Proof of Theorem i3lem3
Step | Hyp | Ref
| Expression |
1 | | omlan 448 |
. 2
(b⊥ ∩ (b ∪ (b⊥ ∩ a⊥ ))) = (b⊥ ∩ a⊥ ) |
2 | | ancom 74 |
. . 3
((a⊥ ∪ b) ∩ b⊥ ) = (b⊥ ∩ (a⊥ ∪ b)) |
3 | | ax-a2 31 |
. . . . 5
(a⊥ ∪ b) = (b ∪
a⊥ ) |
4 | | ax-a3 32 |
. . . . . . 7
((b ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = (b ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) |
5 | 4 | ax-r1 35 |
. . . . . 6
(b ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = ((b ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) |
6 | | i3lem.1 |
. . . . . . . 8
(a →3 b) = 1 |
7 | 6 | i3lem1 504 |
. . . . . . 7
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = a⊥ |
8 | 7 | lor 70 |
. . . . . 6
(b ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = (b ∪ a⊥ ) |
9 | | ancom 74 |
. . . . . . . . 9
(a⊥ ∩ b) = (b ∩
a⊥ ) |
10 | 9 | lor 70 |
. . . . . . . 8
(b ∪ (a⊥ ∩ b)) = (b ∪
(b ∩ a⊥ )) |
11 | | orabs 120 |
. . . . . . . 8
(b ∪ (b ∩ a⊥ )) = b |
12 | 10, 11 | ax-r2 36 |
. . . . . . 7
(b ∪ (a⊥ ∩ b)) = b |
13 | | ancom 74 |
. . . . . . 7
(a⊥ ∩ b⊥ ) = (b⊥ ∩ a⊥ ) |
14 | 12, 13 | 2or 72 |
. . . . . 6
((b ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = (b ∪ (b⊥ ∩ a⊥ )) |
15 | 5, 8, 14 | 3tr2 64 |
. . . . 5
(b ∪ a⊥ ) = (b ∪ (b⊥ ∩ a⊥ )) |
16 | 3, 15 | ax-r2 36 |
. . . 4
(a⊥ ∪ b) = (b ∪
(b⊥ ∩ a⊥ )) |
17 | 16 | lan 77 |
. . 3
(b⊥ ∩ (a⊥ ∪ b)) = (b⊥ ∩ (b ∪ (b⊥ ∩ a⊥ ))) |
18 | 2, 17 | ax-r2 36 |
. 2
((a⊥ ∪ b) ∩ b⊥ ) = (b⊥ ∩ (b ∪ (b⊥ ∩ a⊥ ))) |
19 | 1, 18, 13 | 3tr1 63 |
1
((a⊥ ∪ b) ∩ b⊥ ) = (a⊥ ∩ b⊥ ) |