Proof of Theorem nom13
Step | Hyp | Ref
| Expression |
1 | | oran 87 |
. . . . . . . . 9
(a ∪ (a ∩ b)) =
(a⊥ ∩ (a ∩ b)⊥
)⊥ |
2 | 1 | ax-r1 35 |
. . . . . . . 8
(a⊥ ∩ (a ∩ b)⊥ )⊥ = (a ∪ (a ∩
b)) |
3 | | orabs 120 |
. . . . . . . 8
(a ∪ (a ∩ b)) =
a |
4 | 2, 3 | ax-r2 36 |
. . . . . . 7
(a⊥ ∩ (a ∩ b)⊥ )⊥ = a |
5 | 4 | con3 68 |
. . . . . 6
(a⊥ ∩ (a ∩ b)⊥ ) = a⊥ |
6 | 5 | lor 70 |
. . . . 5
((a⊥ ∩
(a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ )) = ((a⊥ ∩ (a ∩ b))
∪ a⊥
) |
7 | | lea 160 |
. . . . . 6
(a⊥ ∩ (a ∩ b)) ≤
a⊥ |
8 | 7 | df-le2 131 |
. . . . 5
((a⊥ ∩
(a ∩ b)) ∪ a⊥ ) = a⊥ |
9 | 6, 8 | ax-r2 36 |
. . . 4
((a⊥ ∩
(a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ )) = a⊥ |
10 | 9 | ax-r5 38 |
. . 3
(((a⊥ ∩
(a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ )) ∪ (a ∩ (a⊥ ∪ (a ∩ b)))) =
(a⊥ ∪ (a ∩ (a⊥ ∪ (a ∩ b)))) |
11 | | womaa 222 |
. . 3
(a⊥ ∪ (a ∩ (a⊥ ∪ (a ∩ b)))) =
(a⊥ ∪ (a ∩ b)) |
12 | 10, 11 | ax-r2 36 |
. 2
(((a⊥ ∩
(a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ )) ∪ (a ∩ (a⊥ ∪ (a ∩ b)))) =
(a⊥ ∪ (a ∩ b)) |
13 | | df-i3 46 |
. 2
(a →3 (a ∩ b)) =
(((a⊥ ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)⊥ )) ∪ (a ∩ (a⊥ ∪ (a ∩ b)))) |
14 | | df-i1 44 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
15 | 12, 13, 14 | 3tr1 63 |
1
(a →3 (a ∩ b)) =
(a →1 b) |