Proof of Theorem nom15
Step | Hyp | Ref
| Expression |
1 | | anass 76 |
. . . . . . . 8
((a ∩ a) ∩ b) =
(a ∩ (a ∩ b)) |
2 | 1 | ax-r1 35 |
. . . . . . 7
(a ∩ (a ∩ b)) =
((a ∩ a) ∩ b) |
3 | | anidm 111 |
. . . . . . . 8
(a ∩ a) = a |
4 | 3 | ran 78 |
. . . . . . 7
((a ∩ a) ∩ b) =
(a ∩ b) |
5 | 2, 4 | ax-r2 36 |
. . . . . 6
(a ∩ (a ∩ b)) =
(a ∩ b) |
6 | 5 | ax-r5 38 |
. . . . 5
((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b))) =
((a ∩ b) ∪ (a⊥ ∩ (a ∩ b))) |
7 | | ax-a2 31 |
. . . . 5
((a ∩ b) ∪ (a⊥ ∩ (a ∩ b))) =
((a⊥ ∩ (a ∩ b))
∪ (a ∩ b)) |
8 | | lear 161 |
. . . . . 6
(a⊥ ∩ (a ∩ b)) ≤
(a ∩ b) |
9 | 8 | df-le2 131 |
. . . . 5
((a⊥ ∩
(a ∩ b)) ∪ (a
∩ b)) = (a ∩ b) |
10 | 6, 7, 9 | 3tr 65 |
. . . 4
((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b))) =
(a ∩ b) |
11 | | oran3 93 |
. . . . . . 7
(a⊥ ∪ b⊥ ) = (a ∩ b)⊥ |
12 | 11 | lan 77 |
. . . . . 6
(a⊥ ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∩ (a ∩ b)⊥ ) |
13 | 12 | ax-r1 35 |
. . . . 5
(a⊥ ∩ (a ∩ b)⊥ ) = (a⊥ ∩ (a⊥ ∪ b⊥ )) |
14 | | anabs 121 |
. . . . 5
(a⊥ ∩ (a⊥ ∪ b⊥ )) = a⊥ |
15 | 13, 14 | ax-r2 36 |
. . . 4
(a⊥ ∩ (a ∩ b)⊥ ) = a⊥ |
16 | 10, 15 | 2or 72 |
. . 3
(((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)))
∪ (a⊥ ∩ (a ∩ b)⊥ )) = ((a ∩ b) ∪
a⊥ ) |
17 | | ax-a2 31 |
. . 3
((a ∩ b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b)) |
18 | 16, 17 | ax-r2 36 |
. 2
(((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)))
∪ (a⊥ ∩ (a ∩ b)⊥ )) = (a⊥ ∪ (a ∩ b)) |
19 | | df-i5 48 |
. 2
(a →5 (a ∩ b)) =
(((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)))
∪ (a⊥ ∩ (a ∩ b)⊥ )) |
20 | | df-i1 44 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
21 | 18, 19, 20 | 3tr1 63 |
1
(a →5 (a ∩ b)) =
(a →1 b) |