Proof of Theorem u2lemonb
Step | Hyp | Ref
| Expression |
1 | | df-i2 45 |
. . 3
(a →2 b) = (b ∪
(a⊥ ∩ b⊥ )) |
2 | 1 | ax-r5 38 |
. 2
((a →2 b) ∪ b⊥ ) = ((b ∪ (a⊥ ∩ b⊥ )) ∪ b⊥ ) |
3 | | or32 82 |
. . 3
((b ∪ (a⊥ ∩ b⊥ )) ∪ b⊥ ) = ((b ∪ b⊥ ) ∪ (a⊥ ∩ b⊥ )) |
4 | | ax-a2 31 |
. . . 4
((b ∪ b⊥ ) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (b ∪ b⊥ )) |
5 | | df-t 41 |
. . . . . . 7
1 = (b ∪ b⊥ ) |
6 | 5 | lor 70 |
. . . . . 6
((a⊥ ∩ b⊥ ) ∪ 1) = ((a⊥ ∩ b⊥ ) ∪ (b ∪ b⊥ )) |
7 | 6 | ax-r1 35 |
. . . . 5
((a⊥ ∩ b⊥ ) ∪ (b ∪ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ 1) |
8 | | or1 104 |
. . . . 5
((a⊥ ∩ b⊥ ) ∪ 1) = 1 |
9 | 7, 8 | ax-r2 36 |
. . . 4
((a⊥ ∩ b⊥ ) ∪ (b ∪ b⊥ )) = 1 |
10 | 4, 9 | ax-r2 36 |
. . 3
((b ∪ b⊥ ) ∪ (a⊥ ∩ b⊥ )) = 1 |
11 | 3, 10 | ax-r2 36 |
. 2
((b ∪ (a⊥ ∩ b⊥ )) ∪ b⊥ ) = 1 |
12 | 2, 11 | ax-r2 36 |
1
((a →2 b) ∪ b⊥ ) = 1 |