Proof of Theorem u5lemnonb
Step | Hyp | Ref
| Expression |
1 | | u5lemab 614 |
. . . 4
((a →5 b) ∩ b) =
((a ∩ b) ∪ (a⊥ ∩ b)) |
2 | | ax-a2 31 |
. . . . 5
((a ∩ b) ∪ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∪ (a
∩ b)) |
3 | | anor2 89 |
. . . . . 6
(a⊥ ∩ b) = (a ∪
b⊥
)⊥ |
4 | | df-a 40 |
. . . . . 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 →5 b) ∩ b) =
((a ∪ b⊥ )⊥ ∪
(a⊥ ∪ b⊥ )⊥
) |
8 | | df-a 40 |
. . 3
((a →5 b) ∩ b) =
((a →5 b)⊥ ∪ b⊥
)⊥ |
9 | | oran3 93 |
. . 3
((a ∪ b⊥ )⊥ ∪
(a⊥ ∪ b⊥ )⊥ ) =
((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥
))⊥ |
10 | 7, 8, 9 | 3tr2 64 |
. 2
((a →5 b)⊥ ∪ b⊥ )⊥ = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥
))⊥ |
11 | 10 | con1 66 |
1
((a →5 b)⊥ ∪ b⊥ ) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) |