Theorem dp15lemg 1160

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))
dp15lemg.4	c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
dp15lemg.5	c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))

Assertion

Ref	Expression
dp15lemg	(((a1 ∪ a2) ∩ (b1 ∪ b2)) ∪ (((a0 ∪ a2) ∩ (b0 ∪ b2)) ∪ (b1 ∩ (a0 ∪ a1)))) = ((c0 ∪ c1) ∪ (b1 ∩ (a0 ∪ a1)))

Proof of Theorem dp15lemg

Step	Hyp	Ref	Expression
1		dp15lemg.4	. . . 4 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
2		dp15lemg.5	. . . . 5 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
3	2	ror 71	. . . 4 (c1 ∪ (b1 ∩ (a0 ∪ a1))) = (((a0 ∪ a2) ∩ (b0 ∪ b2)) ∪ (b1 ∩ (a0 ∪ a1)))
4	1, 3	2or 72	. . 3 (c0 ∪ (c1 ∪ (b1 ∩ (a0 ∪ a1)))) = (((a1 ∪ a2) ∩ (b1 ∪ b2)) ∪ (((a0 ∪ a2) ∩ (b0 ∪ b2)) ∪ (b1 ∩ (a0 ∪ a1))))
5	4	cm 61	. 2 (((a1 ∪ a2) ∩ (b1 ∪ b2)) ∪ (((a0 ∪ a2) ∩ (b0 ∪ b2)) ∪ (b1 ∩ (a0 ∪ a1)))) = (c0 ∪ (c1 ∪ (b1 ∩ (a0 ∪ a1))))
6		orass 75	. . 3 ((c0 ∪ c1) ∪ (b1 ∩ (a0 ∪ a1))) = (c0 ∪ (c1 ∪ (b1 ∩ (a0 ∪ a1))))
7	6	cm 61	. 2 (c0 ∪ (c1 ∪ (b1 ∩ (a0 ∪ a1)))) = ((c0 ∪ c1) ∪ (b1 ∩ (a0 ∪ a1)))
8	5, 7	tr 62	1 (((a1 ∪ a2) ∩ (b1 ∪ b2)) ∪ (((a0 ∪ a2) ∩ (b0 ∪ b2)) ∪ (b1 ∩ (a0 ∪ a1)))) = ((c0 ∪ c1) ∪ (b1 ∩ (a0 ∪ a1)))

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

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

This theorem is referenced by:  dp15lemh  1161