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!!!