Theorem lem4.6.2e1 1082

Description: Equation 4.10 of [MegPav2000] p. 23. This is the first part of the equation. Note that Lemma 4.6.1 is u1lemaa 600. (Contributed by Roy F. Longton, 29-Jun-2005.) (Revised by Roy F. Longton, 3-Jul-2005.)

Assertion

Ref	Expression
lem4.6.2e1	((a →1 b) ∩ (a⊥ →1 b)) = ((a →1 b) ∩ b)

Proof of Theorem lem4.6.2e1

Step	Hyp	Ref	Expression
1		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2		df-i1 44	. . 3 (a⊥ →1 b) = (a⊥ ⊥ ∪ (a⊥ ∩ b))
3	1, 2	2an 79	. 2 ((a →1 b) ∩ (a⊥ →1 b)) = ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ⊥ ∪ (a⊥ ∩ b)))
4		ax-a1 30	. . . . . 6 a = a⊥ ⊥
5	4	ax-r1 35	. . . . 5 a⊥ ⊥ = a
6	5	ax-r5 38	. . . 4 (a⊥ ⊥ ∪ (a⊥ ∩ b)) = (a ∪ (a⊥ ∩ b))
7	6	lan 77	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ⊥ ∪ (a⊥ ∩ b))) = ((a⊥ ∪ (a ∩ b)) ∩ (a ∪ (a⊥ ∩ b)))
8		comorr 184	. . . . . 6 a⊥ C (a⊥ ∪ (a ∩ b))
9	8	comcom6 459	. . . . 5 a C (a⊥ ∪ (a ∩ b))
10	9	comcom 453	. . . 4 (a⊥ ∪ (a ∩ b)) C a
11		leao1 162	. . . . . 6 (a⊥ ∩ b) ≤ (a⊥ ∪ (a ∩ b))
12	11	lecom 180	. . . . 5 (a⊥ ∩ b) C (a⊥ ∪ (a ∩ b))
13	12	comcom 453	. . . 4 (a⊥ ∪ (a ∩ b)) C (a⊥ ∩ b)
14	10, 13	fh1 469	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ (a ∪ (a⊥ ∩ b))) = (((a⊥ ∪ (a ∩ b)) ∩ a) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b)))
15		ancom 74	. . . . 5 ((a⊥ ∪ (a ∩ b)) ∩ a) = (a ∩ (a⊥ ∪ (a ∩ b)))
16	15	ax-r5 38	. . . 4 (((a⊥ ∪ (a ∩ b)) ∩ a) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b))) = ((a ∩ (a⊥ ∪ (a ∩ b))) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b)))
17		omla 447	. . . . 5 (a ∩ (a⊥ ∪ (a ∩ b))) = (a ∩ b)
18	17	ax-r5 38	. . . 4 ((a ∩ (a⊥ ∪ (a ∩ b))) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b))) = ((a ∩ b) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b)))
19		ancom 74	. . . . . 6 ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∩ (a⊥ ∪ (a ∩ b)))
20	19	lor 70	. . . . 5 ((a ∩ b) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b))) = ((a ∩ b) ∪ ((a⊥ ∩ b) ∩ (a⊥ ∪ (a ∩ b))))
21		coman1 185	. . . . . . 7 (a⊥ ∩ b) C a⊥
22	21	comcom7 460	. . . . . . . 8 (a⊥ ∩ b) C a
23		coman2 186	. . . . . . . 8 (a⊥ ∩ b) C b
24	22, 23	com2an 484	. . . . . . 7 (a⊥ ∩ b) C (a ∩ b)
25	21, 24	fh1 469	. . . . . 6 ((a⊥ ∩ b) ∩ (a⊥ ∪ (a ∩ b))) = (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))
26	25	lor 70	. . . . 5 ((a ∩ b) ∪ ((a⊥ ∩ b) ∩ (a⊥ ∪ (a ∩ b)))) = ((a ∩ b) ∪ (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))))
27		ancom 74	. . . . . . . . 9 (a⊥ ∩ b) = (b ∩ a⊥ )
28	27	ran 78	. . . . . . . 8 ((a⊥ ∩ b) ∩ a⊥ ) = ((b ∩ a⊥ ) ∩ a⊥ )
29	28	ax-r5 38	. . . . . . 7 (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))) = (((b ∩ a⊥ ) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))
30	29	lor 70	. . . . . 6 ((a ∩ b) ∪ (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))) = ((a ∩ b) ∪ (((b ∩ a⊥ ) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))))
31		anass 76	. . . . . . . 8 ((b ∩ a⊥ ) ∩ a⊥ ) = (b ∩ (a⊥ ∩ a⊥ ))
