Proof of Theorem i3abs1
Step | Hyp | Ref
| Expression |
1 | | orordi 112 |
. . . . 5
(a⊥ ∪
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = ((a⊥ ∪ (a⊥ ∩ b)) ∪ (a⊥ ∪ (a⊥ ∩ b⊥ ))) |
2 | | orabs 120 |
. . . . . . 7
(a⊥ ∪ (a⊥ ∩ b)) = a⊥ |
3 | | orabs 120 |
. . . . . . 7
(a⊥ ∪ (a⊥ ∩ b⊥ )) = a⊥ |
4 | 2, 3 | 2or 72 |
. . . . . 6
((a⊥ ∪
(a⊥ ∩ b)) ∪ (a⊥ ∪ (a⊥ ∩ b⊥ ))) = (a⊥ ∪ a⊥ ) |
5 | | oridm 110 |
. . . . . 6
(a⊥ ∪ a⊥ ) = a⊥ |
6 | 4, 5 | ax-r2 36 |
. . . . 5
((a⊥ ∪
(a⊥ ∩ b)) ∪ (a⊥ ∪ (a⊥ ∩ b⊥ ))) = a⊥ |
7 | 1, 6 | ax-r2 36 |
. . . 4
(a⊥ ∪
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = a⊥ |
8 | 7 | ax-r5 38 |
. . 3
((a⊥ ∪
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ (a ∩ (a⊥ ∪ b))) |
9 | | ax-a3 32 |
. . 3
((a⊥ ∪
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))) |
10 | | omln 446 |
. . 3
(a⊥ ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ b) |
11 | 8, 9, 10 | 3tr2 64 |
. 2
(a⊥ ∪
(((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))) = (a⊥ ∪ b) |
12 | | lem4 511 |
. . 3
(a →3 (a →3 (a →3 b))) = (a⊥ ∪ (a →3 b)) |
13 | | df-i3 46 |
. . . 4
(a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) |
14 | 13 | lor 70 |
. . 3
(a⊥ ∪ (a →3 b)) = (a⊥ ∪ (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))) |
15 | 12, 14 | ax-r2 36 |
. 2
(a →3 (a →3 (a →3 b))) = (a⊥ ∪ (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))) |
16 | | lem4 511 |
. 2
(a →3 (a →3 b)) = (a⊥ ∪ b) |
17 | 11, 15, 16 | 3tr1 63 |
1
(a →3 (a →3 (a →3 b))) = (a
→3 (a →3
b)) |