Proof of Theorem dfi4b
| Step | Hyp | Ref
| Expression |
| 1 | | i4i3 263 |
. 2
(a →4 b) = (b⊥ →3 a⊥ ) |
| 2 | | dfi3b 481 |
. . 3
(b⊥ →3
a⊥ ) = ((b⊥ ⊥ ∪
a⊥ ) ∩ ((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ ))) |
| 3 | | ax-a2 30 |
. . . . . 6
(a⊥ ∪ b) = (b ∪
a⊥ ) |
| 4 | | ax-a1 29 |
. . . . . . 7
b = b⊥ ⊥ |
| 5 | 4 | ax-r5 37 |
. . . . . 6
(b ∪ a⊥ ) = (b⊥ ⊥ ∪
a⊥ ) |
| 6 | 3, 5 | ax-r2 35 |
. . . . 5
(a⊥ ∪ b) = (b⊥ ⊥ ∪
a⊥ ) |
| 7 | 4 | ran 71 |
. . . . . . . 8
(b ∩ a⊥ ) = (b⊥ ⊥ ∩
a⊥ ) |
| 8 | 7 | lor 66 |
. . . . . . 7
(b⊥ ∪ (b ∩ a⊥ )) = (b⊥ ∪ (b⊥ ⊥ ∩
a⊥ )) |
| 9 | | ax-a1 29 |
. . . . . . . 8
a = a⊥ ⊥ |
| 10 | 4, 9 | 2an 72 |
. . . . . . 7
(b ∩ a) = (b⊥ ⊥ ∩
a⊥ ⊥
) |
| 11 | 8, 10 | 2or 67 |
. . . . . 6
((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a)) =
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ )) ∪ (b⊥ ⊥ ∩
a⊥ ⊥
)) |
| 12 | | or32 75 |
. . . . . 6
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ )) ∪ (b⊥ ⊥ ∩
a⊥ ⊥ )) =
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ )) |
| 13 | 11, 12 | ax-r2 35 |
. . . . 5
((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a)) =
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ )) |
| 14 | 6, 13 | 2an 72 |
. . . 4
((a⊥ ∪ b) ∩ ((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a))) =
((b⊥ ⊥
∪ a⊥ ) ∩
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ ))) |
| 15 | 14 | ax-r1 34 |
. . 3
((b⊥ ⊥
∪ a⊥ ) ∩
((b⊥ ∪ (b⊥ ⊥ ∩
a⊥ ⊥ ))
∪ (b⊥
⊥ ∩ a⊥ ))) = ((a⊥ ∪ b) ∩ ((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a))) |
| 16 | 2, 15 | ax-r2 35 |
. 2
(b⊥ →3
a⊥ ) = ((a⊥ ∪ b) ∩ ((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a))) |
| 17 | 1, 16 | ax-r2 35 |
1
(a →4 b) = ((a⊥ ∪ b) ∩ ((b⊥ ∪ (b ∩ a⊥ )) ∪ (b ∩ a))) |