Proof of Theorem u3lemona
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 | | or32 82 |
. . 3
((((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∪ a⊥ ) = ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ a⊥ ) ∪ (a ∩ (a⊥ ∪ b))) |
4 | | lea 160 |
. . . . . . 7
(a⊥ ∩ b) ≤ a⊥ |
5 | | lea 160 |
. . . . . . 7
(a⊥ ∩ b⊥ ) ≤ a⊥ |
6 | 4, 5 | lel2or 170 |
. . . . . 6
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ≤ a⊥ |
7 | 6 | df-le2 131 |
. . . . 5
(((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ a⊥ ) = a⊥ |
8 | 7 | ax-r5 38 |
. . . 4
((((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ a⊥ ) ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ (a ∩ (a⊥ ∪ b))) |
9 | | omln 446 |
. . . 4
(a⊥ ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ b) |
10 | 8, 9 | ax-r2 36 |
. . 3
((((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ a⊥ ) ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ b) |
11 | 3, 10 | ax-r2 36 |
. 2
((((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∪ a⊥ ) = (a⊥ ∪ b) |
12 | 2, 11 | ax-r2 36 |
1
((a →3 b) ∪ a⊥ ) = (a⊥ ∪ b) |