Proof of Theorem nom23
| Step | Hyp | Ref
| Expression |
| 1 | | le1 138 |
. . . 4
(a⊥ ∪ (a ∩ b)) ≤
1 |
| 2 | | df-t 40 |
. . . . 5
1 = (a ∪ a⊥ ) |
| 3 | | a5c 113 |
. . . . . . . 8
(a⊥ ∩ (a⊥ ∪ b⊥ )) = a⊥ |
| 4 | 3 | ax-r1 34 |
. . . . . . 7
a⊥ = (a⊥ ∩ (a⊥ ∪ b⊥ )) |
| 5 | | oran3 85 |
. . . . . . . 8
(a⊥ ∪ b⊥ ) = (a ∩ b)⊥ |
| 6 | 5 | lan 70 |
. . . . . . 7
(a⊥ ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∩ (a ∩ b)⊥ ) |
| 7 | 4, 6 | ax-r2 35 |
. . . . . 6
a⊥ = (a⊥ ∩ (a ∩ b)⊥ ) |
| 8 | 7 | lor 66 |
. . . . 5
(a ∪ a⊥ ) = (a ∪ (a⊥ ∩ (a ∩ b)⊥ )) |
| 9 | 2, 8 | ax-r2 35 |
. . . 4
1 = (a ∪ (a⊥ ∩ (a ∩ b)⊥ )) |
| 10 | 1, 9 | lbtr 131 |
. . 3
(a⊥ ∪ (a ∩ b)) ≤
(a ∪ (a⊥ ∩ (a ∩ b)⊥ )) |
| 11 | 10 | df2le2 128 |
. 2
((a⊥ ∪ (a ∩ b))
∩ (a ∪ (a⊥ ∩ (a ∩ b)⊥ ))) = (a⊥ ∪ (a ∩ b)) |
| 12 | | df-id3 51 |
. 2
(a ≡3 (a ∩ b)) =
((a⊥ ∪ (a ∩ b))
∩ (a ∪ (a⊥ ∩ (a ∩ b)⊥ ))) |
| 13 | | df-i1 43 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
| 14 | 11, 12, 13 | 3tr1 60 |
1
(a ≡3 (a ∩ b)) =
(a →1 b) |