Proof of Theorem nom12
| Step | Hyp | Ref
| Expression |
| 1 | | oran 87 |
. . . . . . 7
(a ∪ (a ∩ b)) =
(a⊥ ∩ (a ∩ b)⊥ )⊥ |
| 2 | 1 | ax-r1 35 |
. . . . . 6
(a⊥ ∩ (a ∩ b)⊥ )⊥ = (a ∪ (a ∩
b)) |
| 3 | | a5b 120 |
. . . . . 6
(a ∪ (a ∩ b)) =
a |
| 4 | 2, 3 | ax-r2 36 |
. . . . 5
(a⊥ ∩ (a ∩ b)⊥ )⊥ = a |
| 5 | 4 | con3 68 |
. . . 4
(a⊥ ∩ (a ∩ b)⊥ ) = a⊥ |
| 6 | 5 | lor 70 |
. . 3
((a ∩ b) ∪ (a⊥ ∩ (a ∩ b)⊥ )) = ((a ∩ b) ∪
a⊥ ) |
| 7 | | ax-a2 31 |
. . 3
((a ∩ b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b)) |
| 8 | 6, 7 | ax-r2 36 |
. 2
((a ∩ b) ∪ (a⊥ ∩ (a ∩ b)⊥ )) = (a⊥ ∪ (a ∩ b)) |
| 9 | | df-i2 45 |
. 2
(a →2 (a ∩ b)) =
((a ∩ b) ∪ (a⊥ ∩ (a ∩ b)⊥ )) |
| 10 | | df-i1 44 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
| 11 | 8, 9, 10 | 3tr1 63 |
1
(a →2 (a ∩ b)) =
(a →1 b) |