Theorem vneulem13 1143

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
vneulem13	((c ∩ d) ∪ ((a ∪ b) ∩ ((c ∪ d) ∪ (a ∩ b)))) = ((c ∩ d) ∪ (a ∩ b))

Proof of Theorem vneulem13

Step	Hyp	Ref	Expression
1		leao1 162	. . . . . . 7 (a ∩ b) ≤ (a ∪ b)
2		leid 148	. . . . . . 7 (a ∩ b) ≤ (a ∩ b)
3	1, 2	ler2an 173	. . . . . 6 (a ∩ b) ≤ ((a ∪ b) ∩ (a ∩ b))
4		lear 161	. . . . . 6 ((a ∪ b) ∩ (a ∩ b)) ≤ (a ∩ b)
5	3, 4	lebi 145	. . . . 5 (a ∩ b) = ((a ∪ b) ∩ (a ∩ b))
6	5	lor 70	. . . 4 ((c ∪ d) ∪ (a ∩ b)) = ((c ∪ d) ∪ ((a ∪ b) ∩ (a ∩ b)))
7	6	lan 77	. . 3 ((a ∪ b) ∩ ((c ∪ d) ∪ (a ∩ b))) = ((a ∪ b) ∩ ((c ∪ d) ∪ ((a ∪ b) ∩ (a ∩ b))))
8		mldual 1124	. . 3 ((a ∪ b) ∩ ((c ∪ d) ∪ ((a ∪ b) ∩ (a ∩ b)))) = (((a ∪ b) ∩ (c ∪ d)) ∪ ((a ∪ b) ∩ (a ∩ b)))
9		vneulem13.1	. . . . 5 ((a ∪ b) ∩ (c ∪ d)) = 0
10	4, 3	lebi 145	. . . . 5 ((a ∪ b) ∩ (a ∩ b)) = (a ∩ b)
11	9, 10	2or 72	. . . 4 (((a ∪ b) ∩ (c ∪ d)) ∪ ((a ∪ b) ∩ (a ∩ b))) = (0 ∪ (a ∩ b))
12		or0r 103	. . . 4 (0 ∪ (a ∩ b)) = (a ∩ b)
13	11, 12	tr 62	. . 3 (((a ∪ b) ∩ (c ∪ d)) ∪ ((a ∪ b) ∩ (a ∩ b))) = (a ∩ b)
14	7, 8, 13	3tr 65	. 2 ((a ∪ b) ∩ ((c ∪ d) ∪ (a ∩ b))) = (a ∩ b)
15	14	lor 70	1 ((c ∩ d) ∪ ((a ∪ b) ∩ ((c ∪ d) ∪ (a ∩ b)))) = ((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:  vneulem14  1144