Proof of Theorem mi
Step | Hyp | Ref
| Expression |
1 | | dfb 94 |
. 2
((a ∪ b) ≡ b) =
(((a ∪ b) ∩ b)
∪ ((a ∪ b)⊥ ∩ b⊥ )) |
2 | | ancom 74 |
. . . 4
((a ∪ b) ∩ b) =
(b ∩ (a ∪ b)) |
3 | | ax-a2 31 |
. . . . . 6
(a ∪ b) = (b ∪
a) |
4 | 3 | lan 77 |
. . . . 5
(b ∩ (a ∪ b)) =
(b ∩ (b ∪ a)) |
5 | | anabs 121 |
. . . . 5
(b ∩ (b ∪ a)) =
b |
6 | 4, 5 | ax-r2 36 |
. . . 4
(b ∩ (a ∪ b)) =
b |
7 | 2, 6 | ax-r2 36 |
. . 3
((a ∪ b) ∩ b) =
b |
8 | | oran 87 |
. . . . . . 7
(a ∪ b) = (a⊥ ∩ b⊥
)⊥ |
9 | 8 | con2 67 |
. . . . . 6
(a ∪ b)⊥ = (a⊥ ∩ b⊥ ) |
10 | 9 | ran 78 |
. . . . 5
((a ∪ b)⊥ ∩ b⊥ ) = ((a⊥ ∩ b⊥ ) ∩ b⊥ ) |
11 | | anass 76 |
. . . . 5
((a⊥ ∩ b⊥ ) ∩ b⊥ ) = (a⊥ ∩ (b⊥ ∩ b⊥ )) |
12 | 10, 11 | ax-r2 36 |
. . . 4
((a ∪ b)⊥ ∩ b⊥ ) = (a⊥ ∩ (b⊥ ∩ b⊥ )) |
13 | | anidm 111 |
. . . . 5
(b⊥ ∩ b⊥ ) = b⊥ |
14 | 13 | lan 77 |
. . . 4
(a⊥ ∩ (b⊥ ∩ b⊥ )) = (a⊥ ∩ b⊥ ) |
15 | 12, 14 | ax-r2 36 |
. . 3
((a ∪ b)⊥ ∩ b⊥ ) = (a⊥ ∩ b⊥ ) |
16 | 7, 15 | 2or 72 |
. 2
(((a ∪ b) ∩ b)
∪ ((a ∪ b)⊥ ∩ b⊥ )) = (b ∪ (a⊥ ∩ b⊥ )) |
17 | 1, 16 | ax-r2 36 |
1
((a ∪ b) ≡ b) =
(b ∪ (a⊥ ∩ b⊥ )) |