Proof of Theorem i3lem4
| Step | Hyp | Ref
| Expression |
| 1 | | i3lem.1 |
. . . . 5
(a →3 b) = 1 |
| 2 | 1 | i3lem1 486 |
. . . 4
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = a⊥ |
| 3 | 2 | ax-r5 37 |
. . 3
(((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ (a ∩ (a⊥ ∪ b))) |
| 4 | 3 | ax-r1 34 |
. 2
(a⊥ ∪ (a ∩ (a⊥ ∪ b))) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) |
| 5 | | omln 428 |
. 2
(a⊥ ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ b) |
| 6 | | df-i3 45 |
. . . 4
(a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) |
| 7 | 6 | ax-r1 34 |
. . 3
(((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) = (a
→3 b) |
| 8 | 7, 1 | ax-r2 35 |
. 2
(((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) = 1 |
| 9 | 4, 5, 8 | 3tr2 61 |
1
(a⊥ ∪ b) = 1 |