Proof of Theorem oml5
Step | Hyp | Ref
| Expression |
1 | | oml 445 |
. . 3
((a ∩ b) ∪ ((a
∩ b)⊥ ∩ ((a ∩ b) ∪
(b ∪ c)))) = ((a
∩ b) ∪ (b ∪ c)) |
2 | | ax-a3 32 |
. . . . . 6
((b ∪ (a ∩ b))
∪ c) = (b ∪ ((a
∩ b) ∪ c)) |
3 | | ancom 74 |
. . . . . . . . 9
(a ∩ b) = (b ∩
a) |
4 | 3 | lor 70 |
. . . . . . . 8
(b ∪ (a ∩ b)) =
(b ∪ (b ∩ a)) |
5 | | orabs 120 |
. . . . . . . 8
(b ∪ (b ∩ a)) =
b |
6 | 4, 5 | ax-r2 36 |
. . . . . . 7
(b ∪ (a ∩ b)) =
b |
7 | 6 | ax-r5 38 |
. . . . . 6
((b ∪ (a ∩ b))
∪ c) = (b ∪ c) |
8 | | or12 80 |
. . . . . 6
(b ∪ ((a ∩ b) ∪
c)) = ((a ∩ b) ∪
(b ∪ c)) |
9 | 2, 7, 8 | 3tr2 64 |
. . . . 5
(b ∪ c) = ((a ∩
b) ∪ (b ∪ c)) |
10 | 9 | lan 77 |
. . . 4
((a ∩ b)⊥ ∩ (b ∪ c)) =
((a ∩ b)⊥ ∩ ((a ∩ b) ∪
(b ∪ c))) |
11 | 10 | lor 70 |
. . 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) |