Proof of Theorem u4lemnoa
Step | Hyp | Ref
| Expression |
1 | | u4lemana 608 |
. . . 4
((a →4 b) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |
2 | | ax-a2 31 |
. . . . 5
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b)) |
3 | | anor3 90 |
. . . . . 6
(a⊥ ∩ b⊥ ) = (a ∪ b)⊥ |
4 | | anor2 89 |
. . . . . 6
(a⊥ ∩ b) = (a ∪
b⊥
)⊥ |
5 | 3, 4 | 2or 72 |
. . . . 5
((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b)) = ((a ∪
b)⊥ ∪ (a ∪ b⊥ )⊥
) |
6 | 2, 5 | ax-r2 36 |
. . . 4
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥
) |
7 | 1, 6 | ax-r2 36 |
. . 3
((a →4 b) ∩ a⊥ ) = ((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥
) |
8 | | anor1 88 |
. . 3
((a →4 b) ∩ a⊥ ) = ((a →4 b)⊥ ∪ a)⊥ |
9 | | oran3 93 |
. . 3
((a ∪ b)⊥ ∪ (a ∪ b⊥ )⊥ ) =
((a ∪ b) ∩ (a
∪ b⊥
))⊥ |
10 | 7, 8, 9 | 3tr2 64 |
. 2
((a →4 b)⊥ ∪ a)⊥ = ((a ∪ b) ∩
(a ∪ b⊥
))⊥ |
11 | 10 | con1 66 |
1
((a →4 b)⊥ ∪ a) = ((a ∪
b) ∩ (a ∪ b⊥ )) |