Theorem dp41lemc 1185

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
dp41lemc	((c2 ∩ ((a0 ∪ b0) ∪ b1)) ∩ ((a0 ∪ a1) ∪ b1)) ≤ (c2 ∩ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1))))

Proof of Theorem dp41lemc

Step	Hyp	Ref	Expression
1		anass 76	. 2 ((c2 ∩ ((a0 ∪ b0) ∪ b1)) ∩ ((a0 ∪ a1) ∪ b1)) = (c2 ∩ (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1)))
2		dp41lem.1	. . . . 5 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
3		dp41lem.2	. . . . 5 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
4		dp41lem.3	. . . . 5 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
5		dp41lem.4	. . . . 5 p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
6		dp41lem.5	. . . . 5 p2 = ((a0 ∪ b0) ∩ (a1 ∪ b1))
7		dp41lem.6	. . . . 5 p2 ≤ (a2 ∪ b2)
8	2, 3, 4, 5, 6, 7	dp41lemc0 1184	. . . 4 (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1)) = ((a0 ∪ b1) ∪ ((a0 ∪ b0) ∩ (a1 ∪ b1)))
9		leo 158	. . . . 5 (a0 ∪ b1) ≤ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))
10	2, 3, 4, 5, 6, 7	dp41lema 1182	. . . . 5 ((a0 ∪ b0) ∩ (a1 ∪ b1)) ≤ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))
11	9, 10	lel2or 170	. . . 4 ((a0 ∪ b1) ∪ ((a0 ∪ b0) ∩ (a1 ∪ b1))) ≤ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))
12	8, 11	bltr 138	. . 3 (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1)) ≤ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))
13	12	lelan 167	. 2 (c2 ∩ (((a0 ∪ b0) ∪ b1) ∩ ((a0 ∪ a1) ∪ b1))) ≤ (c2 ∩ ((a0 ∪ b1) ∪ (c2 ∩ (c0 ∪ c1))))
14	1, 13	bltr 138	1 ((c2 ∩ ((a0 ∪ b0) ∪ b1)) ∩ ((a0 ∪ a1) ∪ b1)) ≤ (c2 ∩ ((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:  dp41lemm  1194