Theorem dp41lemd 1186

Description: Part of proof (4)=>(1) in Day/Pickering 1982. (Contributed by NM, 3-Apr-2012.)

Hypotheses

Ref	Expression
dp41lem.1	c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
dp41lem.2	c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
dp41lem.3	c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
dp41lem.4	p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
dp41lem.5	p2 = ((a0 ∪ b0) ∩ (a1 ∪ b1))
dp41lem.6	p2 ≤ (a2 ∪ b2)

Assertion

Ref	Expression
dp41lemd	(c2 ∩ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))) = (c2 ∩ ((c0 ∪ c1) ∪ (c2 ∩ (a0 ∪ b1))))

Proof of Theorem dp41lemd

Step	Hyp	Ref	Expression
1		mldual 1124	. 2 (c2 ∩ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))) = ((c2 ∩ (a0 ∪ b1)) ∪ (c2 ∩ (c0 ∪ c1)))
2		ancom 74	. . 3 (c2 ∩ (c0 ∪ c1)) = ((c0 ∪ c1) ∩ c2)
3	2	lor 70	. 2 ((c2 ∩ (a0 ∪ b1)) ∪ (c2 ∩ (c0 ∪ c1))) = ((c2 ∩ (a0 ∪ b1)) ∪ ((c0 ∪ c1) ∩ c2))
4		lea 160	. . . 4 (c2 ∩ (a0 ∪ b1)) ≤ c2
5	4	ml2i 1125	. . 3 ((c2 ∩ (a0 ∪ b1)) ∪ ((c0 ∪ c1) ∩ c2)) = (((c2 ∩ (a0 ∪ b1)) ∪ (c0 ∪ c1)) ∩ c2)
6		ancom 74	. . 3 (((c2 ∩ (a0 ∪ b1)) ∪ (c0 ∪ c1)) ∩ c2) = (c2 ∩ ((c2 ∩ (a0 ∪ b1)) ∪ (c0 ∪ c1)))
7		ax-a2 31	. . . 4 ((c2 ∩ (a0 ∪ b1)) ∪ (c0 ∪ c1)) = ((c0 ∪ c1) ∪ (c2 ∩ (a0 ∪ b1)))
8	7	lan 77	. . 3 (c2 ∩ ((c2 ∩ (a0 ∪ b1)) ∪ (c0 ∪ c1))) = (c2 ∩ ((c0 ∪ c1) ∪ (c2 ∩ (a0 ∪ b1))))
9	5, 6, 8	3tr 65	. 2 ((c2 ∩ (a0 ∪ b1)) ∪ ((c0 ∪ c1) ∩ c2)) = (c2 ∩ ((c0 ∪ c1) ∪ (c2 ∩ (a0 ∪ b1))))
10	1, 3, 9	3tr 65	1 (c2 ∩ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))) = (c2 ∩ ((c0 ∪ c1) ∪ (c2 ∩ (a0 ∪ b1))))

Syntax hints:   = wb 1   ≤ wle 2   ∪ wo 6   ∩ wa 7

This theorem was proved from 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 theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131

This theorem is referenced by:  dp41lemm  1194