Presentation: "n ->I(m) and m,n RAA(k)“ Introductory Logic PHI 120 Homework • Get Proofs handout (online) 1. Identify and Solve first two ->I problems on handout. 2. Solve S14* : ~P -> Q, ~Q ⊢ P TAs may collect this assignment • Read pp.28-9 "double turnstile“ • Study this presentation at home – esp. S14 • All 10 rules committed to memory!!! The 10 Primitive Rules • You should have the following in hand: – “The Rules” Handout • See bottom of handout Two Rules of Importance • Arrow – Introduction: ->I n ->I(m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . One premise rule • Reductio ad absurdum: RAA m, n RAA(k) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Two premise rule Discharging assumption n ->I(m) m, n RAA(k) Two Rules of Importance • Arrow – Introduction: ->I • Reductio ad absurdum: RAA Discharging assumption n ->I(m) m, n RAA(k) Arrow - Introduction n ->I(m) n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q A 2 (2) Q -> R A (3) ??? n ->I(m) -> R” is not in the premises. S16: P -> Q, Q -> R ⊢ P -> R “P Hence, we have to make it. 1 (1) P -> Q A 2 (2) Q -> R A (3) ??? Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) ⊢ P -> R Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q A 2 (2) Q -> R A (3) ??? Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q A 2 (2) Q -> R A (3) ??? possible premise of an ->E possible premise of an ->E Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q A 2 (2) Q -> R A 3 (3) P A (step 1 in strategy of ->I) Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q A 2 (2) Q -> R A 3 (3) P A (step 1 in strategy of ->I) Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q A 2 (2) Q -> R A 3 (3) P A (4) Read the problem properly! (1) What kind of statement is “R” (the consequent)? (2) Where is it located in premises? Step 2 (often more than one line) Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q A 2 (2) Q -> R A 3 (3) P A (4) antecedent of (1) Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P (4) A A A 1,3 ->E Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P (4) Q A A A 1,3 ->E Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P 1,3 (4) Q A A A 1,3 ->E Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional 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 antecedent of (2) (5) Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional 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 (5) 2,4 ->E Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional 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 (5) R 2,4 ->E Strategy to make an ->I: 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional 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) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional 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: (6) 1) 2) 3) Assume the antecedent of conclusion Solve for the consequent (i.e., as a conclusion) Apply ->I rule to generate the conditional n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P 1,3 (4) Q 1,2,3 (5) R (6) P -> R A A A 1,3 ->E 2,4 ->E n ->I(m) Step 3 n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P 1,3 (4) Q 1,2,3 (5) R (6) P -> R A A A 1,3 ->E 2,4 ->E 5 ->I(3) n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P 1,3 (4) Q 1,2,3 (5) R (6) P -> R A A A 1,3 ->E 2,4 ->E 5 ->I(3) n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P 1,3 (4) Q 1,2,3 (5) R (6) P -> R A A A 1,3 ->E 2,4 ->E 5 ->I(3) This must be an assumption n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P 1,3 (4) Q 1,2,3 (5) R (6) P -> R A A A 1,3 ->E 2,4 ->E 5 ->I(3) n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P 1,3 (4) Q 1,2,3 (5) R 1,2 (6) P -> R A A A 1,3 ->E 2,4 ->E 5 ->I(3) n ->I(m) S16: P -> Q, Q -> R ⊢ P -> R 1 (1) P -> Q 2 (2) Q -> R 3 (3) P 1,3 (4) Q 1,2,3 (5) R 1,2 (6) P -> R A A A 1,3 ->E 2,4 ->E 5 ->I(3) The Two Questions (i) Is (6) the conclusion of the sequent? (ii) Are the assumptions correct? must be an assumption n ->I(m) Any kind of wff (will be the consequent) Any kind of wff (will be the antecedent) Reductio ad absurdum m,n RAA(k) 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) P & Q, ~P ⊢ ~R 1 2 (1) P & Q A (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) P & Q, ~P ⊢ ~R 1 2 (1) P & Q A (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) P & Q, ~P ⊢ ~R 1 2 (1) P & Q A (2) ~P A (3) ?? Strategy of RAA: 1) Assume the denial of the conclusion 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) P & Q, ~P ⊢ ~R 1 2 (1) P & Q A (2) ~P A (3) ?? Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 (1) P & Q A (2) ~P A (3) A Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 (1) P & Q A (2) ~P A (3) R A Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 (1) P & Q A (2) ~P A (3) R A Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 (1) P & Q A (2) ~P A (3) R A Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 (1) P & Q (2) ~P (3) R (4) A A A ??? Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 (1) P & Q (2) ~P (3) R (4) A A A 1 &E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 (1) (2) (3) (4) P& Q ~P R P A A A 1 &E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) P& Q ~P R P A A A 1 &E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) P& Q ~P R P A A A 1 &E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) P& Q ~P R P A A A 1 &E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction. 