Proof of Theorem u5lem1n
Step | Hyp | Ref
| Expression |
1 | | u5lem1 738 |
. . 3
((a →5 b) →5 a) = ((a ∪
b) ∩ (a ∪ b⊥ )) |
2 | | ancom 74 |
. . . 4
((a ∪ b) ∩ (a
∪ b⊥ )) = ((a ∪ b⊥ ) ∩ (a ∪ b)) |
3 | | df-a 40 |
. . . . 5
((a ∪ b⊥ ) ∩ (a ∪ b)) =
((a ∪ b⊥ )⊥ ∪
(a ∪ b)⊥
)⊥ |
4 | | anor2 89 |
. . . . . . . 8
(a⊥ ∩ b) = (a ∪
b⊥
)⊥ |
5 | | anor3 90 |
. . . . . . . 8
(a⊥ ∩ b⊥ ) = (a ∪ b)⊥ |
6 | 4, 5 | 2or 72 |
. . . . . . 7
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a ∪ b⊥ )⊥ ∪
(a ∪ b)⊥ ) |
7 | 6 | ax-r4 37 |
. . . . . 6
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ =
((a ∪ b⊥ )⊥ ∪
(a ∪ b)⊥
)⊥ |
8 | 7 | ax-r1 35 |
. . . . 5
((a ∪ b⊥ )⊥ ∪
(a ∪ b)⊥ )⊥ =
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥
))⊥ |
9 | 3, 8 | ax-r2 36 |
. . . 4
((a ∪ b⊥ ) ∩ (a ∪ b)) =
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥
))⊥ |
10 | 2, 9 | ax-r2 36 |
. . 3
((a ∪ b) ∩ (a
∪ b⊥ )) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥
))⊥ |
11 | 1, 10 | ax-r2 36 |
. 2
((a →5 b) →5 a) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥
))⊥ |
12 | 11 | con2 67 |
1
((a →5 b) →5 a)⊥ = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |