Proof of Theorem lei3
Step | Hyp | Ref
| Expression |
1 | | ax-a3 32 |
. . 3
(((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) = ((a⊥ ∩ b) ∪ ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))) |
2 | | ax-a2 31 |
. . . . 5
(b⊥ ∪ a) = (a ∪
b⊥ ) |
3 | | ancom 74 |
. . . . . . 7
(a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ ) |
4 | | lei3.1 |
. . . . . . . . 9
a ≤ b |
5 | 4 | lecon 154 |
. . . . . . . 8
b⊥ ≤ a⊥ |
6 | 5 | df2le2 136 |
. . . . . . 7
(b⊥ ∩ a⊥ ) = b⊥ |
7 | 3, 6 | ax-r2 36 |
. . . . . 6
(a⊥ ∩ b⊥ ) = b⊥ |
8 | 4 | sklem 230 |
. . . . . . . 8
(a⊥ ∪ b) = 1 |
9 | 8 | lan 77 |
. . . . . . 7
(a ∩ (a⊥ ∪ b)) = (a ∩
1) |
10 | | an1 106 |
. . . . . . 7
(a ∩ 1) = a |
11 | 9, 10 | ax-r2 36 |
. . . . . 6
(a ∩ (a⊥ ∪ b)) = a |
12 | 7, 11 | 2or 72 |
. . . . 5
((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b))) = (b⊥ ∪ a) |
13 | | anor2 89 |
. . . . . 6
(a⊥ ∩ b) = (a ∪
b⊥
)⊥ |
14 | 13 | con2 67 |
. . . . 5
(a⊥ ∩ b)⊥ = (a ∪ b⊥ ) |
15 | 2, 12, 14 | 3tr1 63 |
. . . 4
((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∩ b)⊥ |
16 | 15 | lor 70 |
. . 3
((a⊥ ∩ b) ∪ ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b)⊥ ) |
17 | 1, 16 | ax-r2 36 |
. 2
(((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b)⊥ ) |
18 | | df-i3 46 |
. 2
(a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) |
19 | | df-t 41 |
. 2
1 = ((a⊥ ∩
b) ∪ (a⊥ ∩ b)⊥ ) |
20 | 17, 18, 19 | 3tr1 63 |
1
(a →3 b) = 1 |