Proof of Theorem nom25
Step | Hyp | Ref
| Expression |
1 | | anass 76 |
. . . . . 6
((a ∩ a) ∩ b) =
(a ∩ (a ∩ b)) |
2 | 1 | ax-r1 35 |
. . . . 5
(a ∩ (a ∩ b)) =
((a ∩ a) ∩ b) |
3 | | anidm 111 |
. . . . . 6
(a ∩ a) = a |
4 | 3 | ran 78 |
. . . . 5
((a ∩ a) ∩ b) =
(a ∩ b) |
5 | 2, 4 | ax-r2 36 |
. . . 4
(a ∩ (a ∩ b)) =
(a ∩ b) |
6 | | oran3 93 |
. . . . . . 7
(a⊥ ∪ b⊥ ) = (a ∩ b)⊥ |
7 | 6 | lan 77 |
. . . . . 6
(a⊥ ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∩ (a ∩ b)⊥ ) |
8 | 7 | ax-r1 35 |
. . . . 5
(a⊥ ∩ (a ∩ b)⊥ ) = (a⊥ ∩ (a⊥ ∪ b⊥ )) |
9 | | anabs 121 |
. . . . 5
(a⊥ ∩ (a⊥ ∪ b⊥ )) = a⊥ |
10 | 8, 9 | ax-r2 36 |
. . . 4
(a⊥ ∩ (a ∩ b)⊥ ) = a⊥ |
11 | 5, 10 | 2or 72 |
. . 3
((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)⊥ )) = ((a ∩ b) ∪
a⊥ ) |
12 | | ax-a2 31 |
. . 3
((a ∩ b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b)) |
13 | 11, 12 | ax-r2 36 |
. 2
((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)⊥ )) = (a⊥ ∪ (a ∩ b)) |
14 | | dfb 94 |
. 2
(a ≡ (a ∩ b)) =
((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)⊥ )) |
15 | | df-i1 44 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
16 | 13, 14, 15 | 3tr1 63 |
1
(a ≡ (a ∩ b)) =
(a →1 b) |