Proof of Theorem vneulem11
Step | Hyp | Ref
| Expression |
1 | | ax-a3 32 |
. . . 4
((b ∪ c) ∪ d) =
(b ∪ (c ∪ d)) |
2 | | orcom 73 |
. . . 4
(b ∪ (c ∪ d)) =
((c ∪ d) ∪ b) |
3 | 1, 2 | tr 62 |
. . 3
((b ∪ c) ∪ d) =
((c ∪ d) ∪ b) |
4 | | ax-a2 31 |
. . . . 5
(a ∪ c) = (c ∪
a) |
5 | 4 | ror 71 |
. . . 4
((a ∪ c) ∪ d) =
((c ∪ a) ∪ d) |
6 | | or32 82 |
. . . 4
((c ∪ a) ∪ d) =
((c ∪ d) ∪ a) |
7 | 5, 6 | tr 62 |
. . 3
((a ∪ c) ∪ d) =
((c ∪ d) ∪ a) |
8 | 3, 7 | 2an 79 |
. 2
(((b ∪ c) ∪ d)
∩ ((a ∪ c) ∪ d)) =
(((c ∪ d) ∪ b)
∩ ((c ∪ d) ∪ a)) |
9 | | ancom 74 |
. . . 4
((c ∪ d) ∩ (a
∪ b)) = ((a ∪ b) ∩
(c ∪ d)) |
10 | | vneulem6.1 |
. . . 4
((a ∪ b) ∩ (c
∪ d)) = 0 |
11 | 9, 10 | tr 62 |
. . 3
((c ∪ d) ∩ (a
∪ b)) = 0 |
12 | 11 | vneulem9 1139 |
. 2
(((c ∪ d) ∪ b)
∩ ((c ∪ d) ∪ a)) =
((a ∩ b) ∪ (c
∪ d)) |
13 | | orcom 73 |
. 2
((a ∩ b) ∪ (c
∪ d)) = ((c ∪ d) ∪
(a ∩ b)) |
14 | 8, 12, 13 | 3tr 65 |
1
(((b ∪ c) ∪ d)
∩ ((a ∪ c) ∪ d)) =
((c ∪ d) ∪ (a
∩ b)) |