Proof of Theorem nomcon2
| Step | Hyp | Ref
| Expression |
| 1 | | ax-a2 30 |
. . . 4
(a ∪ b⊥ ) = (b⊥ ∪ a) |
| 2 | | ax-a1 29 |
. . . . 5
a = a⊥ ⊥ |
| 3 | 2 | lor 66 |
. . . 4
(b⊥ ∪ a) = (b⊥ ∪ a⊥ ⊥ ) |
| 4 | 1, 3 | ax-r2 35 |
. . 3
(a ∪ b⊥ ) = (b⊥ ∪ a⊥ ⊥ ) |
| 5 | | ax-a1 29 |
. . . 4
b = b⊥ ⊥ |
| 6 | | ancom 68 |
. . . 4
(a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ ) |
| 7 | 5, 6 | 2or 67 |
. . 3
(b ∪ (a⊥ ∩ b⊥ )) = (b⊥ ⊥ ∪
(b⊥ ∩ a⊥ )) |
| 8 | 4, 7 | 2an 72 |
. 2
((a ∪ b⊥ ) ∩ (b ∪ (a⊥ ∩ b⊥ ))) = ((b⊥ ∪ a⊥ ⊥ ) ∩
(b⊥ ⊥
∪ (b⊥ ∩ a⊥ ))) |
| 9 | | df-id2 50 |
. 2
(a ≡2 b) = ((a ∪
b⊥ ) ∩ (b ∪ (a⊥ ∩ b⊥ ))) |
| 10 | | df-id1 49 |
. 2
(b⊥ ≡1
a⊥ ) = ((b⊥ ∪ a⊥ ⊥ ) ∩
(b⊥ ⊥
∪ (b⊥ ∩ a⊥ ))) |
| 11 | 8, 9, 10 | 3tr1 60 |
1
(a ≡2 b) = (b⊥ ≡1 a⊥ ) |