Proof of Theorem oal2
| Step | Hyp | Ref
| Expression |
| 1 | | ax-3oa 998 |
. 2
((b⊥ →1
a⊥ ) ∩
((b⊥ ∩ c⊥ ) ∪ ((b⊥ →1 a⊥ ) ∩ (c⊥ →1 a⊥ )))) ≤ (c⊥ →1 a⊥ ) |
| 2 | | i2i1 267 |
. . 3
(a →2 b) = (b⊥ →1 a⊥ ) |
| 3 | | anor3 90 |
. . . . 5
(b⊥ ∩ c⊥ ) = (b ∪ c)⊥ |
| 4 | 3 | ax-r1 35 |
. . . 4
(b ∪ c)⊥ = (b⊥ ∩ c⊥ ) |
| 5 | | i2i1 267 |
. . . . 5
(a →2 c) = (c⊥ →1 a⊥ ) |
| 6 | 2, 5 | 2an 79 |
. . . 4
((a →2 b) ∩ (a
→2 c)) = ((b⊥ →1 a⊥ ) ∩ (c⊥ →1 a⊥ )) |
| 7 | 4, 6 | 2or 72 |
. . 3
((b ∪ c)⊥ ∪ ((a →2 b) ∩ (a
→2 c))) = ((b⊥ ∩ c⊥ ) ∪ ((b⊥ →1 a⊥ ) ∩ (c⊥ →1 a⊥ ))) |
| 8 | 2, 7 | 2an 79 |
. 2
((a →2 b) ∩ ((b
∪ c)⊥ ∪ ((a →2 b) ∩ (a
→2 c)))) = ((b⊥ →1 a⊥ ) ∩ ((b⊥ ∩ c⊥ ) ∪ ((b⊥ →1 a⊥ ) ∩ (c⊥ →1 a⊥ )))) |
| 9 | 1, 8, 5 | le3tr1 140 |
1
((a →2 b) ∩ ((b
∪ c)⊥ ∪ ((a →2 b) ∩ (a
→2 c)))) ≤ (a →2 c) |