First we prove a theorem known as identity:
1. Proposition
(P ® P) (theorem1)
| 1 | (P ® (Q ® P)) | add axiom axiom1 |
| 2 | (P ® ((P ® P) ® P)) | replace proposition variable Q by (P ® P) in 1 |
| 3 | ((P ® (Q ® A)) ® ((P ® Q) ® (P ® A))) | add axiom axiom2 |
| 4 | ((P ® ((P ® P) ® A)) ® ((P ® (P ® P)) ® (P ® A))) | replace proposition variable Q by (P ® P) in 3 |
| 5 | ((P ® ((P ® P) ® P)) ® ((P ® (P ® P)) ® (P ® P))) | replace proposition variable A by P in 4 |
| 6 | ((P ® (P ® P)) ® (P ® P)) | modus ponens with 2, 5 |
| 7 | (P ® (P ® P)) | replace proposition variable Q by P in 1 |
| 8 | (P ® P) | modus ponens with 7, 6 |
| qed |
Now we prove a theorem known as (one form of) hypothetical syllogism:
2. Proposition
((Q ® A) ® ((P ® Q) ® (P ® A))) (theorem2)
| 1 | (P ® (Q ® P)) | add axiom axiom1 |
| 2 | (P ® ((Q ® A) ® P)) | replace proposition variable Q by (Q ® A) in 1 |
| 3 | (((P ® (Q ® A)) ® ((P ® Q) ® (P ® A))) ® ((Q ® A) ® ((P ® (Q ® A)) ® ((P ® Q) ® (P ® A))))) | replace proposition variable P by ((P ® (Q ® A)) ® ((P ® Q) ® (P ® A))) in 2 |
| 4 | ((P ® (Q ® A)) ® ((P ® Q) ® (P ® A))) | add axiom axiom2 |
| 5 | ((Q ® A) ® ((P ® (Q ® A)) ® ((P ® Q) ® (P ® A)))) | modus ponens with 4, 3 |
| 6 | ((B ® (Q ® A)) ® ((B ® Q) ® (B ® A))) | replace proposition variable P by B in 4 |
| 7 | ((B ® (C ® A)) ® ((B ® C) ® (B ® A))) | replace proposition variable Q by C in 6 |
| 8 | ((B ® (C ® D)) ® ((B ® C) ® (B ® D))) | replace proposition variable A by D in 7 |
| 9 | (((Q ® A) ® (C ® D)) ® (((Q ® A) ® C) ® ((Q ® A) ® D))) | replace proposition variable B by (Q ® A) in 8 |
| 10 | (((Q ® A) ® ((P ® (Q ® A)) ® D)) ® (((Q ® A) ® (P ® (Q ® A))) ® ((Q ® A) ® D))) | replace proposition variable C by (P ® (Q ® A)) in 9 |
| 11 | (((Q ® A) ® ((P ® (Q ® A)) ® ((P ® Q) ® (P ® A)))) ® (((Q ® A) ® (P ® (Q ® A))) ® ((Q ® A) ® ((P ® Q) ® (P ® A))))) | replace proposition variable D by ((P ® Q) ® (P ® A)) in 10 |
| 12 | (((Q ® A) ® (P ® (Q ® A))) ® ((Q ® A) ® ((P ® Q) ® (P ® A)))) | modus ponens with 5, 11 |
| 13 | (P ® (B ® P)) | replace proposition variable Q by B in 1 |
| 14 | ((Q ® A) ® (B ® (Q ® A))) | replace proposition variable P by (Q ® A) in 13 |
| 15 | ((Q ® A) ® (P ® (Q ® A))) | replace proposition variable B by P in 14 |
| 16 | ((Q ® A) ® ((P ® Q) ® (P ® A))) | modus ponens with 15, 12 |
| qed |