⊢ ~R 1 (1) P & Q A 2 (2) ~P A (3) ?? The Basic Assumptions • If the proof contains incompatible premises, you are allowed to deny any assumption within the proof. m, n RAA(k) Premises: denials of one another Conclusion: will be the denial of some assumption (k)

  35. P & Q, ~P ⊢ ~R 1 (1) P & Q A 2 (2) ~P A (3) ?? Elimination won’t work Introduction won’t work RAA • If the proof contains incompatible premises, you are allowed to deny any assumption within the proof m, n RAA(k) Premises: denials of one another Conclusion: will be the denial of some assumption (k)

  36. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A (3) ?? 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption • If the proof contains incompatible premises, you are allowed to deny any assumption within the proof m, n RAA(k) Premises: denials of one another Conclusion: will be the denial of some assumption (k)

  37. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A (3) ?? 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  38. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A (3) A 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  39. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A (3) R A 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  40. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A 3 (3) R A 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  41. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A 3 (3) R A 2) Derive a contradiction. 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  42. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A 3 (3) R A (4) ??? 2) Derive a contradiction. 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  43. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A 3 (3) R A (4) 1 &E 2) Derive a contradiction. 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  44. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A 3 (3) R A (4) P 1 &E 2) Derive a contradiction. 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  45. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A 3 (3) R A 1 (4) P 1 &E 2) Derive a contradiction. 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  46. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A 3 (3) R A 1 (4) P 1 &E 2) Derive a contradiction. 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption

  47. P & Q, ~P ⊢ ~R Strategy of RAA: 1) Assume the denial of the conclusion 1 (1) P & Q A 2 (2) ~P A 3 (3) R A 1 (4) P 1 &E 2) Derive a contradiction. 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption 3) Use RAA to deny/discharge an assumption

  48. P & Q, ~P ⊢ ~R 1 (1) P & Q A 2 (2) ~P A 3 (3) R A 1 (4) P 1 &E (5) m, n RAA(k) Premises: denials of one another

  49. P & Q, ~P ⊢ ~R 1 (1) P & Q A 2 (2) ~P A 3 (3) R A 1 (4) P 1 &E (5) 2, 4 RAA(k) Conclusion: will be the denial of some assumption (k)

  50. P & Q, ~P ⊢ ~R 1 (1) P & Q A 2 (2) ~P A 3 (3) R A 1 (4) P 1 &E (5) 2, 4 RAA(k) The Basic Assumptions Conclusion: will be the denial of some assumption (k)

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