Theorem ud5lem3a 591

Description: Lemma for unified disjunction. (Contributed by NM, 27-Nov-1997.)

Assertion

Ref	Expression
ud5lem3a	((a →5 b) ∩ (a ∪ (a⊥ ∩ b))) = ((a ∩ b) ∪ (a⊥ ∩ b))

Proof of Theorem ud5lem3a

Step	Hyp	Ref	Expression
1		df-i5 48	. . 3 (a →5 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))
2	1	ran 78	. 2 ((a →5 b) ∩ (a ∪ (a⊥ ∩ b))) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ (a⊥ ∩ b)))
3		comanr1 464	. . . . . 6 a C (a ∩ b)
4		comanr1 464	. . . . . . 7 a⊥ C (a⊥ ∩ b)
5	4	comcom6 459	. . . . . 6 a C (a⊥ ∩ b)
6	3, 5	com2or 483	. . . . 5 a C ((a ∩ b) ∪ (a⊥ ∩ b))
7		comanr1 464	. . . . . 6 a⊥ C (a⊥ ∩ b⊥ )
8	7	comcom6 459	. . . . 5 a C (a⊥ ∩ b⊥ )
9	6, 8	com2or 483	. . . 4 a C (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))
10	9, 5	fh2 470	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ (a⊥ ∩ b))) = (((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a) ∪ ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ (a⊥ ∩ b)))
11	6, 8	fh1r 473	. . . . . 6 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a) ∪ ((a⊥ ∩ b⊥ ) ∩ a))
12	3, 5	fh1r 473	. . . . . . . . 9 (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a) = (((a ∩ b) ∩ a) ∪ ((a⊥ ∩ b) ∩ a))
13		an32 83	. . . . . . . . . . . 12 ((a ∩ b) ∩ a) = ((a ∩ a) ∩ b)
14		anidm 111	. . . . . . . . . . . . 13 (a ∩ a) = a
15	14	ran 78	. . . . . . . . . . . 12 ((a ∩ a) ∩ b) = (a ∩ b)
16	13, 15	ax-r2 36	. . . . . . . . . . 11 ((a ∩ b) ∩ a) = (a ∩ b)
17		ancom 74	. . . . . . . . . . . 12 ((a⊥ ∩ b) ∩ a) = (a ∩ (a⊥ ∩ b))
18		anass 76	. . . . . . . . . . . . . 14 ((a ∩ a⊥ ) ∩ b) = (a ∩ (a⊥ ∩ b))
19	18	ax-r1 35	. . . . . . . . . . . . 13 (a ∩ (a⊥ ∩ b)) = ((a ∩ a⊥ ) ∩ b)
20		dff 101	. . . . . . . . . . . . . . . 16 0 = (a ∩ a⊥ )
21	20	ax-r1 35	. . . . . . . . . . . . . . 15 (a ∩ a⊥ ) = 0
22	21	ran 78	. . . . . . . . . . . . . 14 ((a ∩ a⊥ ) ∩ b) = (0 ∩ b)
23		an0r 109	. . . . . . . . . . . . . 14 (0 ∩ b) = 0
24	22, 23	ax-r2 36	. . . . . . . . . . . . 13 ((a ∩ a⊥ ) ∩ b) = 0
25	19, 24	ax-r2 36	. . . . . . . . . . . 12 (a ∩ (a⊥ ∩ b)) = 0
26	17, 25	ax-r2 36	. . . . . . . . . . 11 ((a⊥ ∩ b) ∩ a) = 0
27	16, 26	2or 72	. . . . . . . . . 10 (((a ∩ b) ∩ a) ∪ ((a⊥ ∩ b) ∩ a)) = ((a ∩ b) ∪ 0)
28		or0 102	. . . . . . . . . 10 ((a ∩ b) ∪ 0) = (a ∩ b)
29	27, 28	ax-r2 36	. . . . . . . . 9 (((a ∩ b) ∩ a) ∪ ((a⊥ ∩ b) ∩ a)) = (a ∩ b)
30	12, 29	ax-r2 36	. . . . . . . 8 (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a) = (a ∩ b)
31		ancom 74	. . . . . . . . 9 ((a⊥ ∩ b⊥ ) ∩ a) = (a ∩ (a⊥ ∩ b⊥ ))
32		anass 76	. . . . . . . . . . 11 ((a ∩ a⊥ ) ∩ b⊥ ) = (a ∩ (a⊥ ∩ b⊥ ))
33	32	ax-r1 35	. . . . . . . . . 10 (a ∩ (a⊥ ∩ b⊥ )) = ((a ∩ a⊥ ) ∩ b⊥ )
34	21	ran 78	. . . . . . . . . . 11 ((a ∩ a⊥ ) ∩ b⊥ ) = (0 ∩ b⊥ )
35		an0r 109	. . . . . . . . . . 11 (0 ∩ b⊥ ) = 0
36	34, 35	ax-r2 36	. . . . . . . . . 10 ((a ∩ a⊥ ) ∩ b⊥ ) = 0
37	33, 36	ax-r2 36	. . . . . . . . 9 (a ∩ (a⊥ ∩ b⊥ )) = 0
38	31, 37	ax-r2 36	. . . . . . . 8 ((a⊥ ∩ b⊥ ) ∩ a) = 0
39	30, 38	2or 72	. . . . . . 7 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a) ∪ ((a⊥ ∩ b⊥ ) ∩ a)) = ((a ∩ b) ∪ 0)
