Proof of Theorem u2lem1
Step | Hyp | Ref
| Expression |
1 | | df-i2 45 |
. 2
((a →2 b) →2 a) = (a ∪
((a →2 b)⊥ ∩ a⊥ )) |
2 | | ud2lem0c 278 |
. . . . . 6
(a →2 b)⊥ = (b⊥ ∩ (a ∪ b)) |
3 | 2 | ran 78 |
. . . . 5
((a →2 b)⊥ ∩ a⊥ ) = ((b⊥ ∩ (a ∪ b))
∩ a⊥
) |
4 | | an32 83 |
. . . . . 6
((b⊥ ∩
(a ∪ b)) ∩ a⊥ ) = ((b⊥ ∩ a⊥ ) ∩ (a ∪ b)) |
5 | | ax-a2 31 |
. . . . . . . . 9
(a ∪ b) = (b ∪
a) |
6 | | oran 87 |
. . . . . . . . 9
(b ∪ a) = (b⊥ ∩ a⊥
)⊥ |
7 | 5, 6 | ax-r2 36 |
. . . . . . . 8
(a ∪ b) = (b⊥ ∩ a⊥
)⊥ |
8 | 7 | lan 77 |
. . . . . . 7
((b⊥ ∩ a⊥ ) ∩ (a ∪ b)) =
((b⊥ ∩ a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥
) |
9 | | dff 101 |
. . . . . . . 8
0 = ((b⊥ ∩
a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥
) |
10 | 9 | ax-r1 35 |
. . . . . . 7
((b⊥ ∩ a⊥ ) ∩ (b⊥ ∩ a⊥ )⊥ ) =
0 |
11 | 8, 10 | ax-r2 36 |
. . . . . 6
((b⊥ ∩ a⊥ ) ∩ (a ∪ b)) =
0 |
12 | 4, 11 | ax-r2 36 |
. . . . 5
((b⊥ ∩
(a ∪ b)) ∩ a⊥ ) = 0 |
13 | 3, 12 | ax-r2 36 |
. . . 4
((a →2 b)⊥ ∩ a⊥ ) = 0 |
14 | 13 | lor 70 |
. . 3
(a ∪ ((a →2 b)⊥ ∩ a⊥ )) = (a ∪ 0) |
15 | | or0 102 |
. . 3
(a ∪ 0) = a |
16 | 14, 15 | ax-r2 36 |
. 2
(a ∪ ((a →2 b)⊥ ∩ a⊥ )) = a |
17 | 1, 16 | ax-r2 36 |
1
((a →2 b) →2 a) = a |