Theorem vneulem6 1136

Description: Part of von Neumann's lemma. Lemma 9, Kalmbach p. 96. (Contributed by NM, 15-Mar-2010.) (Revised by NM, 31-Mar-2011.)

Hypothesis

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

Assertion

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

Proof of Theorem vneulem6

Step	Hyp	Ref	Expression
1		orcom 73	. . . . . . 7 (a ∪ b) = (b ∪ a)
2	1	ror 71	. . . . . 6 ((a ∪ b) ∪ d) = ((b ∪ a) ∪ d)
3		or32 82	. . . . . 6 ((b ∪ a) ∪ d) = ((b ∪ d) ∪ a)
4	2, 3	tr 62	. . . . 5 ((a ∪ b) ∪ d) = ((b ∪ d) ∪ a)
5		or32 82	. . . . 5 ((b ∪ c) ∪ d) = ((b ∪ d) ∪ c)
6	4, 5	2an 79	. . . 4 (((a ∪ b) ∪ d) ∩ ((b ∪ c) ∪ d)) = (((b ∪ d) ∪ a) ∩ ((b ∪ d) ∪ c))
7		vneulem5 1135	. . . 4 (((b ∪ d) ∪ a) ∩ ((b ∪ d) ∪ c)) = ((b ∪ d) ∪ (((b ∪ d) ∪ a) ∩ c))
8	6, 7	ax-r2 36	. . 3 (((a ∪ b) ∪ d) ∩ ((b ∪ c) ∪ d)) = ((b ∪ d) ∪ (((b ∪ d) ∪ a) ∩ c))
9		leor 159	. . . 4 (b ∪ d) ≤ ((c ∩ a) ∪ (b ∪ d))
10		or32 82	. . . . . . 7 ((b ∪ d) ∪ a) = ((b ∪ a) ∪ d)
11	10	ran 78	. . . . . 6 (((b ∪ d) ∪ a) ∩ c) = (((b ∪ a) ∪ d) ∩ c)
12		ax-a2 31	. . . . . . . . 9 (b ∪ a) = (a ∪ b)
13		ax-a2 31	. . . . . . . . 9 (d ∪ c) = (c ∪ d)
14	12, 13	2an 79	. . . . . . . 8 ((b ∪ a) ∩ (d ∪ c)) = ((a ∪ b) ∩ (c ∪ d))
15		vneulem6.1	. . . . . . . 8 ((a ∪ b) ∩ (c ∪ d)) = 0
16	14, 15	tr 62	. . . . . . 7 ((b ∪ a) ∩ (d ∪ c)) = 0
17	16	vneulem4 1134	. . . . . 6 (((b ∪ a) ∪ d) ∩ c) = (d ∩ c)
18	11, 17	tr 62	. . . . 5 (((b ∪ d) ∪ a) ∩ c) = (d ∩ c)
19		leao3 164	. . . . . 6 (d ∩ c) ≤ (b ∪ d)
20	19	lerr 150	. . . . 5 (d ∩ c) ≤ ((c ∩ a) ∪ (b ∪ d))
21	18, 20	bltr 138	. . . 4 (((b ∪ d) ∪ a) ∩ c) ≤ ((c ∩ a) ∪ (b ∪ d))
22	9, 21	lel2or 170	. . 3 ((b ∪ d) ∪ (((b ∪ d) ∪ a) ∩ c)) ≤ ((c ∩ a) ∪ (b ∪ d))
23	8, 22	bltr 138	. 2 (((a ∪ b) ∪ d) ∩ ((b ∪ c) ∪ d)) ≤ ((c ∩ a) ∪ (b ∪ d))
24		leao2 163	. . . . 5 (c ∩ a) ≤ (a ∪ b)
25	24	ler 149	. . . 4 (c ∩ a) ≤ ((a ∪ b) ∪ d)
26		leor 159	. . . . 5 b ≤ (a ∪ b)
27	26	leror 152	. . . 4 (b ∪ d) ≤ ((a ∪ b) ∪ d)
28	25, 27	lel2or 170	. . 3 ((c ∩ a) ∪ (b ∪ d)) ≤ ((a ∪ b) ∪ d)
29		leao3 164	. . . . 5 (c ∩ a) ≤ (b ∪ c)
30	29	ler 149	. . . 4 (c ∩ a) ≤ ((b ∪ c) ∪ d)
31		leo 158	. . . . 5 b ≤ (b ∪ c)
32	31	leror 152	. . . 4 (b ∪ d) ≤ ((b ∪ c) ∪ d)
33	30, 32	lel2or 170	. . 3 ((c ∩ a) ∪ (b ∪ d)) ≤ ((b ∪ c) ∪ d)
34	28, 33	ler2an 173	. 2 ((c ∩ a) ∪ (b ∪ d)) ≤ (((a ∪ b) ∪ d) ∩ ((b ∪ c) ∪ d))
35	23, 34	lebi 145	1 (((a ∪ b) ∪ d) ∩ ((b ∪ c) ∪ d)) = ((c ∩ a) ∪ (b ∪ 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:  vneulem8  1138