Theorem dp15lemd 1157

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
dp15lemd	(((a0 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ (b1 ∪ b2))) = (((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ (((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))) ∩ (b1 ∪ b2)))

Proof of Theorem dp15lemd

Step	Hyp	Ref	Expression
1		or12 80	. . . 4 (a0 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) = (a2 ∪ (a0 ∪ (a0 ∩ (a1 ∪ b1))))
2		orabs 120	. . . . 5 (a0 ∪ (a0 ∩ (a1 ∪ b1))) = a0
3	2	lor 70	. . . 4 (a2 ∪ (a0 ∪ (a0 ∩ (a1 ∪ b1)))) = (a2 ∪ a0)
4		orcom 73	. . . 4 (a2 ∪ a0) = (a0 ∪ a2)
5	1, 3, 4	3tr 65	. . 3 (a0 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) = (a0 ∪ a2)
6	5	ran 78	. 2 ((a0 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) = ((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2))
7		orass 75	. . . 4 ((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))) = (a1 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1))))
8	7	ran 78	. . 3 (((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))) ∩ (b1 ∪ b2)) = ((a1 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ (b1 ∪ b2))
9	8	cm 61	. 2 ((a1 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ (b1 ∪ b2)) = (((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))) ∩ (b1 ∪ b2))
10	6, 9	2or 72	1 (((a0 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ ((a1 ∪ (a2 ∪ (a0 ∩ (a1 ∪ b1)))) ∩ (b1 ∪ b2))) = (((a0 ∪ a2) ∩ ((b0 ∩ (a0 ∪ p0)) ∪ b2)) ∪ (((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))) ∩ (b1 ∪ b2)))

Syntax hints:   = wb 1   ∪ wo 6   ∩ wa 7

This theorem was proved from axioms:  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40

This theorem is referenced by:  dp15lemh  1161