Proof of Theorem li3
| Step | Hyp | Ref
| Expression |
| 1 | | li3.1 |
. . . . 5
a = b |
| 2 | 1 | lan 70 |
. . . 4
(c⊥ ∩ a) = (c⊥ ∩ b) |
| 3 | 1 | ax-r4 36 |
. . . . 5
a⊥ = b⊥ |
| 4 | 3 | lan 70 |
. . . 4
(c⊥ ∩ a⊥ ) = (c⊥ ∩ b⊥ ) |
| 5 | 2, 4 | 2or 67 |
. . 3
((c⊥ ∩ a) ∪ (c⊥ ∩ a⊥ )) = ((c⊥ ∩ b) ∪ (c⊥ ∩ b⊥ )) |
| 6 | 1 | lor 66 |
. . . 4
(c⊥ ∪ a) = (c⊥ ∪ b) |
| 7 | 6 | lan 70 |
. . 3
(c ∩ (c⊥ ∪ a)) = (c ∩
(c⊥ ∪ b)) |
| 8 | 5, 7 | 2or 67 |
. 2
(((c⊥ ∩ a) ∪ (c⊥ ∩ a⊥ )) ∪ (c ∩ (c⊥ ∪ a))) = (((c⊥ ∩ b) ∪ (c⊥ ∩ b⊥ )) ∪ (c ∩ (c⊥ ∪ b))) |
| 9 | | df-i3 45 |
. 2
(c →3 a) = (((c⊥ ∩ a) ∪ (c⊥ ∩ a⊥ )) ∪ (c ∩ (c⊥ ∪ a))) |
| 10 | | df-i3 45 |
. 2
(c →3 b) = (((c⊥ ∩ b) ∪ (c⊥ ∩ b⊥ )) ∪ (c ∩ (c⊥ ∪ b))) |
| 11 | 8, 9, 10 | 3tr1 60 |
1
(c →3 a) = (c
→3 b) |