Theorem dp35lema 1178

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
dp35lema	(b1 ∪ (b0 ∩ (a0 ∪ p0))) ≤ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1)))

Proof of Theorem dp35lema

Step	Hyp	Ref	Expression
1		leo 158	. 2 b1 ≤ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1)))
2		dp35lem.1	. . . 4 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
3		dp35lem.2	. . . 4 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
4		dp35lem.3	. . . 4 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
5		dp35lem.4	. . . 4 p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
6		dp35lem.5	. . . 4 p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
7	2, 3, 4, 5, 6	dp35lembb 1177	. . 3 (b0 ∩ (a0 ∪ p0)) ≤ (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1))))
8		lear 161	. . 3 (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1)))) ≤ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1)))
9	7, 8	letr 137	. 2 (b0 ∩ (a0 ∪ p0)) ≤ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1)))
10	1, 9	lel2or 170	1 (b1 ∪ (b0 ∩ (a0 ∪ p0))) ≤ (b1 ∪ ((a0 ∪ a1) ∩ (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:  dp35lem0  1179