I. Derivations for the following tautologies:
A. P v ~P
1. P Assumption
2. P 1, R
3. P -> P 1-2, CP
4. ~P v P 3, Impl.
5. P v ~P 4, Comm.
B. ~(P & ~P)
1. P Assumption
2. P 1, R
3. P -> P 1-2, CP
4. ~P v P 3, Impl.
5. ~P v ~~P 4, DN
6. ~(P & ~P) 5, DM
C. P <-> (~P -> P) (Note: pay attention to this one. It will
form the basis for the next lesson.)
1. P Assumption
2. ~P Assumption
3. P 1, R
4. ~P -> P 2-3, CP
5. P -> (~P -> P) 1-4, CP
6. ~P -> P Assumption
7. ~~P v P 6, Impl.
8. P v P 7, DN
9. P 8, Taut.
10. (~P -> P) -> P 6-9, CP
11. [P -> (~P -> P)] & [(~P -> P) -> P]
5, 10, Conj.
12. P <-> (~P -> P) 11, Equiv.
or
1. P Assumption
2. P v P 1, Taut.
3. ~~P v P 2, DN
4. ~P -> P 3, Impl.
5. P -> (~P -> P) 1-4, CP
6. ~P -> P Assumption
7. ~~P v P 6, Impl.
8. P v P 7, DN
9. P 8, Taut.
10. (~P -> P) -> P 6-9, CP
11. [P -> (~P -> P)] & [(~P -> P) -> P]
5, 10, Conj.
12. P <-> (~P -> P) 11, Equiv.
II. Without using Repetition (R), derive P -> P.
1. P Assumption
2. P & P 1, Taut.
3. P 2, Simp.
4. P -> P 1-3, CP
or
1. P Assumption
2. P v P 1, Taut.
3. P 2, Taut.
4. P -> P 1-3, CP
III. Without using Material Implication (Impl.), derive (~P v Q) ->
(P -> Q). (Hint: Use one conditional proof inside another.)
1. ~P v Q Assumption
2. P Assumption
3. ~~P 2, DN
4. Q 1, 3, DS
5. P -> Q 2-4, CP
6. (~P v Q) -> (P -> Q) 1-5, CP
IV. Without using Exportation (Exp.), derive [(P & Q) -> R] -> [P
-> (Q -> R)].
1. (P & Q) -> R Assumption
2. P Assumption
3. Q Assumption
4. P & Q 2, 3, Conj.
5. R 1, 4, MP
6. Q -> R 3-5, CP
7. P -> (Q -> R) 2-6, CP
8. [(P & Q) -> R] -> [P -> (Q -> R)]
1-7, CP
V. Without using Hypothetical Syllogism (HS), derive [(P -> Q) &
(Q -> R)] -> (P -> R).
1. (P -> Q) & (Q -> R) Assumption
2. P Assumption
3. P -> Q 1, Simp.
4. Q 2, 3, MP
5. (Q -> R) & (P -> Q) 1, Comm.
6. Q -> R 5, Simp.
7. R 4, 6, MP
8. P -> R 2-7, CP
9. [(P -> Q) & (Q -> R)] -> (P -> R)
1-8, CP