Proof of Theorem oml6
Step | Hyp | Ref
| Expression |
1 | | comor1 461 |
. . . 4
(a⊥ ∪ b⊥ ) C a⊥ |
2 | 1 | comcom7 460 |
. . 3
(a⊥ ∪ b⊥ ) C a |
3 | | comor2 462 |
. . . 4
(a⊥ ∪ b⊥ ) C b⊥ |
4 | 3 | comcom7 460 |
. . 3
(a⊥ ∪ b⊥ ) C b |
5 | 2, 4 | fh4c 478 |
. 2
(a ∪ (b ∩ (a⊥ ∪ b⊥ ))) = ((a ∪ b) ∩
(a ∪ (a⊥ ∪ b⊥ ))) |
6 | | df-t 41 |
. . . . . 6
1 = (a ∪ a⊥ ) |
7 | 6 | ax-r5 38 |
. . . . 5
(1 ∪ b⊥ ) =
((a ∪ a⊥ ) ∪ b⊥ ) |
8 | | ax-a2 31 |
. . . . . 6
(1 ∪ b⊥ ) =
(b⊥ ∪
1) |
9 | | or1 104 |
. . . . . 6
(b⊥ ∪ 1) =
1 |
10 | 8, 9 | ax-r2 36 |
. . . . 5
(1 ∪ b⊥ ) =
1 |
11 | | ax-a3 32 |
. . . . 5
((a ∪ a⊥ ) ∪ b⊥ ) = (a ∪ (a⊥ ∪ b⊥ )) |
12 | 7, 10, 11 | 3tr2 64 |
. . . 4
1 = (a ∪ (a⊥ ∪ b⊥ )) |
13 | 12 | ax-r1 35 |
. . 3
(a ∪ (a⊥ ∪ b⊥ )) = 1 |
14 | 13 | lan 77 |
. 2
((a ∪ b) ∩ (a
∪ (a⊥ ∪ b⊥ ))) = ((a ∪ b) ∩
1) |
15 | | an1 106 |
. 2
((a ∪ b) ∩ 1) = (a
∪ b) |
16 | 5, 14, 15 | 3tr 65 |
1
(a ∪ (b ∩ (a⊥ ∪ b⊥ ))) = (a ∪ b) |