Proof of Theorem vneulem16
Step | Hyp | Ref
| Expression |
1 | | ancom 74 |
. 2
((((a ∪ b) ∪ c)
∩ ((a ∪ c) ∪ d))
∩ (((a ∪ b) ∪ d)
∩ ((b ∪ c) ∪ d))) =
((((a ∪ b) ∪ d)
∩ ((b ∪ c) ∪ d))
∩ (((a ∪ b) ∪ c)
∩ ((a ∪ c) ∪ d))) |
2 | | an4 86 |
. . 3
((((a ∪ b) ∪ d)
∩ ((b ∪ c) ∪ d))
∩ (((a ∪ b) ∪ c)
∩ ((a ∪ c) ∪ d))) =
((((a ∪ b) ∪ d)
∩ ((a ∪ b) ∪ c))
∩ (((b ∪ c) ∪ d)
∩ ((a ∪ c) ∪ d))) |
3 | | vneulem13.1 |
. . . . 5
((a ∪ b) ∩ (c
∪ d)) = 0 |
4 | 3 | vneulem9 1139 |
. . . 4
(((a ∪ b) ∪ d)
∩ ((a ∪ b) ∪ c)) =
((c ∩ d) ∪ (a
∪ b)) |
5 | 3 | vneulem11 1141 |
. . . 4
(((b ∪ c) ∪ d)
∩ ((a ∪ c) ∪ d)) =
((c ∪ d) ∪ (a
∩ b)) |
6 | 4, 5 | 2an 79 |
. . 3
((((a ∪ b) ∪ d)
∩ ((a ∪ b) ∪ c))
∩ (((b ∪ c) ∪ d)
∩ ((a ∪ c) ∪ d))) =
(((c ∩ d) ∪ (a
∪ b)) ∩ ((c ∪ d) ∪
(a ∩ b))) |
7 | 2, 6 | tr 62 |
. 2
((((a ∪ b) ∪ d)
∩ ((b ∪ c) ∪ d))
∩ (((a ∪ b) ∪ c)
∩ ((a ∪ c) ∪ d))) =
(((c ∩ d) ∪ (a
∪ b)) ∩ ((c ∪ d) ∪
(a ∩ b))) |
8 | 3 | vneulem14 1144 |
. . 3
(((c ∩ d) ∪ (a
∪ b)) ∩ ((c ∪ d) ∪
(a ∩ b))) = ((c ∩
d) ∪ (a ∩ b)) |
9 | | orcom 73 |
. . 3
((c ∩ d) ∪ (a
∩ b)) = ((a ∩ b) ∪
(c ∩ d)) |
10 | 8, 9 | tr 62 |
. 2
(((c ∩ d) ∪ (a
∪ b)) ∩ ((c ∪ d) ∪
(a ∩ b))) = ((a ∩
b) ∪ (c ∩ d)) |
11 | 1, 7, 10 | 3tr 65 |
1
((((a ∪ b) ∪ c)
∩ ((a ∪ c) ∪ d))
∩ (((a ∪ b) ∪ d)
∩ ((b ∪ c) ∪ d))) =
((a ∩ b) ∪ (c
∩ d)) |