Proof of Theorem i3th5
Step | Hyp | Ref
| Expression |
1 | | ax-a2 31 |
. . . . . 6
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b)) |
2 | | lea 160 |
. . . . . . 7
(a⊥ ∩ b⊥ ) ≤ a⊥ |
3 | | lear 161 |
. . . . . . 7
(a⊥ ∩ b) ≤ b |
4 | 2, 3 | le2or 168 |
. . . . . 6
((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b)) ≤ (a⊥ ∪ b) |
5 | 1, 4 | bltr 138 |
. . . . 5
((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ≤ (a⊥ ∪ b) |
6 | | lear 161 |
. . . . 5
(a ∩ (a⊥ ∪ b)) ≤ (a⊥ ∪ b) |
7 | 5, 6 | le2or 168 |
. . . 4
(((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ≤ ((a⊥ ∪ b) ∪ (a⊥ ∪ b)) |
8 | | oridm 110 |
. . . 4
((a⊥ ∪ b) ∪ (a⊥ ∪ b)) = (a⊥ ∪ b) |
9 | 7, 8 | lbtr 139 |
. . 3
(((a⊥ ∩
b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ≤ (a⊥ ∪ b) |
10 | | df-i3 46 |
. . 3
(a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) |
11 | | lem4 511 |
. . 3
(a →3 (a →3 b)) = (a⊥ ∪ b) |
12 | 9, 10, 11 | le3tr1 140 |
. 2
(a →3 b) ≤ (a
→3 (a →3
b)) |
13 | 12 | lei3 246 |
1
((a →3 b) →3 (a →3 (a →3 b))) = 1 |