Proof of Theorem oml6
| Step | Hyp | Ref
| Expression |
| 1 | | comor1 443 |
. . . 4
(a⊥ ∪ b⊥ ) C a⊥ |
| 2 | 1 | comcom7 442 |
. . 3
(a⊥ ∪ b⊥ ) C a |
| 3 | | comor2 444 |
. . . 4
(a⊥ ∪ b⊥ ) C b⊥ |
| 4 | 3 | comcom7 442 |
. . 3
(a⊥ ∪ b⊥ ) C b |
| 5 | 2, 4 | fh4c 460 |
. 2
(a ∪ (b ∩ (a⊥ ∪ b⊥ ))) = ((a ∪ b) ∩
(a ∪ (a⊥ ∪ b⊥ ))) |
| 6 | | df-t 40 |
. . . . . 6
1 = (a ∪ a⊥ ) |
| 7 | 6 | ax-r5 37 |
. . . . 5
(1 ∪ b⊥ ) = ((a ∪ a⊥ ) ∪ b⊥ ) |
| 8 | | ax-a2 30 |
. . . . . 6
(1 ∪ b⊥ ) = (b⊥ ∪ 1) |
| 9 | | or1 96 |
. . . . . 6
(b⊥ ∪ 1) = 1 |
| 10 | 8, 9 | ax-r2 35 |
. . . . 5
(1 ∪ b⊥ ) =
1 |
| 11 | | ax-a3 31 |
. . . . 5
((a ∪ a⊥ ) ∪ b⊥ ) = (a ∪ (a⊥ ∪ b⊥ )) |
| 12 | 7, 10, 11 | 3tr2 61 |
. . . 4
1 = (a ∪ (a⊥ ∪ b⊥ )) |
| 13 | 12 | ax-r1 34 |
. . 3
(a ∪ (a⊥ ∪ b⊥ )) = 1 |
| 14 | 13 | lan 70 |
. 2
((a ∪ b) ∩ (a
∪ (a⊥ ∪ b⊥ ))) = ((a ∪ b) ∩
1) |
| 15 | | an1 98 |
. 2
((a ∪ b) ∩ 1) = (a
∪ b) |
| 16 | 5, 14, 15 | 3tr 62 |
1
(a ∪ (b ∩ (a⊥ ∪ b⊥ ))) = (a ∪ b) |