Proof of Theorem nom24
Step | Hyp | Ref
| Expression |
1 | | leo 158 |
. . . . 5
a⊥ ≤ (a⊥ ∪ b⊥ ) |
2 | 1 | leror 152 |
. . . 4
(a⊥ ∪ (a ∩ b)) ≤
((a⊥ ∪ b⊥ ) ∪ (a ∩ b)) |
3 | | oran3 93 |
. . . . 5
(a⊥ ∪ b⊥ ) = (a ∩ b)⊥ |
4 | | anidm 111 |
. . . . . . . 8
(a ∩ a) = a |
5 | 4 | ran 78 |
. . . . . . 7
((a ∩ a) ∩ b) =
(a ∩ b) |
6 | 5 | ax-r1 35 |
. . . . . 6
(a ∩ b) = ((a ∩
a) ∩ b) |
7 | | anass 76 |
. . . . . 6
((a ∩ a) ∩ b) =
(a ∩ (a ∩ b)) |
8 | 6, 7 | ax-r2 36 |
. . . . 5
(a ∩ b) = (a ∩
(a ∩ b)) |
9 | 3, 8 | 2or 72 |
. . . 4
((a⊥ ∪ b⊥ ) ∪ (a ∩ b)) =
((a ∩ b)⊥ ∪ (a ∩ (a ∩
b))) |
10 | 2, 9 | lbtr 139 |
. . 3
(a⊥ ∪ (a ∩ b)) ≤
((a ∩ b)⊥ ∪ (a ∩ (a ∩
b))) |
11 | 10 | df2le2 136 |
. 2
((a⊥ ∪
(a ∩ b)) ∩ ((a
∩ b)⊥ ∪ (a ∩ (a ∩
b)))) = (a⊥ ∪ (a ∩ b)) |
12 | | df-id4 53 |
. 2
(a ≡4 (a ∩ b)) =
((a⊥ ∪ (a ∩ b))
∩ ((a ∩ b)⊥ ∪ (a ∩ (a ∩
b)))) |
13 | | df-i1 44 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
14 | 11, 12, 13 | 3tr1 63 |
1
(a ≡4 (a ∩ b)) =
(a →1 b) |