Proof of Theorem go1
Step | Hyp | Ref
| Expression |
1 | | df-i1 44 |
. . 3
(a →1 b⊥ ) = (a⊥ ∪ (a ∩ b⊥ )) |
2 | 1 | lan 77 |
. 2
((a ∩ b) ∩ (a
→1 b⊥ )) =
((a ∩ b) ∩ (a⊥ ∪ (a ∩ b⊥ ))) |
3 | | lear 161 |
. . . . . 6
(a ∩ b⊥ ) ≤ b⊥ |
4 | 3 | lelor 166 |
. . . . 5
(a⊥ ∪ (a ∩ b⊥ )) ≤ (a⊥ ∪ b⊥ ) |
5 | 4 | lelan 167 |
. . . 4
((a ∩ b) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ≤ ((a ∩ b) ∩
(a⊥ ∪ b⊥ )) |
6 | | oran3 93 |
. . . . . 6
(a⊥ ∪ b⊥ ) = (a ∩ b)⊥ |
7 | 6 | lan 77 |
. . . . 5
((a ∩ b) ∩ (a⊥ ∪ b⊥ )) = ((a ∩ b) ∩
(a ∩ b)⊥ ) |
8 | | dff 101 |
. . . . . 6
0 = ((a ∩ b) ∩ (a
∩ b)⊥
) |
9 | 8 | ax-r1 35 |
. . . . 5
((a ∩ b) ∩ (a
∩ b)⊥ ) =
0 |
10 | 7, 9 | ax-r2 36 |
. . . 4
((a ∩ b) ∩ (a⊥ ∪ b⊥ )) = 0 |
11 | 5, 10 | lbtr 139 |
. . 3
((a ∩ b) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ≤ 0 |
12 | | le0 147 |
. . 3
0 ≤ ((a ∩ b) ∩ (a⊥ ∪ (a ∩ b⊥ ))) |
13 | 11, 12 | lebi 145 |
. 2
((a ∩ b) ∩ (a⊥ ∪ (a ∩ b⊥ ))) = 0 |
14 | 2, 13 | ax-r2 36 |
1
((a ∩ b) ∩ (a
→1 b⊥ )) =
0 |