Proof of Theorem nom20
Step | Hyp | Ref
| Expression |
1 | | lea 160 |
. . . . . 6
(a ∩ b) ≤ a |
2 | | leor 159 |
. . . . . 6
a ≤ (b⊥ ∪ a) |
3 | 1, 2 | letr 137 |
. . . . 5
(a ∩ b) ≤ (b⊥ ∪ a) |
4 | 3 | lelor 166 |
. . . 4
(a⊥ ∪ (a ∩ b)) ≤
(a⊥ ∪ (b⊥ ∪ a)) |
5 | | ax-a3 32 |
. . . . . 6
((a⊥ ∪ b⊥ ) ∪ a) = (a⊥ ∪ (b⊥ ∪ a)) |
6 | 5 | ax-r1 35 |
. . . . 5
(a⊥ ∪ (b⊥ ∪ a)) = ((a⊥ ∪ b⊥ ) ∪ a) |
7 | | oran3 93 |
. . . . . 6
(a⊥ ∪ b⊥ ) = (a ∩ b)⊥ |
8 | 7 | ax-r5 38 |
. . . . 5
((a⊥ ∪ b⊥ ) ∪ a) = ((a ∩
b)⊥ ∪ a) |
9 | 6, 8 | ax-r2 36 |
. . . 4
(a⊥ ∪ (b⊥ ∪ a)) = ((a ∩
b)⊥ ∪ a) |
10 | 4, 9 | lbtr 139 |
. . 3
(a⊥ ∪ (a ∩ b)) ≤
((a ∩ b)⊥ ∪ a) |
11 | 10 | df2le2 136 |
. 2
((a⊥ ∪
(a ∩ b)) ∩ ((a
∩ b)⊥ ∪ a)) = (a⊥ ∪ (a ∩ b)) |
12 | | df-id0 49 |
. 2
(a ≡0 (a ∩ b)) =
((a⊥ ∪ (a ∩ b))
∩ ((a ∩ b)⊥ ∪ a)) |
13 | | df-i1 44 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
14 | 11, 12, 13 | 3tr1 63 |
1
(a ≡0 (a ∩ b)) =
(a →1 b) |