Theorem dp35leme 1173

Description: Part of proof (3)=>(5) in Day/Pickering 1982. (Contributed by NM, 12-Apr-2012.)

Hypotheses

Ref	Expression
dp35lem.1	c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
dp35lem.2	c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
dp35lem.3	c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
dp35lem.4	p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
dp35lem.5	p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))

Assertion

Ref	Expression
dp35leme	(b0 ∩ (a0 ∪ p0)) ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))

Proof of Theorem dp35leme

Step	Hyp	Ref	Expression
1		leor 159	. . 3 b0 ≤ (a0 ∪ b0)
2		dp35lem.4	. . . . 5 p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
3	2	lor 70	. . . 4 (a0 ∪ p0) = (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
4	3	bile 142	. . 3 (a0 ∪ p0) ≤ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
5	1, 4	le2an 169	. 2 (b0 ∩ (a0 ∪ p0)) ≤ ((a0 ∪ b0) ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))
6		ancom 74	. . . . . 6 (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∩ (a0 ∪ b0)) = ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
7		anass 76	. . . . . . 7 (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2)) = ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
8	7	cm 61	. . . . . 6 ((a0 ∪ b0) ∩ ((a1 ∪ b1) ∩ (a2 ∪ b2))) = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
9	6, 8	tr 62	. . . . 5 (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∩ (a0 ∪ b0)) = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
10	9	lor 70	. . . 4 (a0 ∪ (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∩ (a0 ∪ b0))) = (a0 ∪ (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2)))
11		ancom 74	. . . . 5 ((a0 ∪ b0) ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))) = ((a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))) ∩ (a0 ∪ b0))
12		leo 158	. . . . . 6 a0 ≤ (a0 ∪ b0)
13	12	mlduali 1128	. . . . 5 ((a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))) ∩ (a0 ∪ b0)) = (a0 ∪ (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∩ (a0 ∪ b0)))
14	11, 13	tr 62	. . . 4 ((a0 ∪ b0) ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))) = (a0 ∪ (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∩ (a0 ∪ b0)))
15		dp35lem.5	. . . . 5 p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
16	15	lor 70	. . . 4 (a0 ∪ p) = (a0 ∪ (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2)))
17	10, 14, 16	3tr1 63	. . 3 ((a0 ∪ b0) ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))) = (a0 ∪ p)
18		dp35lem.1	. . . 4 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
19		dp35lem.2	. . . 4 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
20		dp35lem.3	. . . 4 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
21	18, 19, 20, 2, 15	dp35lemf 1172	. . 3 (a0 ∪ p) ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))
22	17, 21	bltr 138	. 2 ((a0 ∪ b0) ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))) ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))
23	5, 22	letr 137	1 (b0 ∩ (a0 ∪ p0)) ≤ (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:  dp35lemd  1174