Theorem dp41lemb 1183

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
dp41lemb	c2 = ((c2 ∩ ((a0 ∪ b0) ∪ b1)) ∩ ((a0 ∪ a1) ∪ b1))

Proof of Theorem dp41lemb

Step	Hyp	Ref	Expression
1		dp41lem.3	. . . . . 6 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
2		ancom 74	. . . . . 6 ((a0 ∪ a1) ∩ (b0 ∪ b1)) = ((b0 ∪ b1) ∩ (a0 ∪ a1))
3	1, 2	tr 62	. . . . 5 c2 = ((b0 ∪ b1) ∩ (a0 ∪ a1))
4		leor 159	. . . . . . 7 b0 ≤ (a0 ∪ b0)
5	4	leror 152	. . . . . 6 (b0 ∪ b1) ≤ ((a0 ∪ b0) ∪ b1)
6		leo 158	. . . . . 6 (a0 ∪ a1) ≤ ((a0 ∪ a1) ∪ b1)
7	5, 6	le2an 169	. . . . 5 ((b0 ∪ b1) ∩ (a0 ∪ a1)) ≤ (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1))
8	3, 7	bltr 138	. . . 4 c2 ≤ (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1))
9	8	df2le2 136	. . 3 (c2 ∩ (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1))) = c2
10	9	cm 61	. 2 c2 = (c2 ∩ (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1)))
11		anass 76	. . 3 ((c2 ∩ ((a0 ∪ b0) ∪ b1)) ∩ ((a0 ∪ a1) ∪ b1)) = (c2 ∩ (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1)))
12	11	cm 61	. 2 (c2 ∩ (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1))) = ((c2 ∩ ((a0 ∪ b0) ∪ b1)) ∩ ((a0 ∪ a1) ∪ b1))
13	10, 12	tr 62	1 c2 = ((c2 ∩ ((a0 ∪ b0) ∪ b1)) ∩ ((a0 ∪ a1) ∪ b1))

This proof depends on syntax axioms:   = wb 1   ≤ wle 2   ∪ wo 6   ∩ wa 7

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

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:  dp41lemm  1194