Proof of Theorem u3lem14mp
| Step | Hyp | Ref
| Expression |
| 1 | | lear 153 |
. . . 4
(((a ∪ b⊥ ⊥ ) ∩
(a ∪ b⊥ )) ∩ (a⊥ ∪ (a ∩ b⊥ ⊥ ))) ≤
(a⊥ ∪ (a ∩ b⊥ ⊥ )) |
| 2 | | lear 153 |
. . . . . 6
(a ∩ b⊥ ⊥ ) ≤
b⊥
⊥ |
| 3 | | ax-a1 29 |
. . . . . . 7
b = b⊥ ⊥ |
| 4 | 3 | ax-r1 34 |
. . . . . 6
b⊥ ⊥ =
b |
| 5 | 2, 4 | lbtr 131 |
. . . . 5
(a ∩ b⊥ ⊥ ) ≤
b |
| 6 | 5 | lelor 158 |
. . . 4
(a⊥ ∪ (a ∩ b⊥ ⊥ )) ≤
(a⊥ ∪ b) |
| 7 | 1, 6 | letr 129 |
. . 3
(((a ∪ b⊥ ⊥ ) ∩
(a ∪ b⊥ )) ∩ (a⊥ ∪ (a ∩ b⊥ ⊥ ))) ≤
(a⊥ ∪ b) |
| 8 | | ud3lem0c 271 |
. . 3
(a →3 b⊥ )⊥ =
(((a ∪ b⊥ ⊥ ) ∩
(a ∪ b⊥ )) ∩ (a⊥ ∪ (a ∩ b⊥ ⊥ ))) |
| 9 | | u3lem5 745 |
. . 3
(a →3 (a →3 b)) = (a⊥ ∪ b) |
| 10 | 7, 8, 9 | le3tr1 132 |
. 2
(a →3 b⊥ )⊥ ≤
(a →3 (a →3 b)) |
| 11 | 10 | u3lemle1 694 |
1
((a →3 b⊥ )⊥ →3
(a →3 (a →3 b))) = 1 |