Theorem u3lemaa 584

Description: Lemma for Kalmbach implication study.

Assertion

Ref	Expression
u3lemaa	((a →3 b) ∩ a) = (a ∩ (a⊥ ∪ b))

Proof of Theorem u3lemaa

Step	Hyp	Ref	Expression
1		df-i3 45	. . 3 (a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))
2	1	ran 71	. 2 ((a →3 b) ∩ a) = ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∩ a)
3		comanr1 446	. . . . . 6 a⊥ C (a⊥ ∩ b)
4	3	comcom6 441	. . . . 5 a C (a⊥ ∩ b)
5		comanr1 446	. . . . . 6 a⊥ C (a⊥ ∩ b⊥ )
6	5	comcom6 441	. . . . 5 a C (a⊥ ∩ b⊥ )
7	4, 6	com2or 465	. . . 4 a C ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))
8		comid 179	. . . . 5 a C a
9		comorr 176	. . . . . 6 a⊥ C (a⊥ ∪ b)
10	9	comcom6 441	. . . . 5 a C (a⊥ ∪ b)
11	8, 10	com2an 466	. . . 4 a C (a ∩ (a⊥ ∪ b))
12	7, 11	fh1r 455	. . 3 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∩ a) = ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ a) ∪ ((a ∩ (a⊥ ∪ b)) ∩ a))
13	4, 6	fh1r 455	. . . . . 6 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ a) = (((a⊥ ∩ b) ∩ a) ∪ ((a⊥ ∩ b⊥ ) ∩ a))
14		ancom 68	. . . . . . . . 9 ((a⊥ ∩ b) ∩ a) = (a ∩ (a⊥ ∩ b))
15		anass 69	. . . . . . . . . . 11 ((a ∩ a⊥ ) ∩ b) = (a ∩ (a⊥ ∩ b))
16	15	ax-r1 34	. . . . . . . . . 10 (a ∩ (a⊥ ∩ b)) = ((a ∩ a⊥ ) ∩ b)
17		ancom 68	. . . . . . . . . . 11 ((a ∩ a⊥ ) ∩ b) = (b ∩ (a ∩ a⊥ ))
18		dff 93	. . . . . . . . . . . . . 14 0 = (a ∩ a⊥ )
19	18	ax-r1 34	. . . . . . . . . . . . 13 (a ∩ a⊥ ) = 0
20	19	lan 70	. . . . . . . . . . . 12 (b ∩ (a ∩ a⊥ )) = (b ∩ 0)
21		an0 100	. . . . . . . . . . . 12 (b ∩ 0) = 0
22	20, 21	ax-r2 35	. . . . . . . . . . 11 (b ∩ (a ∩ a⊥ )) = 0
23	17, 22	ax-r2 35	. . . . . . . . . 10 ((a ∩ a⊥ ) ∩ b) = 0
24	16, 23	ax-r2 35	. . . . . . . . 9 (a ∩ (a⊥ ∩ b)) = 0
25	14, 24	ax-r2 35	. . . . . . . 8 ((a⊥ ∩ b) ∩ a) = 0
26		ancom 68	. . . . . . . . 9 ((a⊥ ∩ b⊥ ) ∩ a) = (a ∩ (a⊥ ∩ b⊥ ))
27		anass 69	. . . . . . . . . . 11 ((a ∩ a⊥ ) ∩ b⊥ ) = (a ∩ (a⊥ ∩ b⊥ ))
28	27	ax-r1 34	. . . . . . . . . 10 (a ∩ (a⊥ ∩ b⊥ )) = ((a ∩ a⊥ ) ∩ b⊥ )
29		ancom 68	. . . . . . . . . . 11 ((a ∩ a⊥ ) ∩ b⊥ ) = (b⊥ ∩ (a ∩ a⊥ ))
30	19	lan 70	. . . . . . . . . . . 12 (b⊥ ∩ (a ∩ a⊥ )) = (b⊥ ∩ 0)
31		an0 100	. . . . . . . . . . . 12 (b⊥ ∩ 0) = 0
32	30, 31	ax-r2 35	. . . . . . . . . . 11 (b⊥ ∩ (a ∩ a⊥ )) = 0
33	29, 32	ax-r2 35	. . . . . . . . . 10 ((a ∩ a⊥ ) ∩ b⊥ ) = 0
34	28, 33	ax-r2 35	. . . . . . . . 9 (a ∩ (a⊥ ∩ b⊥ )) = 0
35	26, 34	ax-r2 35	. . . . . . . 8 ((a⊥ ∩ b⊥ ) ∩ a) = 0
36	25, 35	2or 67	. . . . . . 7 (((a⊥ ∩ b) ∩ a) ∪ ((a⊥ ∩ b⊥ ) ∩ a)) = (0 ∪ 0)
37		or0 94	. . . . . . 7 (0 ∪ 0) = 0
38	36, 37	ax-r2 35	. . . . . 6 (((a⊥ ∩ b) ∩ a) ∪ ((a⊥ ∩ b⊥ ) ∩ a)) = 0
39	13, 38	ax-r2 35	. . . . 5 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ a) = 0
40		an32 76	. . . . . 6 ((a ∩ (a⊥ ∪ b)) ∩ a) = ((a ∩ a) ∩ (a⊥ ∪ b))
41		anidm 103	. . . . . . 7 (a ∩ a) = a
42	41	ran 71	. . . . . 6 ((a ∩ a) ∩ (a⊥ ∪ b)) = (a ∩ (a⊥ ∪ b))
43	40, 42	ax-r2 35	. . . . 5 ((a ∩ (a⊥ ∪ b)) ∩ a) = (a ∩ (a⊥ ∪ b))
44	39, 43	2or 67	. . . 4 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ a) ∪ ((a ∩ (a⊥ ∪ b)) ∩ a)) = (0 ∪ (a ∩ (a⊥ ∪ b)))
45		ax-a2 30	. . . . 5 (0 ∪ (a ∩ (a⊥ ∪ b))) = ((a ∩ (a⊥ ∪ b)) ∪ 0)
46		or0 94	. . . . 5 ((a ∩ (a⊥ ∪ b)) ∪ 0) = (a ∩ (a⊥ ∪ b))
47	45, 46	ax-r2 35	. . . 4 (0 ∪ (a ∩ (a⊥ ∪ b))) = (a ∩ (a⊥ ∪ b))
48	44, 47	ax-r2 35	. . 3 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ a) ∪ ((a ∩ (a⊥ ∪ b)) ∩ a)) = (a ∩ (a⊥ ∪ b))
49	12, 48	ax-r2 35	. 2 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∩ a) = (a ∩ (a⊥ ∪ b))
50	2, 49	ax-r2 35	1 ((a →3 b) ∩ a) = (a ∩ (a⊥ ∪ b))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 10   →₃ wi3 15

This theorem is referenced by:  u3lemnona 649  u3lem13b 772

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i3 45  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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