Theorem dp41lemg 1189

Description: Part of proof (4)=>(1) in Day/Pickering 1982. (Contributed by NM, 3-Apr-2012.)

Hypotheses

Ref	Expression
dp41lem.1	c0 = ((a1 ∪ a2) ∩ (b1 ∪ b2))
dp41lem.2	c1 = ((a0 ∪ a2) ∩ (b0 ∪ b2))
dp41lem.3	c2 = ((a0 ∪ a1) ∩ (b0 ∪ b1))
dp41lem.4	p = (((a0 ∪ b0) ∩ (a1 ∪ b1)) ∩ (a2 ∪ b2))
dp41lem.5	p2 = ((a0 ∪ b0) ∩ (a1 ∪ b1))
dp41lem.6	p2 ≤ (a2 ∪ b2)

Assertion

Ref	Expression
dp41lemg	(((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (b1 ∩ (a0 ∪ a1)))) ∪ ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (a0 ∩ (b0 ∪ b1))))) = (((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1)))) ∪ ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (b1 ∩ (a0 ∪ b0)))))

Proof of Theorem dp41lemg

Step	Hyp	Ref	Expression
1		or32 82	. . . 4 ((a1 ∪ a2) ∪ (b1 ∩ (a0 ∪ a1))) = ((a1 ∪ (b1 ∩ (a0 ∪ a1))) ∪ a2)
2		ml3 1130	. . . . . 6 (a1 ∪ (b1 ∩ (a0 ∪ a1))) = (a1 ∪ (a0 ∩ (b1 ∪ a1)))
3		orcom 73	. . . . . . . 8 (b1 ∪ a1) = (a1 ∪ b1)
4	3	lan 77	. . . . . . 7 (a0 ∩ (b1 ∪ a1)) = (a0 ∩ (a1 ∪ b1))
5	4	lor 70	. . . . . 6 (a1 ∪ (a0 ∩ (b1 ∪ a1))) = (a1 ∪ (a0 ∩ (a1 ∪ b1)))
6	2, 5	tr 62	. . . . 5 (a1 ∪ (b1 ∩ (a0 ∪ a1))) = (a1 ∪ (a0 ∩ (a1 ∪ b1)))
7	6	ror 71	. . . 4 ((a1 ∪ (b1 ∩ (a0 ∪ a1))) ∪ a2) = ((a1 ∪ (a0 ∩ (a1 ∪ b1))) ∪ a2)
8		or32 82	. . . 4 ((a1 ∪ (a0 ∩ (a1 ∪ b1))) ∪ a2) = ((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1)))
9	1, 7, 8	3tr 65	. . 3 ((a1 ∪ a2) ∪ (b1 ∩ (a0 ∪ a1))) = ((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1)))
10	9	lan 77	. 2 ((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (b1 ∩ (a0 ∪ a1)))) = ((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1))))
11		or32 82	. . . 4 ((b0 ∪ b2) ∪ (a0 ∩ (b0 ∪ b1))) = ((b0 ∪ (a0 ∩ (b0 ∪ b1))) ∪ b2)
12		orcom 73	. . . . . . . 8 (b0 ∪ b1) = (b1 ∪ b0)
13	12	lan 77	. . . . . . 7 (a0 ∩ (b0 ∪ b1)) = (a0 ∩ (b1 ∪ b0))
14	13	lor 70	. . . . . 6 (b0 ∪ (a0 ∩ (b0 ∪ b1))) = (b0 ∪ (a0 ∩ (b1 ∪ b0)))
15		ml3 1130	. . . . . 6 (b0 ∪ (a0 ∩ (b1 ∪ b0))) = (b0 ∪ (b1 ∩ (a0 ∪ b0)))
16	14, 15	tr 62	. . . . 5 (b0 ∪ (a0 ∩ (b0 ∪ b1))) = (b0 ∪ (b1 ∩ (a0 ∪ b0)))
17	16	ror 71	. . . 4 ((b0 ∪ (a0 ∩ (b0 ∪ b1))) ∪ b2) = ((b0 ∪ (b1 ∩ (a0 ∪ b0))) ∪ b2)
18		or32 82	. . . 4 ((b0 ∪ (b1 ∩ (a0 ∪ b0))) ∪ b2) = ((b0 ∪ b2) ∪ (b1 ∩ (a0 ∪ b0)))
19	11, 17, 18	3tr 65	. . 3 ((b0 ∪ b2) ∪ (a0 ∩ (b0 ∪ b1))) = ((b0 ∪ b2) ∪ (b1 ∩ (a0 ∪ b0)))
20	19	lan 77	. 2 ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (a0 ∩ (b0 ∪ b1)))) = ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (b1 ∩ (a0 ∪ b0))))
21	10, 20	2or 72	1 (((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (b1 ∩ (a0 ∪ a1)))) ∪ ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (a0 ∩ (b0 ∪ b1))))) = (((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (a0 ∩ (a1 ∪ b1)))) ∪ ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (b1 ∩ (a0 ∪ b0)))))

This proof depends on syntax axioms:   = wb 1   ≤ wle 2   ∪ wo 6   ∩ wa 7

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-ml 1122

This proof depends on definitions:  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131

This theorem is used by:  dp41lemm  1194