Proof of Theorem ska13
| Step | Hyp | Ref
| Expression |
| 1 | | ledio 168 |
. . . . 5
((a ∩ b) ∪ (a⊥ ∩ b⊥ )) ≤ (((a ∩ b) ∪
a⊥ ) ∩ ((a ∩ b) ∪
b⊥ )) |
| 2 | | lea 152 |
. . . . 5
(((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪
b⊥ )) ≤ ((a ∩ b) ∪
a⊥ ) |
| 3 | 1, 2 | letr 129 |
. . . 4
((a ∩ b) ∪ (a⊥ ∩ b⊥ )) ≤ ((a ∩ b) ∪
a⊥ ) |
| 4 | | ancom 68 |
. . . . . 6
(a ∩ b) = (b ∩
a) |
| 5 | | lea 152 |
. . . . . 6
(b ∩ a) ≤ b |
| 6 | 4, 5 | bltr 130 |
. . . . 5
(a ∩ b) ≤ b |
| 7 | 6 | leror 144 |
. . . 4
((a ∩ b) ∪ a⊥ ) ≤ (b ∪ a⊥ ) |
| 8 | 3, 7 | letr 129 |
. . 3
((a ∩ b) ∪ (a⊥ ∩ b⊥ )) ≤ (b ∪ a⊥ ) |
| 9 | | dfb 86 |
. . 3
(a ≡ b) = ((a ∩
b) ∪ (a⊥ ∩ b⊥ )) |
| 10 | | ax-a2 30 |
. . 3
(a⊥ ∪ b) = (b ∪
a⊥ ) |
| 11 | 8, 9, 10 | le3tr1 132 |
. 2
(a ≡ b) ≤ (a⊥ ∪ b) |
| 12 | 11 | sklem 222 |
1
((a ≡ b)⊥ ∪ (a⊥ ∪ b)) = 1 |