Theorem dp41lema 1182

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
dp41lema	((a0 ∪ b0) ∩ (a1 ∪ b1)) ≤ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))

Proof of Theorem dp41lema

Step	Hyp	Ref	Expression
1		dp41lem.5	. . . . . . 7 p2 = ((a0 ∪ b0) ∩ (a1 ∪ b1))
2	1	cm 61	. . . . . 6 ((a0 ∪ b0) ∩ (a1 ∪ b1)) = p2
3		dp41lem.6	. . . . . 6 p2 ≤ (a2 ∪ b2)
4	2, 3	bltr 138	. . . . 5 ((a0 ∪ b0) ∩ (a1 ∪ b1)) ≤ (a2 ∪ b2)
5	4	df2le2 136	. . . 4 (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2)) = ((a0 ∪ b0) ∩ (a1 ∪ b1))
6	5	cm 61	. . 3 ((a0 ∪ b0) ∩ (a1 ∪ b1)) = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
7		dp41lem.4	. . . 4 p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
8	7	cm 61	. . 3 (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2)) = p
9	6, 8	tr 62	. 2 ((a0 ∪ b0) ∩ (a1 ∪ b1)) = p
10		dp41lem.1	. . 3 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
11		dp41lem.2	. . 3 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
12		dp41lem.3	. . 3 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
13	10, 11, 12, 7	dp34 1181	. 2 p ≤ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))
14	9, 13	bltr 138	1 ((a0 ∪ b0) ∩ (a1 ∪ b1)) ≤ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))

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-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-ml 1122  ax-arg 1153

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:  dp41lemc  1185