Theorem vneulem12 1142

Description: Part of von Neumann's lemma. Lemma 9, Kalmbach p. 96. (Contributed by NM, 31-Mar-2011.)

Assertion

Ref	Expression
vneulem12	(((c ∩ d) ∪ (a ∪ b)) ∩ ((c ∪ d) ∪ (a ∩ b))) = ((c ∩ d) ∪ ((a ∪ b) ∩ ((c ∪ d) ∪ (a ∩ b))))

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))))

This proof depends on syntax axioms:   = wb 1   ∪ wo 6   ∩ wa 7

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-ml 1122

This proof depends on definitions:  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131

This theorem is used by:  vneulem14  1144