Theorem dp35lembb 1177

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
dp35lembb	(b0 ∩ (a0 ∪ p0)) ≤ (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1))))

Proof of Theorem dp35lembb

Step	Hyp	Ref	Expression
1		dp35lem.1	. . 3 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
2		dp35lem.2	. . 3 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
3		dp35lem.3	. . 3 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
4		dp35lem.4	. . 3 p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
5		dp35lem.5	. . 3 p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
6	1, 2, 3, 4, 5	dp35lemd 1174	. 2 (b0 ∩ (a0 ∪ p0)) ≤ (b0 ∩ (((a0 ∩ b0) ∪ b1) ∪ (c2 ∩ (c0 ∪ c1))))
7	1, 2, 3, 4, 5	dp35lemc 1175	. . 3 (b0 ∩ (((a0 ∩ b0) ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))) = (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1))))
8	1, 2, 3, 4, 5	dp35lemb 1176	. . 3 (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))) = (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1))))
9	7, 8	tr 62	. 2 (b0 ∩ (((a0 ∩ b0) ∪ b1) ∪ (c2 ∩ (c0 ∪ c1)))) = (b0 ∩ (b1 ∪ ((a0 ∪ a1) ∩ (c0 ∪ c1))))
10	6, 9	lbtr 139	1 (b0 ∩ (a0 ∪ p0)) ≤ (b0 ∩ (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:  dp35lema  1178