Proof of Theorem nom53
Step | Hyp | Ref
| Expression |
1 | | ancom 74 |
. . . . . . . 8
(b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ ) |
2 | | anor3 90 |
. . . . . . . 8
(a⊥ ∩ b⊥ ) = (a ∪ b)⊥ |
3 | 1, 2 | ax-r2 36 |
. . . . . . 7
(b⊥ ∩ a⊥ ) = (a ∪ b)⊥ |
4 | 3 | ax-r1 35 |
. . . . . 6
(a ∪ b)⊥ = (b⊥ ∩ a⊥ ) |
5 | 4 | lor 70 |
. . . . 5
(b⊥
⊥ ∪ (a ∪ b)⊥ ) = (b⊥ ⊥ ∪
(b⊥ ∩ a⊥ )) |
6 | 4 | ax-r4 37 |
. . . . . 6
(a ∪ b)⊥ ⊥ = (b⊥ ∩ a⊥
)⊥ |
7 | 4 | lan 77 |
. . . . . 6
(b⊥ ∩ (a ∪ b)⊥ ) = (b⊥ ∩ (b⊥ ∩ a⊥ )) |
8 | 6, 7 | 2or 72 |
. . . . 5
((a ∪ b)⊥ ⊥ ∪
(b⊥ ∩ (a ∪ b)⊥ )) = ((b⊥ ∩ a⊥ )⊥ ∪
(b⊥ ∩ (b⊥ ∩ a⊥ ))) |
9 | 5, 8 | 2an 79 |
. . . 4
((b⊥
⊥ ∪ (a ∪ b)⊥ ) ∩ ((a ∪ b)⊥ ⊥ ∪
(b⊥ ∩ (a ∪ b)⊥ ))) = ((b⊥ ⊥ ∪
(b⊥ ∩ a⊥ )) ∩ ((b⊥ ∩ a⊥ )⊥ ∪
(b⊥ ∩ (b⊥ ∩ a⊥ )))) |
10 | | df-id4 53 |
. . . 4
(b⊥ ≡4
(a ∪ b)⊥ ) = ((b⊥ ⊥ ∪
(a ∪ b)⊥ ) ∩ ((a ∪ b)⊥ ⊥ ∪
(b⊥ ∩ (a ∪ b)⊥ ))) |
11 | | df-id4 53 |
. . . 4
(b⊥ ≡4
(b⊥ ∩ a⊥ )) = ((b⊥ ⊥ ∪
(b⊥ ∩ a⊥ )) ∩ ((b⊥ ∩ a⊥ )⊥ ∪
(b⊥ ∩ (b⊥ ∩ a⊥ )))) |
12 | 9, 10, 11 | 3tr1 63 |
. . 3
(b⊥ ≡4
(a ∪ b)⊥ ) = (b⊥ ≡4 (b⊥ ∩ a⊥ )) |
13 | | nom24 317 |
. . 3
(b⊥ ≡4
(b⊥ ∩ a⊥ )) = (b⊥ →1 a⊥ ) |
14 | 12, 13 | ax-r2 36 |
. 2
(b⊥ ≡4
(a ∪ b)⊥ ) = (b⊥ →1 a⊥ ) |
15 | | nomcon3 304 |
. 2
((a ∪ b) ≡3 b) = (b⊥ ≡4 (a ∪ b)⊥ ) |
16 | | i2i1 267 |
. 2
(a →2 b) = (b⊥ →1 a⊥ ) |
17 | 14, 15, 16 | 3tr1 63 |
1
((a ∪ b) ≡3 b) = (a
→2 b) |