Theorem dp53lemd 1166

Description: Part of proof (5)=>(3) in Day/Pickering 1982. (Contributed by NM, 3-Apr-2012.)

Hypotheses

Ref	Expression
dp53lem.1	c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
dp53lem.2	c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
dp53lem.3	c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
dp53lem.4	p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
dp53lem.5	p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))

Assertion

Ref	Expression
dp53lemd	(b0 ∩ (a0 ∪ p0)) ≤ (b0 ∩ (((a0 ∩ b0) ∪ b1) ∪ (c2 ∩ (c0 ∪ c1))))

Proof of Theorem dp53lemd

Step	Hyp	Ref	Expression
1		lea 160	. . 3 (b0 ∩ (a0 ∪ p0)) ≤ b0
2		leor 159	. . . 4 (b0 ∩ (a0 ∪ p0)) ≤ (b1 ∪ (b0 ∩ (a0 ∪ p0)))
3		dp53lem.1	. . . . 5 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
4		dp53lem.2	. . . . 5 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
5		dp53lem.3	. . . . 5 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
6		dp53lem.4	. . . . 5 p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
7		dp53lem.5	. . . . 5 p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
8	3, 4, 5, 6, 7	dp53lema 1163	. . . 4 (b1 ∪ (b0 ∩ (a0 ∪ p0))) ≤ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1)))
9	2, 8	letr 137	. . 3 (b0 ∩ (a0 ∪ p0)) ≤ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1)))
10	1, 9	ler2an 173	. 2 (b0 ∩ (a0 ∪ p0)) ≤ (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1))))
11	3, 4, 5, 6, 7	dp53lemc 1165	. . . 4 (b0 ∩ (((a0 ∩ b0) ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))) = (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1))))
12	3, 4, 5, 6, 7	dp53lemb 1164	. . . 4 (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))) = (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1))))
13	11, 12	tr 62	. . 3 (b0 ∩ (((a0 ∩ b0) ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))) = (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1))))
14	13	cm 61	. 2 (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1)))) = (b0 ∩ (((a0 ∩ b0) ∪ b1) ∪ (c2 ∩ (c0 ∪ c1))))
15	10, 14	lbtr 139	1 (b0 ∩ (a0 ∪ p0)) ≤ (b0 ∩ (((a0 ∩ b0) ∪ 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:  dp53leme  1167