Theorem dp15lemb 1155

Description: Part of proof (1)=>(5) in Day/Pickering 1982. (Contributed by NM, 1-Apr-2012.)

Hypotheses

Ref	Expression
dp15lema.1	d = (a2 ∪ (a0 ∩ (a1 ∪ b1)))
dp15lema.2	p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
dp15lema.3	e = (b0 ∩ (a0 ∪ p0))

Assertion

Ref	Expression
dp15lemb	((a0 ∪ a1) ∩ (e ∪ b1)) ≤ (((a0 ∪ d) ∩ (e ∪ b2)) ∪ ((a1 ∪ d) ∩ (b1 ∪ b2)))

Proof of Theorem dp15lemb

Step	Hyp	Ref	Expression
1		dp15lema.1	. . 3 d = (a2 ∪ (a0 ∩ (a1 ∪ b1)))
2		dp15lema.2	. . 3 p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
3		dp15lema.3	. . 3 e = (b0 ∩ (a0 ∪ p0))
4	1, 2, 3	dp15lema 1154	. 2 ((a0 ∪ e) ∩ (a1 ∪ b1)) ≤ (d ∪ b2)
5	4	ax-arg 1153	1 ((a0 ∪ a1) ∩ (e ∪ b1)) ≤ (((a0 ∪ d) ∩ (e ∪ b2)) ∪ ((a1 ∪ d) ∩ (b1 ∪ b2)))

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:  dp15lemc  1156