Theorem dp15lema 1154

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
dp15lema	((a0 ∪ e) ∩ (a1 ∪ b1)) ≤ (d ∪ b2)

Proof of Theorem dp15lema

Step	Hyp	Ref	Expression
1		dp15lema.3	. . . . 5 e = (b0 ∩ (a0 ∪ p0))
2		dp15lema.2	. . . . . . 7 p0 = ((a1 ∪ b1) ∩ (a2 ∪ b2))
3	2	lor 70	. . . . . 6 (a0 ∪ p0) = (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
4	3	lan 77	. . . . 5 (b0 ∩ (a0 ∪ p0)) = (b0 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))
5	1, 4	tr 62	. . . 4 e = (b0 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))
6	5	lor 70	. . 3 (a0 ∪ e) = (a0 ∪ (b0 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))))
7	6	ran 78	. 2 ((a0 ∪ e) ∩ (a1 ∪ b1)) = ((a0 ∪ (b0 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ∩ (a1 ∪ b1))
8		le1 146	. . . . . 6 b0 ≤ 1
9	8	leran 153	. . . . 5 (b0 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))) ≤ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))
10	9	lelor 166	. . . 4 (a0 ∪ (b0 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ≤ (a0 ∪ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))))
11	10	leran 153	. . 3 ((a0 ∪ (b0 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ∩ (a1 ∪ b1)) ≤ ((a0 ∪ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ∩ (a1 ∪ b1))
12		an1r 107	. . . . . . . . 9 (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))) = (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
13	12	lor 70	. . . . . . . 8 (a0 ∪ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) = (a0 ∪ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))
14		orass 75	. . . . . . . . 9 ((a0 ∪ a0) ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))) = (a0 ∪ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))
15	14	cm 61	. . . . . . . 8 (a0 ∪ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))) = ((a0 ∪ a0) ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
16		oridm 110	. . . . . . . . . 10 (a0 ∪ a0) = a0
17	16	ror 71	. . . . . . . . 9 ((a0 ∪ a0) ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))) = (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2)))
18		orcom 73	. . . . . . . . 9 (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))) = (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∪ a0)
19	17, 18	tr 62	. . . . . . . 8 ((a0 ∪ a0) ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))) = (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∪ a0)
20	13, 15, 19	3tr 65	. . . . . . 7 (a0 ∪ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) = (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∪ a0)
21	20	ran 78	. . . . . 6 ((a0 ∪ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ∩ (a1 ∪ b1)) = ((((a1 ∪ b1) ∩ (a2 ∪ b2)) ∪ a0) ∩ (a1 ∪ b1))
22		lea 160	. . . . . . 7 ((a1 ∪ b1) ∩ (a2 ∪ b2)) ≤ (a1 ∪ b1)
23	22	mlduali 1128	. . . . . 6 ((((a1 ∪ b1) ∩ (a2 ∪ b2)) ∪ a0) ∩ (a1 ∪ b1)) = (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∪ (a0 ∩ (a1 ∪ b1)))
24	21, 23	tr 62	. . . . 5 ((a0 ∪ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ∩ (a1 ∪ b1)) = (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∪ (a0 ∩ (a1 ∪ b1)))
25		lear 161	. . . . . 6 ((a1 ∪ b1) ∩ (a2 ∪ b2)) ≤ (a2 ∪ b2)
26	25	leror 152	. . . . 5 (((a1 ∪ b1) ∩ (a2 ∪ b2)) ∪ (a0 ∩ (a1 ∪ b1))) ≤ ((a2 ∪ b2) ∪ (a0 ∩ (a1 ∪ b1)))
27	24, 26	bltr 138	. . . 4 ((a0 ∪ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ∩ (a1 ∪ b1)) ≤ ((a2 ∪ b2) ∪ (a0 ∩ (a1 ∪ b1)))
28		or32 82	. . . . 5 ((a2 ∪ b2) ∪ (a0 ∩ (a1 ∪ b1))) = ((a2 ∪ (a0 ∩ (a1 ∪ b1))) ∪ b2)
29		dp15lema.1	. . . . . . 7 d = (a2 ∪ (a0 ∩ (a1 ∪ b1)))
30	29	ror 71	. . . . . 6 (d ∪ b2) = ((a2 ∪ (a0 ∩ (a1 ∪ b1))) ∪ b2)
31	30	cm 61	. . . . 5 ((a2 ∪ (a0 ∩ (a1 ∪ b1))) ∪ b2) = (d ∪ b2)
32	28, 31	tr 62	. . . 4 ((a2 ∪ b2) ∪ (a0 ∩ (a1 ∪ b1))) = (d ∪ b2)
33	27, 32	lbtr 139	. . 3 ((a0 ∪ (1 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ∩ (a1 ∪ b1)) ≤ (d ∪ b2)
34	11, 33	letr 137	. 2 ((a0 ∪ (b0 ∩ (a0 ∪ ((a1 ∪ b1) ∩ (a2 ∪ b2))))) ∩ (a1 ∪ b1)) ≤ (d ∪ b2)
35	7, 34	bltr 138	1 ((a0 ∪ e) ∩ (a1 ∪ b1)) ≤ (d ∪ b2)

Syntax hints:   = wb 1   ≤ wle 2   ∪ wo 6   ∩ wa 7  1wt 8

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

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:  dp15lemb  1155