Proof of Theorem id5leid0
| Step | Hyp | Ref
| Expression |
| 1 | | ax-a2 31 |
. . 3
((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ b)) |
| 2 | | lea 160 |
. . . . 5
(a⊥ ∩ b⊥ ) ≤ a⊥ |
| 3 | | lear 161 |
. . . . 5
(a ∩ b) ≤ b |
| 4 | 2, 3 | le2or 168 |
. . . 4
((a⊥ ∩ b⊥ ) ∪ (a ∩ b)) ≤
(a⊥ ∪ b) |
| 5 | | lear 161 |
. . . . 5
(a⊥ ∩ b⊥ ) ≤ b⊥ |
| 6 | | lea 160 |
. . . . 5
(a ∩ b) ≤ a |
| 7 | 5, 6 | le2or 168 |
. . . 4
((a⊥ ∩ b⊥ ) ∪ (a ∩ b)) ≤
(b⊥ ∪ a) |
| 8 | 4, 7 | ler2an 173 |
. . 3
((a⊥ ∩ b⊥ ) ∪ (a ∩ b)) ≤
((a⊥ ∪ b) ∩ (b⊥ ∪ a)) |
| 9 | 1, 8 | bltr 138 |
. 2
((a ∩ b) ∪ (a⊥ ∩ b⊥ )) ≤ ((a⊥ ∪ b) ∩ (b⊥ ∪ a)) |
| 10 | | dfb 94 |
. 2
(a ≡ b) = ((a ∩
b) ∪ (a⊥ ∩ b⊥ )) |
| 11 | | df-id0 49 |
. 2
(a ≡0 b) = ((a⊥ ∪ b) ∩ (b⊥ ∪ a)) |
| 12 | 9, 10, 11 | le3tr1 140 |
1
(a ≡ b) ≤ (a
≡0 b) |