40	39, 28	ax-r2 36	. . . . . 6 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a) ∪ ((a⊥ ∩ b⊥ ) ∩ a)) = (a ∩ b)
41	11, 40	ax-r2 36	. . . . 5 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a) = (a ∩ b)
42		coman1 185	. . . . . . . . 9 (a⊥ ∩ b) C a⊥
43	42	comcom7 460	. . . . . . . 8 (a⊥ ∩ b) C a
44		coman2 186	. . . . . . . 8 (a⊥ ∩ b) C b
45	43, 44	com2an 484	. . . . . . 7 (a⊥ ∩ b) C (a ∩ b)
46	42, 44	com2an 484	. . . . . . 7 (a⊥ ∩ b) C (a⊥ ∩ b)
47	45, 46	com2or 483	. . . . . 6 (a⊥ ∩ b) C ((a ∩ b) ∪ (a⊥ ∩ b))
48	44	comcom2 183	. . . . . . 7 (a⊥ ∩ b) C b⊥
49	42, 48	com2an 484	. . . . . 6 (a⊥ ∩ b) C (a⊥ ∩ b⊥ )
50	47, 49	fh1r 473	. . . . 5 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ (a⊥ ∩ b)) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∩ (a⊥ ∩ b)))
51	41, 50	2or 72	. . . 4 (((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a) ∪ ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ (a⊥ ∩ b))) = ((a ∩ b) ∪ ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∩ (a⊥ ∩ b))))
52		ancom 74	. . . . . . . 8 (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∩ ((a ∩ b) ∪ (a⊥ ∩ b)))
53		ax-a2 31	. . . . . . . . . 10 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∪ (a ∩ b))
54	53	lan 77	. . . . . . . . 9 ((a⊥ ∩ b) ∩ ((a ∩ b) ∪ (a⊥ ∩ b))) = ((a⊥ ∩ b) ∩ ((a⊥ ∩ b) ∪ (a ∩ b)))
55		anabs 121	. . . . . . . . 9 ((a⊥ ∩ b) ∩ ((a⊥ ∩ b) ∪ (a ∩ b))) = (a⊥ ∩ b)
56	54, 55	ax-r2 36	. . . . . . . 8 ((a⊥ ∩ b) ∩ ((a ∩ b) ∪ (a⊥ ∩ b))) = (a⊥ ∩ b)
57	52, 56	ax-r2 36	. . . . . . 7 (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ (a⊥ ∩ b)) = (a⊥ ∩ b)
58		an4 86	. . . . . . . 8 ((a⊥ ∩ b⊥ ) ∩ (a⊥ ∩ b)) = ((a⊥ ∩ a⊥ ) ∩ (b⊥ ∩ b))
59		ancom 74	. . . . . . . . . . 11 (b⊥ ∩ b) = (b ∩ b⊥ )
60		dff 101	. . . . . . . . . . . 12 0 = (b ∩ b⊥ )
61	60	ax-r1 35	. . . . . . . . . . 11 (b ∩ b⊥ ) = 0
62	59, 61	ax-r2 36	. . . . . . . . . 10 (b⊥ ∩ b) = 0
63	62	lan 77	. . . . . . . . 9 ((a⊥ ∩ a⊥ ) ∩ (b⊥ ∩ b)) = ((a⊥ ∩ a⊥ ) ∩ 0)
64		an0 108	. . . . . . . . 9 ((a⊥ ∩ a⊥ ) ∩ 0) = 0
65	63, 64	ax-r2 36	. . . . . . . 8 ((a⊥ ∩ a⊥ ) ∩ (b⊥ ∩ b)) = 0
66	58, 65	ax-r2 36	. . . . . . 7 ((a⊥ ∩ b⊥ ) ∩ (a⊥ ∩ b)) = 0
67	57, 66	2or 72	. . . . . 6 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∩ (a⊥ ∩ b))) = ((a⊥ ∩ b) ∪ 0)
68		or0 102	. . . . . 6 ((a⊥ ∩ b) ∪ 0) = (a⊥ ∩ b)
69	67, 68	ax-r2 36	. . . . 5 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∩ (a⊥ ∩ b))) = (a⊥ ∩ b)
70	69	lor 70	. . . 4 ((a ∩ b) ∪ ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ (a⊥ ∩ b)) ∪ ((a⊥ ∩ b⊥ ) ∩ (a⊥ ∩ b)))) = ((a ∩ b) ∪ (a⊥ ∩ b))
71	51, 70	ax-r2 36	. . 3 (((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a) ∪ ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ (a⊥ ∩ b))) = ((a ∩ b) ∪ (a⊥ ∩ b))
72	10, 71	ax-r2 36	. 2 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ (a⊥ ∩ b))) = ((a ∩ b) ∪ (a⊥ ∩ b))
73	2, 72	ax-r2 36	1 ((a →5 b) ∩ (a ∪ (a⊥ ∩ b))) = ((a ∩ b) ∪ (a⊥ ∩ b))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →₅ wi5 16

This proof depends on 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-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i5 48  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  ud5lem3  594