Proof of Theorem nom23
Step | Hyp | Ref
| Expression |
1 | | le1 146 |
. . . 4
(a⊥ ∪ (a ∩ b)) ≤
1 |
2 | | df-t 41 |
. . . . 5
1 = (a ∪ a⊥ ) |
3 | | anabs 121 |
. . . . . . . 8
(a⊥ ∩ (a⊥ ∪ b⊥ )) = a⊥ |
4 | 3 | ax-r1 35 |
. . . . . . 7
a⊥ = (a⊥ ∩ (a⊥ ∪ b⊥ )) |
5 | | oran3 93 |
. . . . . . . 8
(a⊥ ∪ b⊥ ) = (a ∩ b)⊥ |
6 | 5 | lan 77 |
. . . . . . 7
(a⊥ ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∩ (a ∩ b)⊥ ) |
7 | 4, 6 | ax-r2 36 |
. . . . . 6
a⊥ = (a⊥ ∩ (a ∩ b)⊥ ) |
8 | 7 | lor 70 |
. . . . 5
(a ∪ a⊥ ) = (a ∪ (a⊥ ∩ (a ∩ b)⊥ )) |
9 | 2, 8 | ax-r2 36 |
. . . 4
1 = (a ∪ (a⊥ ∩ (a ∩ b)⊥ )) |
10 | 1, 9 | lbtr 139 |
. . 3
(a⊥ ∪ (a ∩ b)) ≤
(a ∪ (a⊥ ∩ (a ∩ b)⊥ )) |
11 | 10 | df2le2 136 |
. 2
((a⊥ ∪
(a ∩ b)) ∩ (a
∪ (a⊥ ∩ (a ∩ b)⊥ ))) = (a⊥ ∪ (a ∩ b)) |
12 | | df-id3 52 |
. 2
(a ≡3 (a ∩ b)) =
((a⊥ ∪ (a ∩ b))
∩ (a ∪ (a⊥ ∩ (a ∩ b)⊥ ))) |
13 | | df-i1 44 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
14 | 11, 12, 13 | 3tr1 63 |
1
(a ≡3 (a ∩ b)) =
(a →1 b) |