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 |