Proof of Theorem u3lemoa
Step | Hyp | Ref
| Expression |
1 | | df-i3 46 |
. . 3
(a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) |
2 | 1 | ax-r5 38 |
. 2
((a →3 b) ∪ a) =
((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∪ a) |
3 | | ax-a3 32 |
. . 3
((((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∪ a) =
(((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ (a⊥ ∪ b)) ∪ a)) |
4 | | lea 160 |
. . . . . 6
(a ∩ (a⊥ ∪ b)) ≤ a |
5 | 4 | df-le2 131 |
. . . . 5
((a ∩ (a⊥ ∪ b)) ∪ a) =
a |
6 | 5 | lor 70 |
. . . 4
(((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ (a⊥ ∪ b)) ∪ a)) =
(((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ a) |
7 | | ax-a2 31 |
. . . 4
(((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ a) = (a ∪
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) |
8 | 6, 7 | ax-r2 36 |
. . 3
(((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ (a⊥ ∪ b)) ∪ a)) =
(a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) |
9 | 3, 8 | ax-r2 36 |
. 2
((((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∪ a) =
(a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) |
10 | 2, 9 | ax-r2 36 |
1
((a →3 b) ∪ a) =
(a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) |