Proof of Theorem lem3.4.6
| Step | Hyp | Ref
| Expression |
| 1 | | lem3.3.6 1056 |
. . . 4
(a →2 (b ∪ c)) =
((a ∪ c) →2 (b ∪ c)) |
| 2 | 1 | ax-r1 35 |
. . 3
((a ∪ c) →2 (b ∪ c)) =
(a →2 (b ∪ c)) |
| 3 | | lem3.4.6.1 |
. . . 4
(a ≡5 b) = 1 |
| 4 | 3 | lem3.4.5 1078 |
. . 3
(a →2 (b ∪ c)) =
1 |
| 5 | 2, 4 | ax-r2 36 |
. 2
((a ∪ c) →2 (b ∪ c)) =
1 |
| 6 | | lem3.3.6 1056 |
. . . 4
(b →2 (a ∪ c)) =
((b ∪ c) →2 (a ∪ c)) |
| 7 | 6 | ax-r1 35 |
. . 3
((b ∪ c) →2 (a ∪ c)) =
(b →2 (a ∪ c)) |
| 8 | | df-id5 1047 |
. . . . 5
(b ≡5 a) = ((b ∩
a) ∪ (b⊥ ∩ a⊥ )) |
| 9 | | ancom 74 |
. . . . . . 7
(b ∩ a) = (a ∩
b) |
| 10 | | ancom 74 |
. . . . . . 7
(b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ ) |
| 11 | 9, 10 | 2or 72 |
. . . . . 6
((b ∩ a) ∪ (b⊥ ∩ a⊥ )) = ((a ∩ b) ∪
(a⊥ ∩ b⊥ )) |
| 12 | | df-id5 1047 |
. . . . . . 7
(a ≡5 b) = ((a ∩
b) ∪ (a⊥ ∩ b⊥ )) |
| 13 | 12 | ax-r1 35 |
. . . . . 6
((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = (a ≡5 b) |
| 14 | 11, 13, 3 | 3tr 65 |
. . . . 5
((b ∩ a) ∪ (b⊥ ∩ a⊥ )) = 1 |
| 15 | 8, 14 | ax-r2 36 |
. . . 4
(b ≡5 a) = 1 |
| 16 | 15 | lem3.4.5 1078 |
. . 3
(b →2 (a ∪ c)) =
1 |
| 17 | 7, 16 | ax-r2 36 |
. 2
((b ∪ c) →2 (a ∪ c)) =
1 |
| 18 | 5, 17 | lem3.4.4 1077 |
1
((a ∪ c) ≡5 (b ∪ c)) =
1 |