Introductory Logic PHI 120

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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 - PowerPoint PPT Presentation

Transcript of Introductory Logic PHI 120

Introductory LogicPHI 120

Presentation: "n ->I(m) and m,n RAA(k)“

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

• All 10 rules committed to memory!!!

TAs may collect this assignment

The 10 Primitive Rules

• You should have the following in hand:– “The Rules” Handout• See bottom of handout

Two Rules of Importance

• Arrow – Introduction: ->In ->I(m)

• Reductio ad absurdum: RAAm, n RAA(k)

Discharging assumptionn ->I(m)m, n RAA(k)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . One premise rule

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Two premise rule

Two Rules of Importance

• Arrow – Introduction: ->I

• Reductio ad absurdum: RAA

Discharging assumptionn ->I(m)m, n RAA(k)Strategy

n ->I(m)Arrow - Introduction

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A

(3) ???

n ->I(m)

S16: P -> Q, Q -> R ⊢ P -> R1 (1) P -> Q A2 (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

n ->I(m)

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

⊢ P -> R⊢ P -> R⊢ P -> R⊢ P -> R

n ->I(m)

S16: P -> Q, Q -> R ⊢ P -> R1 (1) P -> Q A2 (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

n ->I(m)

S16: P -> Q, Q -> R ⊢ P -> R1 (1) P -> Q A2 (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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (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

Apply ->I Strategy

n ->I(m)

S16: P -> Q, Q -> R ⊢ P -> R1 (1) P -> Q A2 (2) Q -> R A3 (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

n ->I(m)

S16: P -> Q, Q -> R ⊢ P -> R1 (1) P -> Q A2 (2) Q -> R A3 (3) P A

(4)

(1) What kind of statement is “R” (the consequent)?

(2) Where is it located in premises?

Step 2(often more than one line)

Read the problem properly!

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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A

(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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E antecedent 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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E

(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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (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

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (5) R 2,4 ->E

(6) P -> R n ->I(m) Step 3

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (5) R 2,4 ->E

(6) P -> R 5 ->I(3)

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (5) R 2,4 ->E

(6) P -> R 5 ->I(3)

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (5) R 2,4 ->E

(6) P -> R 5 ->I(3)

This must be an assumption

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (5) R 2,4 ->E

(6) P -> R 5 ->I(3)

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (5) R 2,4 ->E1,2 (6) P -> R 5 ->I(3)

n ->I(m)

S16: P -> Q, Q -> R P -> R⊢1 (1) P -> Q A2 (2) Q -> R A3 (3) P A1,3 (4) Q 1,3 ->E1,2,3 (5) R 2,4 ->E1,2 (6) P -> R 5 ->I(3)

(i) Is (6) the conclusion of the sequent?

(ii) Are the assumptions correct?

The Two Questions

Any kind of wff(will be the consequent)

Any kind of wff(will be the antecedent)

must be an assumption

n ->I(m)

m,n RAA(k)Reductio ad absurdum

The Key to 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)

Denials

A B

1 (1) P & Q A2 (2) ~P A

(3) ??

TheBasicAssumptions

– 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 (1) P & Q A2 (2) ~P A

(3) ??

– 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⊢

Elimination won’t work

Introduction won’t work

RAA

1 (1) P & Q A2 (2) ~P A

(3) ??

Strategy of RAA: 1) Assume the denial of the

conclusion

– 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⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

Strategy of RAA: 1) Assume the denial of the

conclusion

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A

(3) ??

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

Strategy of RAA: 1) Assume the denial of the

conclusion

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A

(3) A

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

Strategy of RAA: 1) Assume the denial of the

conclusion

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A

(3) R A

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

Strategy of RAA: 1) Assume the denial of the

conclusion

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A3 (3) R A

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A3 (3) R A

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

2) Derive a contradiction.

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A3 (3) R A

(4) ???

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

2) Derive a contradiction.

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A3 (3) R A

(4) 1 &E

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

2) Derive a contradiction.

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A3 (3) R A

(4) P 1 &E

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

2) Derive a contradiction.

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

2) Derive a contradiction.

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

2) Derive a contradiction.

Strategy of RAA: 1) Assume the denial of the

conclusion1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

P & Q, ~P ~R⊢

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

2) Derive a contradiction.

