Theorem vneulem9 1139

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

Hypothesis

Ref	Expression
vneulem6.1	((a ∪ b) ∩ (c ∪ d)) = 0

Assertion

Ref	Expression
vneulem9	(((a ∪ b) ∪ d) ∩ ((a ∪ b) ∪ c)) = ((c ∩ d) ∪ (a ∪ b))

Proof of Theorem vneulem9

Step	Hyp	Ref	Expression
1		ancom 74	. . 3 (((a ∪ b) ∪ d) ∩ ((a ∪ b) ∪ c)) = (((a ∪ b) ∪ c) ∩ ((a ∪ b) ∪ d))
2		vneulem5 1135	. . 3 (((a ∪ b) ∪ c) ∩ ((a ∪ b) ∪ d)) = ((a ∪ b) ∪ (((a ∪ b) ∪ c) ∩ d))
3	1, 2	ax-r2 36	. 2 (((a ∪ b) ∪ d) ∩ ((a ∪ b) ∪ c)) = ((a ∪ b) ∪ (((a ∪ b) ∪ c) ∩ d))
4		orcom 73	. 2 ((a ∪ b) ∪ (((a ∪ b) ∪ c) ∩ d)) = ((((a ∪ b) ∪ c) ∩ d) ∪ (a ∪ b))
5		vneulem6.1	. . . 4 ((a ∪ b) ∩ (c ∪ d)) = 0
6	5	vneulem4 1134	. . 3 (((a ∪ b) ∪ c) ∩ d) = (c ∩ d)
7	6	ror 71	. 2 ((((a ∪ b) ∪ c) ∩ d) ∪ (a ∪ b)) = ((c ∩ d) ∪ (a ∪ b))
8	3, 4, 7	3tr 65	1 (((a ∪ b) ∪ d) ∩ ((a ∪ b) ∪ c)) = ((c ∩ d) ∪ (a ∪ b))

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:  vneulem11  1141  vneulem16  1146