Proof of Theorem oal1
| Step | Hyp | Ref
| Expression |
| 1 | | oal2 979 |
. 2
((c⊥ →2
a⊥ ) ∩
((a⊥ ∪ b⊥ )⊥ ∪
((c⊥ →2
a⊥ ) ∩ (c⊥ →2 b⊥ )))) ≤ (c⊥ →2 b⊥ ) |
| 2 | | i1i2 258 |
. . 3
(a →1 c) = (c⊥ →2 a⊥ ) |
| 3 | | df-a 39 |
. . . 4
(a ∩ b) = (a⊥ ∪ b⊥ )⊥ |
| 4 | | i1i2 258 |
. . . . 5
(b →1 c) = (c⊥ →2 b⊥ ) |
| 5 | 2, 4 | 2an 72 |
. . . 4
((a →1 c) ∩ (b
→1 c)) = ((c⊥ →2 a⊥ ) ∩ (c⊥ →2 b⊥ )) |
| 6 | 3, 5 | 2or 67 |
. . 3
((a ∩ b) ∪ ((a
→1 c) ∩ (b →1 c))) = ((a⊥ ∪ b⊥ )⊥ ∪
((c⊥ →2
a⊥ ) ∩ (c⊥ →2 b⊥ ))) |
| 7 | 2, 6 | 2an 72 |
. 2
((a →1 c) ∩ ((a
∩ b) ∪ ((a →1 c) ∩ (b
→1 c)))) = ((c⊥ →2 a⊥ ) ∩ ((a⊥ ∪ b⊥ )⊥ ∪
((c⊥ →2
a⊥ ) ∩ (c⊥ →2 b⊥ )))) |
| 8 | 1, 7, 4 | le3tr1 132 |
1
((a →1 c) ∩ ((a
∩ b) ∪ ((a →1 c) ∩ (b
→1 c)))) ≤ (b →1 c) |