Proof of Theorem dfi4b
Step | Hyp | Ref
| Expression |
1 | | i4i3 271 |
. 2
(a →4 b) = (b⊥ →3 a⊥ ) |
2 | | dfi3b 499 |
. . 3
(b⊥ →3
a⊥ ) = ((b⊥ ⊥ ∪
a⊥ ) ∩ ((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ ))) |
3 | | ax-a2 31 |
. . . . . 6
(a⊥ ∪ b) = (b ∪
a⊥ ) |
4 | | ax-a1 30 |
. . . . . . 7
b = b⊥
⊥ |
5 | 4 | ax-r5 38 |
. . . . . 6
(b ∪ a⊥ ) = (b⊥ ⊥ ∪
a⊥ ) |
6 | 3, 5 | ax-r2 36 |
. . . . 5
(a⊥ ∪ b) = (b⊥ ⊥ ∪
a⊥ ) |
7 | 4 | ran 78 |
. . . . . . . 8
(b ∩ a⊥ ) = (b⊥ ⊥ ∩
a⊥ ) |
8 | 7 | lor 70 |
. . . . . . 7
(b⊥ ∪ (b ∩ a⊥ )) = (b⊥ ∪ (b⊥ ⊥ ∩
a⊥ )) |
9 | | ax-a1 30 |
. . . . . . . 8
a = a⊥
⊥ |
10 | 4, 9 | 2an 79 |
. . . . . . 7
(b ∩ a) = (b⊥ ⊥ ∩
a⊥ ⊥
) |
11 | 8, 10 | 2or 72 |
. . . . . 6
((b⊥ ∪
(b ∩ a⊥ )) ∪ (b ∩ a)) =
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ )) ∪ (b⊥ ⊥ ∩
a⊥ ⊥
)) |
12 | | or32 82 |
. . . . . 6
((b⊥ ∪
(b⊥ ⊥
∩ a⊥ )) ∪
(b⊥ ⊥
∩ a⊥
⊥ )) = ((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ )) |
13 | 11, 12 | ax-r2 36 |
. . . . 5
((b⊥ ∪
(b ∩ a⊥ )) ∪ (b ∩ a)) =
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ )) |
14 | 6, 13 | 2an 79 |
. . . 4
((a⊥ ∪ b) ∩ ((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a))) =
((b⊥ ⊥
∪ a⊥ ) ∩
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ ))) |
15 | 14 | ax-r1 35 |
. . 3
((b⊥
⊥ ∪ a⊥ ) ∩ ((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ ))) = ((a⊥ ∪ b) ∩ ((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a))) |
16 | 2, 15 | ax-r2 36 |
. 2
(b⊥ →3
a⊥ ) = ((a⊥ ∪ b) ∩ ((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a))) |
17 | 1, 16 | ax-r2 36 |
1
(a →4 b) = ((a⊥ ∪ b) ∩ ((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a))) |