3) Use RAA to deny/discharge an assumption

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) m, n RAA(k)

P & Q, ~P ~R⊢

Premises: denials of one another

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) 2, 4 RAA(k)

P & Q, ~P ~R⊢

Conclusion: will be the denial of some assumption (k)

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) 2, 4 RAA(k)

P & Q, ~P ~R⊢

TheBasicAssumptions

Conclusion: will be the denial of some assumption (k)

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) 2, 4 RAA(k)

P & Q, ~P ~R⊢

Conclusion: will be the denial of some assumption (k)

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) 2, 4 RAA(3)

P & Q, ~P ~R⊢

Conclusion: will be the denial of some assumption (k)

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) ~R 2, 4 RAA(3)

P & Q, ~P ~R⊢

Conclusion: will be the denial of some assumption (k)

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) ~R 2, 4 RAA(3)

P & Q, ~P ~R⊢

Don't forget to define the assumption set!

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) ~R 2, 4 RAA(3)

P & Q, ~P ~R⊢

Don't forget to define the assumption set!

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E

(5) ~R 2, 4 RAA(3)

P & Q, ~P ~R⊢

Don't forget to define the assumption set!

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E1,2 (5) ~R 2, 4 RAA(3)

P & Q, ~P ~R⊢

Don't forget to define the assumption set!

1 (1) P & Q A2 (2) ~P A3 (3) R A1 (4) P 1 &E1,2 (5) ~R 2, 4 RAA(3)

P & Q, ~P ~R⊢

Any kind of wff(will be the consequent)

Any kind of wff(will be the antecedent)

must be an assumption

n ->I(m)

m,n RAA(k)

Premises: denials of one another

Conclusion: will be the denial of assumption: k

SOLVE S14 FOR HOMEWORKm,n RAA (k)

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 A2 (2) ~Q A

(3)

Strategy of RAA: 1) Assume the denial of the conclusion 2) Derive a contradiction 3) Use RAA to deny/discharge an assumption

Note: neither introduction nor elimination strategy will work for “P”

m,n RAA(k)

S14: ~P -> Q, ~Q ⊢ P1 (1) ~P -> Q A2 (2) ~Q A

(3) (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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A

(3) 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A

(3) ~P 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P 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

S14: ~P -> Q, ~Q ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A

Step Back. Read the premises.

m,n RAA(k)

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 ⊢ P1 (1) ~P -> Q A (possible –>E)

2 (2) ~Q A3 (3) ~P 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 ⊢ P1 (1) ~P -> Q A (possible –>E)

2 (2) ~Q A3 (3) ~P A (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 ⊢ P1 (1) ~P -> Q A (possible –>E)

2 (2) ~Q A (denial of consequent)

3 (3) ~P A (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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A

(4) ??

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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A

(4) 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A

(4) Q 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A

(4) Q 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) ??

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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) ??

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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) 2, 4 RAA(?)

Question: which assumption will we discharge?

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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) 2, 4 RAA(?)

Never discharge your basic 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) 2, 4 RAA(3)

The sole remaining assumption

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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) ? 2, 4 RAA(3)

Conclusion of RAA: denial of

discharged assumption

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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) P 2, 4 RAA(3)

Conclusion of RAA: denial of

discharged assumption

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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) P 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E

(5) P 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 ⊢ P1 (1) ~P -> Q A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E1,2 (5) P 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 A2 (2) ~Q A3 (3) ~P A1,3 (4) Q 1, 3 ->E1,2 (5) P 2, 4 RAA(3)

(i) Is (5) the conclusion of the sequent?(ii) Is (5) derived from the basic assumptions given in the sequent?

The Two Questions

Any kind of wff(will be the consequent)

Any kind of wff(will be the antecedent)

must be an assumption

n ->I(m)

m,n RAA(k)

Premises: denials of one another

Conclusion: will be the denial of assumption: k

Any kind of wff(will be the consequent)

Any kind of wff(will be the antecedent)

must be an assumption

n ->I(m)

m,n RAA(k)

Premises: denials of one another

Conclusion: will be the denial of assumption: k

Strategy

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

• All 10 rules committed to memory!!!

TAs may collect this assignment