Proof of Theorem u5lem3
Step | Hyp | Ref
| Expression |
1 | | u5lemc1b 685 |
. . 3
a C (b →5 a) |
2 | 1 | u5lemc4 705 |
. 2
(a →5 (b →5 a)) = (a⊥ ∪ (b →5 a)) |
3 | | ax-a2 31 |
. . 3
(a⊥ ∪ (b →5 a)) = ((b
→5 a) ∪ a⊥ ) |
4 | | u5lemonb 639 |
. . . 4
((b →5 a) ∪ a⊥ ) = (((b ∩ a) ∪
(b⊥ ∩ a)) ∪ a⊥ ) |
5 | | ancom 74 |
. . . . . . 7
(b ∩ a) = (a ∩
b) |
6 | | ancom 74 |
. . . . . . 7
(b⊥ ∩ a) = (a ∩
b⊥ ) |
7 | 5, 6 | 2or 72 |
. . . . . 6
((b ∩ a) ∪ (b⊥ ∩ a)) = ((a ∩
b) ∪ (a ∩ b⊥ )) |
8 | 7 | ax-r5 38 |
. . . . 5
(((b ∩ a) ∪ (b⊥ ∩ a)) ∪ a⊥ ) = (((a ∩ b) ∪
(a ∩ b⊥ )) ∪ a⊥ ) |
9 | | ax-a2 31 |
. . . . 5
(((a ∩ b) ∪ (a
∩ b⊥ )) ∪ a⊥ ) = (a⊥ ∪ ((a ∩ b) ∪
(a ∩ b⊥ ))) |
10 | 8, 9 | ax-r2 36 |
. . . 4
(((b ∩ a) ∪ (b⊥ ∩ a)) ∪ a⊥ ) = (a⊥ ∪ ((a ∩ b) ∪
(a ∩ b⊥ ))) |
11 | 4, 10 | ax-r2 36 |
. . 3
((b →5 a) ∪ a⊥ ) = (a⊥ ∪ ((a ∩ b) ∪
(a ∩ b⊥ ))) |
12 | 3, 11 | ax-r2 36 |
. 2
(a⊥ ∪ (b →5 a)) = (a⊥ ∪ ((a ∩ b) ∪
(a ∩ b⊥ ))) |
13 | 2, 12 | ax-r2 36 |
1
(a →5 (b →5 a)) = (a⊥ ∪ ((a ∩ b) ∪
(a ∩ b⊥ ))) |