Theorem dp53lemf 1168

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
dp53lemf	(a0 ∪ p) ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))

Proof of Theorem dp53lemf

Step	Hyp	Ref	Expression
1		leo 158	. 2 a0 ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))
2		dp53lem.5	. . . . . . 7 p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
3		anass 76	. . . . . . 7 (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2)) = ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
4	2, 3	tr 62	. . . . . 6 p = ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
5		dp53lem.4	. . . . . . . . . 10 p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
6	5	lan 77	. . . . . . . . 9 ((a0 ∪ b0) ∩ p0) = ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
7	6	cm 61	. . . . . . . 8 ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2))) = ((a0 ∪ b0) ∩ p0)
8		leao4 165	. . . . . . . 8 ((a0 ∪ b0) ∩ p0) ≤ (a0 ∪ p0)
9	7, 8	bltr 138	. . . . . . 7 ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2))) ≤ (a0 ∪ p0)
10		lea 160	. . . . . . . 8 ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2))) ≤ (a0 ∪ b0)
11		orcom 73	. . . . . . . 8 (a0 ∪ b0) = (b0 ∪ a0)
12	10, 11	lbtr 139	. . . . . . 7 ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2))) ≤ (b0 ∪ a0)
13	9, 12	ler2an 173	. . . . . 6 ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2))) ≤ ((a0 ∪ p0) ∩ (b0 ∪ a0))
14	4, 13	bltr 138	. . . . 5 p ≤ ((a0 ∪ p0) ∩ (b0 ∪ a0))
15		leo 158	. . . . . . 7 a0 ≤ (a0 ∪ p0)
16	15	mldual2i 1127	. . . . . 6 ((a0 ∪ p0) ∩ (b0 ∪ a0)) = (((a0 ∪ p0) ∩ b0) ∪ a0)
17		ancom 74	. . . . . . 7 ((a0 ∪ p0) ∩ b0) = (b0 ∩ (a0 ∪ p0))
18	17	ror 71	. . . . . 6 (((a0 ∪ p0) ∩ b0) ∪ a0) = ((b0 ∩ (a0 ∪ p0)) ∪ a0)
19	16, 18	tr 62	. . . . 5 ((a0 ∪ p0) ∩ (b0 ∪ a0)) = ((b0 ∩ (a0 ∪ p0)) ∪ a0)
20	14, 19	lbtr 139	. . . 4 p ≤ ((b0 ∩ (a0 ∪ p0)) ∪ a0)
21	1	lelor 166	. . . 4 ((b0 ∩ (a0 ∪ p0)) ∪ a0) ≤ ((b0 ∩ (a0 ∪ p0)) ∪ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1))))))
22	20, 21	letr 137	. . 3 p ≤ ((b0 ∩ (a0 ∪ p0)) ∪ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1))))))
23		dp53lem.1	. . . . 5 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
24		dp53lem.2	. . . . 5 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
25		dp53lem.3	. . . . 5 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
26	23, 24, 25, 5, 2	dp53leme 1167	. . . 4 (b0 ∩ (a0 ∪ p0)) ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))
27	26	df-le2 131	. . 3 ((b0 ∩ (a0 ∪ p0)) ∪ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))) = (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))
28	22, 27	lbtr 139	. 2 p ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))
29	1, 28	lel2or 170	1 (a0 ∪ p) ≤ (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:  dp53lemg  1169