Theorem oadp35lemg 1209

Description: Part of proof (3)=>(5) in Day/Pickering 1982. (Contributed by NM, 12-Jul-2015.)

Hypotheses

Ref	Expression
oadp35lem.1	c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
oadp35lem.2	c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
oadp35lem.3	c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
oadp35lem.4	p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
oadp35lem.5	p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))

Assertion

Ref	Expression
oadp35lemg	p ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (c0 ∪ c1)))))

Proof of Theorem oadp35lemg

Step	Hyp	Ref	Expression
1		oadp35lem.1	. 2 c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
2		oadp35lem.2	. 2 c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
3		oadp35lem.3	. 2 c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
4		oadp35lem.5	. 2 p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
5	1, 2, 3, 4	dp53 1170	1 p ≤ (a0 ∪ (b0 ∩ (b1 ∪ (c2 ∩ (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:  oadp35lemf  1210