3) Use RAA to deny/discharge an assumption P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P A A A 1 &E m, n RAA(k) Premises: denials of one another P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P A A A 1 &E 2, 4 RAA(k) Conclusion: will be the denial of some assumption (k) P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P A The A Basic Assumptions A 1 &E 2, 4 RAA(k) Conclusion: will be the denial of some assumption (k) P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P A A A 1 &E 2, 4 RAA(k) Conclusion: will be the denial of some assumption (k) P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P A A A 1 &E 2, 4 RAA(3) Conclusion: will be the denial of some assumption (k) P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P ~R A A A 1 &E 2, 4 RAA(3) Conclusion: will be the denial of some assumption (k) P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P ~R Don't forget to define the assumption set! A A A 1 &E 2, 4 RAA(3) P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P ~R Don't forget to define the assumption set! A A A 1 &E 2, 4 RAA(3) P & Q, ~P ⊢ ~R 1 2 3 1 (1) (2) (3) (4) (5) P& Q ~P R P ~R Don't forget to define the assumption set! A A A 1 &E 2, 4 RAA(3) P & Q, ~P ⊢ ~R 1 2 3 1 1,2 (1) (2) (3) (4) (5) P& Q ~P R P ~R Don't forget to define the assumption set! A A A 1 &E 2, 4 RAA(3) P & Q, ~P ⊢ ~R 1 2 3 1 1,2 (1) (2) (3) (4) (5) P& Q ~P R P ~R A A A 1 &E 2, 4 RAA(3) must be an assumption n ->I(m) Any kind of wff (will be the consequent) Any kind of wff (will be the antecedent) m,n RAA(k) Premises: denials of one another Conclusion: will be the denial of assumption: k m,n RAA (k) SOLVE S14 FOR HOMEWORK Homework • Get Proofs handout (online) 1. Identify and Solve first two ->I problems on handout. 2. Solve S14 : ~P -> Q, Q ⊢ P • Read pp.28-9 "double turnstile“ • Study this presentation at home – esp. S14 TAs may collect this assignment m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q (3) A A Note: neither introduction nor elimination strategy will work for “P” Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q (3) A A (first step of RAA) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q (3) A A A (assume) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q (3) ~P A A A (denial of conclusion) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P A A A (denial of conclusion) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P A A A Step Back. Read the premises. Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P A A A (possible –>E) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P A A A (possible –>E) (antecedent) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P A A A (possible –>E) (denial of consequent) (antecedent) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P (4) A A A ?? Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P (4) A A A 1, 3 ->E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P (4) Q A A A 1, 3 ->E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P (4) Q A A A 1, 3 ->E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q A A A 1, 3 ->E Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) A A A 1, 3 ->E ?? Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) A A A 1, 3 ->E ?? Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) A A A 1, 3 ->E 2, 4 RAA(?) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) Question: which A assumption will we discharge? A A 1, 3 ->E 2, 4 RAA(?) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) Never discharge your A basic premises! A A 1, 3 ->E 2, 4 RAA(?) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) A A The sole remaining A assumption 1, 3 ->E 2, 4 RAA(3) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) ? Conclusion of RAA: denial of discharged assumption A A A 1, 3 ->E 2, 4 RAA(3) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) P Conclusion of RAA: denial of discharged assumption A A A 1, 3 ->E 2, 4 RAA(3) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) P A A A 1, 3 ->E 2, 4 RAA(3) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q (5) P A A A 1, 3 ->E 2, 4 RAA(3) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q 1,2 (5) P A A A 1, 3 ->E 2, 4 RAA(3) Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption m,n RAA(k) S14: ~P -> Q, ~Q ⊢ P 1 (1) ~P -> Q 2 (2) ~Q 3 (3) ~P 1,3 (4) Q 1,2 (5) P A A A 1, 3 ->E 2, 4 RAA(3) The Two Questions (i) Is (5) the conclusion of the sequent? (ii) Is (5) derived from the basic assumptions given in the sequent? must be an assumption n ->I(m) Any kind of wff (will be the consequent) Any kind of wff (will be the antecedent) m,n RAA(k) Premises: denials of one another Conclusion: will be the denial of assumption: k must be an assumption n ->I(m) Any kind of wff (will be the consequent) Any kind of wff (will be the antecedent) m,n RAA(k) Premises: denials of one another Conclusion: will be the denial of assumption: k Homework • Get Proofs handout (online) 1. Identify and Solve first two ->I problems on handout. 2. Solve S14* : ~P -> Q, ~Q ⊢ P TAs may collect this assignment • Read pp.28-9 "double turnstile“ • Study this presentation at home – esp. S14 • All 10 rules committed to memory!!!