Theorem dp41lemj 1191

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
dp41lemj	(((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (a0 ∩ (a2 ∪ b2)))) ∪ ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (b1 ∩ (a2 ∪ b2))))) = (((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (b2 ∩ (a0 ∪ a2)))) ∪ ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (a2 ∩ (b1 ∪ b2)))))

Proof of Theorem dp41lemj

Step	Hyp	Ref	Expression
1		ax-a2 31	. . . . . . . 8 (a2 ∪ b2) = (b2 ∪ a2)
2	1	lan 77	. . . . . . 7 (a0 ∩ (a2 ∪ b2)) = (a0 ∩ (b2 ∪ a2))
3	2	lor 70	. . . . . 6 (a2 ∪ (a0 ∩ (a2 ∪ b2))) = (a2 ∪ (a0 ∩ (b2 ∪ a2)))
4		ml3 1130	. . . . . 6 (a2 ∪ (a0 ∩ (b2 ∪ a2))) = (a2 ∪ (b2 ∩ (a0 ∪ a2)))
5	3, 4	tr 62	. . . . 5 (a2 ∪ (a0 ∩ (a2 ∪ b2))) = (a2 ∪ (b2 ∩ (a0 ∪ a2)))
6	5	lor 70	. . . 4 (a1 ∪ (a2 ∪ (a0 ∩ (a2 ∪ b2)))) = (a1 ∪ (a2 ∪ (b2 ∩ (a0 ∪ a2))))
7		orass 75	. . . 4 ((a1 ∪ a2) ∪ (a0 ∩ (a2 ∪ b2))) = (a1 ∪ (a2 ∪ (a0 ∩ (a2 ∪ b2))))
8		orass 75	. . . 4 ((a1 ∪ a2) ∪ (b2 ∩ (a0 ∪ a2))) = (a1 ∪ (a2 ∪ (b2 ∩ (a0 ∪ a2))))
9	6, 7, 8	3tr1 63	. . 3 ((a1 ∪ a2) ∪ (a0 ∩ (a2 ∪ b2))) = ((a1 ∪ a2) ∪ (b2 ∩ (a0 ∪ a2)))
10	9	lan 77	. 2 ((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (a0 ∩ (a2 ∪ b2)))) = ((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (b2 ∩ (a0 ∪ a2))))
11		ml3 1130	. . . . 5 (b2 ∪ (b1 ∩ (a2 ∪ b2))) = (b2 ∪ (a2 ∩ (b1 ∪ b2)))
12	11	lor 70	. . . 4 (b0 ∪ (b2 ∪ (b1 ∩ (a2 ∪ b2)))) = (b0 ∪ (b2 ∪ (a2 ∩ (b1 ∪ b2))))
13		orass 75	. . . 4 ((b0 ∪ b2) ∪ (b1 ∩ (a2 ∪ b2))) = (b0 ∪ (b2 ∪ (b1 ∩ (a2 ∪ b2))))
14		orass 75	. . . 4 ((b0 ∪ b2) ∪ (a2 ∩ (b1 ∪ b2))) = (b0 ∪ (b2 ∪ (a2 ∩ (b1 ∪ b2))))
15	12, 13, 14	3tr1 63	. . 3 ((b0 ∪ b2) ∪ (b1 ∩ (a2 ∪ b2))) = ((b0 ∪ b2) ∪ (a2 ∩ (b1 ∪ b2)))
16	15	lan 77	. 2 ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (b1 ∩ (a2 ∪ b2)))) = ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (a2 ∩ (b1 ∪ b2))))
17	10, 16	2or 72	1 (((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (a0 ∩ (a2 ∪ b2)))) ∪ ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (b1 ∩ (a2 ∪ b2))))) = (((b1 ∪ b2) ∩ ((a1 ∪ a2) ∪ (b2 ∩ (a0 ∪ a2)))) ∪ ((a0 ∪ a2) ∩ ((b0 ∪ b2) ∪ (a2 ∩ (b1 ∪ b2)))))

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