Proof of Theorem oaidlem1
| Step | Hyp | Ref
| Expression |
| 1 | | df-i1 43 |
. . 3
(b →1 c) = (b⊥ ∪ (b ∩ c)) |
| 2 | 1 | lor 66 |
. 2
(a⊥ ∪ (b →1 c)) = (a⊥ ∪ (b⊥ ∪ (b ∩ c))) |
| 3 | | oran3 85 |
. . . 4
(a⊥ ∪ b⊥ ) = (a ∩ b)⊥ |
| 4 | 3 | ax-r5 37 |
. . 3
((a⊥ ∪ b⊥ ) ∪ (b ∩ c)) =
((a ∩ b)⊥ ∪ (b ∩ c)) |
| 5 | | ax-a3 31 |
. . 3
((a⊥ ∪ b⊥ ) ∪ (b ∩ c)) =
(a⊥ ∪ (b⊥ ∪ (b ∩ c))) |
| 6 | | lear 153 |
. . . . 5
(a ∩ b) ≤ b |
| 7 | | oaidlem1.1 |
. . . . 5
(a ∩ b) ≤ c |
| 8 | 6, 7 | ler2an 165 |
. . . 4
(a ∩ b) ≤ (b ∩
c) |
| 9 | 8 | sklem 222 |
. . 3
((a ∩ b)⊥ ∪ (b ∩ c)) =
1 |
| 10 | 4, 5, 9 | 3tr2 61 |
. 2
(a⊥ ∪ (b⊥ ∪ (b ∩ c))) =
1 |
| 11 | 2, 10 | ax-r2 35 |
1
(a⊥ ∪ (b →1 c)) = 1 |