32	31	ax-r5 38	. . . . . . 7 (((b ∩ a⊥ ) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))) = ((b ∩ (a⊥ ∩ a⊥ )) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))
33	32	lor 70	. . . . . 6 ((a ∩ b) ∪ (((b ∩ a⊥ ) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))) = ((a ∩ b) ∪ ((b ∩ (a⊥ ∩ a⊥ )) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))))
34		anidm 111	. . . . . . . . . 10 (a⊥ ∩ a⊥ ) = a⊥
35	34	lan 77	. . . . . . . . 9 (b ∩ (a⊥ ∩ a⊥ )) = (b ∩ a⊥ )
36	35	ax-r5 38	. . . . . . . 8 ((b ∩ (a⊥ ∩ a⊥ )) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))) = ((b ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))
37	36	lor 70	. . . . . . 7 ((a ∩ b) ∪ ((b ∩ (a⊥ ∩ a⊥ )) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))) = ((a ∩ b) ∪ ((b ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))))
38		ancom 74	. . . . . . . . 9 (b ∩ a⊥ ) = (a⊥ ∩ b)
39	38	ax-r5 38	. . . . . . . 8 ((b ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))) = ((a⊥ ∩ b) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))
40	39	lor 70	. . . . . . 7 ((a ∩ b) ∪ ((b ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))) = ((a ∩ b) ∪ ((a⊥ ∩ b) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))))
41		orabs 120	. . . . . . . . 9 ((a⊥ ∩ b) ∪ ((a⊥ ∩ b) ∩ (a ∩ b))) = (a⊥ ∩ b)
42	41	lor 70	. . . . . . . 8 ((a ∩ b) ∪ ((a⊥ ∩ b) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))) = ((a ∩ b) ∪ (a⊥ ∩ b))
43		coman1 185	. . . . . . . . . 10 (a ∩ b) C a
44	43	comcom2 183	. . . . . . . . 9 (a ∩ b) C a⊥
45		coman2 186	. . . . . . . . 9 (a ∩ b) C b
46	44, 45	fh3 471	. . . . . . . 8 ((a ∩ b) ∪ (a⊥ ∩ b)) = (((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪ b))
47		ax-a2 31	. . . . . . . . 9 ((a ∩ b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))
48		lear 161	. . . . . . . . . 10 (a ∩ b) ≤ b
49	48	df-le2 131	. . . . . . . . 9 ((a ∩ b) ∪ b) = b
50	47, 49	2an 79	. . . . . . . 8 (((a ∩ b) ∪ a⊥ ) ∩ ((a ∩ b) ∪ b)) = ((a⊥ ∪ (a ∩ b)) ∩ b)
51	42, 46, 50	3tr 65	. . . . . . 7 ((a ∩ b) ∪ ((a⊥ ∩ b) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))) = ((a⊥ ∪ (a ∩ b)) ∩ b)
52	37, 40, 51	3tr 65	. . . . . 6 ((a ∩ b) ∪ ((b ∩ (a⊥ ∩ a⊥ )) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))) = ((a⊥ ∪ (a ∩ b)) ∩ b)
53	30, 33, 52	3tr 65	. . . . 5 ((a ∩ b) ∪ (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ (a ∩ b)))) = ((a⊥ ∪ (a ∩ b)) ∩ b)
54	20, 26, 53	3tr 65	. . . 4 ((a ∩ b) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b))) = ((a⊥ ∪ (a ∩ b)) ∩ b)
55	16, 18, 54	3tr 65	. . 3 (((a⊥ ∪ (a ∩ b)) ∩ a) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ∩ b))) = ((a⊥ ∪ (a ∩ b)) ∩ b)
56	7, 14, 55	3tr 65	. 2 ((a⊥ ∪ (a ∩ b)) ∩ (a⊥ ⊥ ∪ (a⊥ ∩ b))) = ((a⊥ ∪ (a ∩ b)) ∩ b)
57	1	ax-r1 35	. . 3 (a⊥ ∪ (a ∩ b)) = (a →1 b)
58	57	ran 78	. 2 ((a⊥ ∪ (a ∩ b)) ∩ b) = ((a →1 b) ∩ b)
59	3, 56, 58	3tr 65	1 ((a →1 b) ∩ (a⊥ →1 b)) = ((a →1 b) ∩ b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

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-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)