Proof of Theorem wql1
| Step | Hyp | Ref
| Expression |
| 1 | | df-i2 45 |
. 2
(a →2 b) = (b ∪
(a⊥ ∩ b⊥ )) |
| 2 | | anor3 90 |
. . 3
(a⊥ ∩ b⊥ ) = (a ∪ b)⊥ |
| 3 | 2 | lor 70 |
. 2
(b ∪ (a⊥ ∩ b⊥ )) = (b ∪ (a ∪
b)⊥ ) |
| 4 | | ax-a2 31 |
. . 3
(b ∪ (a ∪ b)⊥ ) = ((a ∪ b)⊥ ∪ b) |
| 5 | | wql1.3 |
. . . . . . . . 9
c = b |
| 6 | 5 | lor 70 |
. . . . . . . 8
(b ∪ c) = (b ∪
b) |
| 7 | | oridm 110 |
. . . . . . . 8
(b ∪ b) = b |
| 8 | 6, 7 | ax-r2 36 |
. . . . . . 7
(b ∪ c) = b |
| 9 | 8 | ud1lem0a 255 |
. . . . . 6
((a ∪ c) →1 (b ∪ c)) =
((a ∪ c) →1 b) |
| 10 | 9 | ax-r1 35 |
. . . . 5
((a ∪ c) →1 b) = ((a ∪
c) →1 (b ∪ c)) |
| 11 | 5 | lor 70 |
. . . . . 6
(a ∪ c) = (a ∪
b) |
| 12 | 11 | ud1lem0b 256 |
. . . . 5
((a ∪ c) →1 b) = ((a ∪
b) →1 b) |
| 13 | | wql1.2 |
. . . . 5
((a ∪ c) →1 (b ∪ c)) =
1 |
| 14 | 10, 12, 13 | 3tr2 64 |
. . . 4
((a ∪ b) →1 b) = 1 |
| 15 | 14 | wql1lem 287 |
. . 3
((a ∪ b)⊥ ∪ b) = 1 |
| 16 | 4, 15 | ax-r2 36 |
. 2
(b ∪ (a ∪ b)⊥ ) = 1 |
| 17 | 1, 3, 16 | 3tr 65 |
1
(a →2 b) = 1 |