Proof of Theorem 1b
Step | Hyp | Ref
| Expression |
1 | | dfb 94 |
. 2
(1 ≡ a) = ((1 ∩ a) ∪ (1⊥ ∩ a⊥ )) |
2 | | ancom 74 |
. . . . 5
(1 ∩ a) = (a ∩ 1) |
3 | | ancom 74 |
. . . . . 6
(1⊥ ∩ a⊥ ) = (a⊥ ∩ 1⊥
) |
4 | | df-f 42 |
. . . . . . . 8
0 = 1⊥ |
5 | 4 | ax-r1 35 |
. . . . . . 7
1⊥ = 0 |
6 | 5 | lan 77 |
. . . . . 6
(a⊥ ∩
1⊥ ) = (a⊥ ∩ 0) |
7 | 3, 6 | ax-r2 36 |
. . . . 5
(1⊥ ∩ a⊥ ) = (a⊥ ∩ 0) |
8 | 2, 7 | 2or 72 |
. . . 4
((1 ∩ a) ∪
(1⊥ ∩ a⊥ )) = ((a ∩ 1) ∪ (a⊥ ∩ 0)) |
9 | | an1 106 |
. . . . 5
(a ∩ 1) = a |
10 | | an0 108 |
. . . . 5
(a⊥ ∩ 0) =
0 |
11 | 9, 10 | 2or 72 |
. . . 4
((a ∩ 1) ∪ (a⊥ ∩ 0)) = (a ∪ 0) |
12 | 8, 11 | ax-r2 36 |
. . 3
((1 ∩ a) ∪
(1⊥ ∩ a⊥ )) = (a ∪ 0) |
13 | | or0 102 |
. . 3
(a ∪ 0) = a |
14 | 12, 13 | ax-r2 36 |
. 2
((1 ∩ a) ∪
(1⊥ ∩ a⊥ )) = a |
15 | 1, 14 | ax-r2 36 |
1
(1 ≡ a) = a |