Proof of Theorem oml5a
Step | Hyp | Ref
| Expression |
1 | | omla 447 |
. . 3
((a ∪ b) ∩ ((a
∪ b)⊥ ∪ ((a ∪ b) ∩
(b ∩ c)))) = ((a
∪ b) ∩ (b ∩ c)) |
2 | | anass 76 |
. . . . . 6
((b ∩ (a ∪ b))
∩ c) = (b ∩ ((a
∪ b) ∩ c)) |
3 | | ax-a2 31 |
. . . . . . . . 9
(a ∪ b) = (b ∪
a) |
4 | 3 | lan 77 |
. . . . . . . 8
(b ∩ (a ∪ b)) =
(b ∩ (b ∪ a)) |
5 | | anabs 121 |
. . . . . . . 8
(b ∩ (b ∪ a)) =
b |
6 | 4, 5 | ax-r2 36 |
. . . . . . 7
(b ∩ (a ∪ b)) =
b |
7 | 6 | ran 78 |
. . . . . 6
((b ∩ (a ∪ b))
∩ c) = (b ∩ c) |
8 | | an12 81 |
. . . . . 6
(b ∩ ((a ∪ b) ∩
c)) = ((a ∪ b) ∩
(b ∩ c)) |
9 | 2, 7, 8 | 3tr2 64 |
. . . . 5
(b ∩ c) = ((a ∪
b) ∩ (b ∩ c)) |
10 | 9 | lor 70 |
. . . 4
((a ∪ b)⊥ ∪ (b ∩ c)) =
((a ∪ b)⊥ ∪ ((a ∪ b) ∩
(b ∩ c))) |
11 | 10 | lan 77 |
. . 3
((a ∪ b) ∩ ((a
∪ b)⊥ ∪ (b ∩ c))) =
((a ∪ b) ∩ ((a
∪ b)⊥ ∪ ((a ∪ b) ∩
(b ∩ c)))) |
12 | 2, 8 | ax-r2 36 |
. . 3
((b ∩ (a ∪ b))
∩ c) = ((a ∪ b) ∩
(b ∩ c)) |
13 | 1, 11, 12 | 3tr1 63 |
. 2
((a ∪ b) ∩ ((a
∪ b)⊥ ∪ (b ∩ c))) =
((b ∩ (a ∪ b))
∩ c) |
14 | 13, 7 | ax-r2 36 |
1
((a ∪ b) ∩ ((a
∪ b)⊥ ∪ (b ∩ c))) =
(b ∩ c) |