Proof of Theorem vneulem12
Step | Hyp | Ref
| Expression |
1 | | ml 1123 |
. . 3
((c ∩ d) ∪ ((a
∪ b) ∩ ((c ∩ d) ∪
((c ∪ d) ∪ (a
∩ b))))) = (((c ∩ d) ∪
(a ∪ b)) ∩ ((c
∩ d) ∪ ((c ∪ d) ∪
(a ∩ b)))) |
2 | 1 | cm 61 |
. 2
(((c ∩ d) ∪ (a
∪ b)) ∩ ((c ∩ d) ∪
((c ∪ d) ∪ (a
∩ b)))) = ((c ∩ d) ∪
((a ∪ b) ∩ ((c
∩ d) ∪ ((c ∪ d) ∪
(a ∩ b))))) |
3 | | orass 75 |
. . . . 5
(((c ∩ d) ∪ (c
∪ d)) ∪ (a ∩ b)) =
((c ∩ d) ∪ ((c
∪ d) ∪ (a ∩ b))) |
4 | 3 | cm 61 |
. . . 4
((c ∩ d) ∪ ((c
∪ d) ∪ (a ∩ b))) =
(((c ∩ d) ∪ (c
∪ d)) ∪ (a ∩ b)) |
5 | | leao1 162 |
. . . . . 6
(c ∩ d) ≤ (c ∪
d) |
6 | 5 | df-le2 131 |
. . . . 5
((c ∩ d) ∪ (c
∪ d)) = (c ∪ d) |
7 | 6 | ror 71 |
. . . 4
(((c ∩ d) ∪ (c
∪ d)) ∪ (a ∩ b)) =
((c ∪ d) ∪ (a
∩ b)) |
8 | 4, 7 | tr 62 |
. . 3
((c ∩ d) ∪ ((c
∪ d) ∪ (a ∩ b))) =
((c ∪ d) ∪ (a
∩ b)) |
9 | 8 | lan 77 |
. 2
(((c ∩ d) ∪ (a
∪ b)) ∩ ((c ∩ d) ∪
((c ∪ d) ∪ (a
∩ b)))) = (((c ∩ d) ∪
(a ∪ b)) ∩ ((c
∪ d) ∪ (a ∩ b))) |
10 | 8 | lan 77 |
. . 3
((a ∪ b) ∩ ((c
∩ d) ∪ ((c ∪ d) ∪
(a ∩ b)))) = ((a
∪ b) ∩ ((c ∪ d) ∪
(a ∩ b))) |
11 | 10 | lor 70 |
. 2
((c ∩ d) ∪ ((a
∪ b) ∩ ((c ∩ d) ∪
((c ∪ d) ∪ (a
∩ b))))) = ((c ∩ d) ∪
((a ∪ b) ∩ ((c
∪ d) ∪ (a ∩ b)))) |
12 | 2, 9, 11 | 3tr2 64 |
1
(((c ∩ d) ∪ (a
∪ b)) ∩ ((c ∪ d) ∪
(a ∩ b))) = ((c ∩
d) ∪ ((a ∪ b) ∩
((c ∪ d) ∪ (a
∩ b)))) |