Theorem vneulem16 1146

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

Hypothesis

Ref	Expression
vneulem13.1	((a ∪ b) ∩ (c ∪ d)) = 0

Assertion

Ref	Expression
vneulem16	((((a ∪ b) ∪ c) ∩ ((a ∪ c) ∪ d)) ∩ (((a ∪ b) ∪ d) ∩ ((b ∪ c) ∪ d))) = ((a ∩ b) ∪ (c ∩ d))

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))

This proof depends on syntax axioms:   = wb 1   ∪ wo 6   ∩ wa 7  0wf 9

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:  vneulem  1147