Proof of Theorem i2i1i1
Step | Hyp | Ref
| Expression |
1 | | an1r 107 |
. . 3
(1 ∩ (b ∪ (a⊥ ∩ b⊥ ))) = (b ∪ (a⊥ ∩ b⊥ )) |
2 | 1 | ax-r1 35 |
. 2
(b ∪ (a⊥ ∩ b⊥ )) = (1 ∩ (b ∪ (a⊥ ∩ b⊥ ))) |
3 | | df-i2 45 |
. 2
(a →2 b) = (b ∪
(a⊥ ∩ b⊥ )) |
4 | | anabs 121 |
. . . . . 6
(a ∩ (a ∪ b)) =
a |
5 | 4 | lor 70 |
. . . . 5
(a⊥ ∪ (a ∩ (a ∪
b))) = (a⊥ ∪ a) |
6 | | ax-a2 31 |
. . . . 5
(a⊥ ∪ a) = (a ∪
a⊥ ) |
7 | 5, 6 | ax-r2 36 |
. . . 4
(a⊥ ∪ (a ∩ (a ∪
b))) = (a ∪ a⊥ ) |
8 | | df-i1 44 |
. . . 4
(a →1 (a ∪ b)) =
(a⊥ ∪ (a ∩ (a ∪
b))) |
9 | | df-t 41 |
. . . 4
1 = (a ∪ a⊥ ) |
10 | 7, 8, 9 | 3tr1 63 |
. . 3
(a →1 (a ∪ b)) =
1 |
11 | | df-i1 44 |
. . . 4
((a ∪ b) →1 b) = ((a ∪
b)⊥ ∪ ((a ∪ b) ∩
b)) |
12 | | anor3 90 |
. . . . . 6
(a⊥ ∩ b⊥ ) = (a ∪ b)⊥ |
13 | | leor 159 |
. . . . . . . 8
b ≤ (a ∪ b) |
14 | | leid 148 |
. . . . . . . 8
b ≤ b |
15 | 13, 14 | ler2an 173 |
. . . . . . 7
b ≤ ((a ∪ b) ∩
b) |
16 | | lear 161 |
. . . . . . 7
((a ∪ b) ∩ b) ≤
b |
17 | 15, 16 | lebi 145 |
. . . . . 6
b = ((a ∪ b) ∩
b) |
18 | 12, 17 | 2or 72 |
. . . . 5
((a⊥ ∩ b⊥ ) ∪ b) = ((a ∪
b)⊥ ∪ ((a ∪ b) ∩
b)) |
19 | 18 | ax-r1 35 |
. . . 4
((a ∪ b)⊥ ∪ ((a ∪ b) ∩
b)) = ((a⊥ ∩ b⊥ ) ∪ b) |
20 | | ax-a2 31 |
. . . 4
((a⊥ ∩ b⊥ ) ∪ b) = (b ∪
(a⊥ ∩ b⊥ )) |
21 | 11, 19, 20 | 3tr 65 |
. . 3
((a ∪ b) →1 b) = (b ∪
(a⊥ ∩ b⊥ )) |
22 | 10, 21 | 2an 79 |
. 2
((a →1 (a ∪ b))
∩ ((a ∪ b) →1 b)) = (1 ∩ (b ∪ (a⊥ ∩ b⊥ ))) |
23 | 2, 3, 22 | 3tr1 63 |
1
(a →2 b) = ((a
→1 (a ∪ b)) ∩ ((a
∪ b) →1 b)) |