The three axiom
schemas below, together with the one rule of inference, constitute a
complete set of axioms for sentential logic. The Greek letters stand
for any sentence in the system.
Ax 1 ( ( ))
Ax 2 ((( ( )) (( ) ( )))
Ax 3 ((~ ~ ) ((~ ) )))
Rule of Inference (Modus Ponens): , ( )
(To make life a
little easier I will omit outside parentheses below.)
T1. P P
- P ((P P)
P)
Ax1
- ((P ((P P) P )) ((P
(P P)) (P P ))
Ax2
- (P (P P)) (P P )
1, 2 MP
- P (P P)
Ax1
- P P
3, 4 MP
I. Proof that (~P P) P
1. (~P ~P) ((~P P) P))
Ax 3
2. ~P ~P
Thm 1
3. (~P P) P
1,2 MP
II. Proof that
{(P Q), (Q R)} (P R)
1. (Q R) (P (Q R))
Ax 1
2. Q R
Assumption
3. P (Q R)
1,2 MP
4. ((P (Q R)) ((P Q) (P R))
Ax 2
5. (P Q) (P
R)
3,4 MP
6. P Q
Assumption
7. P R)
5,6 MP
III. Proof that {P, ~P} Q (i.e., that anything follows
from a contradiction; sometimes called the principle of explosion))
1. (P (~Q P))
Ax
1
2. P
Assumption
3. (~Q P)
1,2 MP
4. ~P (~Q ~P)
Ax 1
5. ~P
Assumption
6. ~Q ~P)
4,5 MP
7. (~Q ~P) ((~Q P) Q))
Ax 3
8. (~Q P) Q
6, 7 MP
9. Q
3,8 MP
IV. Proof that
~ ~P
P
1. (~P ~ ~P) [(~P ~P) P]
Ax 3
2. ((~P ~ ~P) [(~P ~P) P])
([(~P ~ ~P) (~P ~P)] [(~P ~ ~P) P])
Ax 2
3. [(~P ~ ~P) (~P ~P)] [(~P ~ ~P) P]
1,2 MP
4. (~P ~P) [(~P ~
~P) (~P ~P)]
Ax 1
5. ~P ~P
Thm
1
6. (~P ~ ~P) (~P
~P)
4, 5 MP
7. (~P ~ ~P) P
3,6 MP
8. ~ ~P (~P ~ ~P)
Ax 1
9. ~ ~P P
8,7 II (above)
Corollary of IV. ~~P P
(Double Negation, Part 1)
Proof: Given ~~P, use that with ~~P P and MP to infer P.
V.
Proof that P Q, ~Q ~P
(Modus Tollens)
1. ~ ~P P
Thm (IV, above)
2. P Q
Assumption
3. ~ ~P Q
1,2 II (above)
4. (~ ~P ~Q) [(~ ~P Q) ~P]
Ax 3
5. ~Q (~ ~P ~Q)
Ax 1
6. ~Q
Assumption
7. (~ ~P ~Q)
5,6 MP
8. (~ ~P Q) ~P
4,7 MP
9. ~P
3,7 MP
VI. Proof that P ~ ~P
(Double Negation, Part 2)
1. (~~~P ~P) [(~~~P P) ~~P]
Ax 3
2. ~~~P ~P
Thm IV (above)
3. (~~~P P) ~~P
1, 2 MP
4. P
Assumption
5. P (~~~P P)
Ax 1
6. ~~~P P
4, 5 MP
7. ~~P
3,6 MP
Thus
the rules of MP, DN and MT all hold. If we had
the
resources for conditional proof we would have all the machinery for a
natural
deduction system. What we require is the following result.
(Weak) Deduction Theorem (Herbrand, 1930): If ,
then ( ).
(It is not hard to show this by induction on the number of steps in the
derivation of from
.
See Mendelson, p. 32.)