Theorem u5lemanb 619

Description: Lemma for relevance implication study. (Contributed by NM, 14-Dec-1997.)

Assertion

Ref	Expression
u5lemanb	((a →5 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

Proof of Theorem u5lemanb

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) ∩ b⊥ ) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ )
3		comanr2 465	. . . . . 6 b C (a ∩ b)
4	3	comcom3 454	. . . . 5 b⊥ C (a ∩ b)
5		comanr2 465	. . . . . 6 b C (a⊥ ∩ b)
6	5	comcom3 454	. . . . 5 b⊥ C (a⊥ ∩ b)
7	4, 6	com2or 483	. . . 4 b⊥ C ((a ∩ b) ∪ (a⊥ ∩ b))
8		comanr2 465	. . . 4 b⊥ C (a⊥ ∩ b⊥ )
9	7, 8	fh1r 473	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ ) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ ))
10		ax-a2 31	. . . 4 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ )) = (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ b⊥ ))
11		anass 76	. . . . . . 7 ((a⊥ ∩ b⊥ ) ∩ b⊥ ) = (a⊥ ∩ (b⊥ ∩ b⊥ ))
12		anidm 111	. . . . . . . 8 (b⊥ ∩ b⊥ ) = b⊥
13	12	lan 77	. . . . . . 7 (a⊥ ∩ (b⊥ ∩ b⊥ )) = (a⊥ ∩ b⊥ )
14	11, 13	ax-r2 36	. . . . . 6 ((a⊥ ∩ b⊥ ) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
15	4, 6	fh1r 473	. . . . . . 7 (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ b⊥ ) = (((a ∩ b) ∩ b⊥ ) ∪ ((a⊥ ∩ b) ∩ b⊥ ))
16		anass 76	. . . . . . . . . 10 ((a ∩ b) ∩ b⊥ ) = (a ∩ (b ∩ b⊥ ))
17		dff 101	. . . . . . . . . . . . 13 0 = (b ∩ b⊥ )
18	17	lan 77	. . . . . . . . . . . 12 (a ∩ 0) = (a ∩ (b ∩ b⊥ ))
19	18	ax-r1 35	. . . . . . . . . . 11 (a ∩ (b ∩ b⊥ )) = (a ∩ 0)
20		an0 108	. . . . . . . . . . 11 (a ∩ 0) = 0
21	19, 20	ax-r2 36	. . . . . . . . . 10 (a ∩ (b ∩ b⊥ )) = 0
22	16, 21	ax-r2 36	. . . . . . . . 9 ((a ∩ b) ∩ b⊥ ) = 0
23		anass 76	. . . . . . . . . 10 ((a⊥ ∩ b) ∩ b⊥ ) = (a⊥ ∩ (b ∩ b⊥ ))
24	17	lan 77	. . . . . . . . . . . 12 (a⊥ ∩ 0) = (a⊥ ∩ (b ∩ b⊥ ))
25	24	ax-r1 35	. . . . . . . . . . 11 (a⊥ ∩ (b ∩ b⊥ )) = (a⊥ ∩ 0)
26		an0 108	. . . . . . . . . . 11 (a⊥ ∩ 0) = 0
27	25, 26	ax-r2 36	. . . . . . . . . 10 (a⊥ ∩ (b ∩ b⊥ )) = 0
28	23, 27	ax-r2 36	. . . . . . . . 9 ((a⊥ ∩ b) ∩ b⊥ ) = 0
29	22, 28	2or 72	. . . . . . . 8 (((a ∩ b) ∩ b⊥ ) ∪ ((a⊥ ∩ b) ∩ b⊥ )) = (0 ∪ 0)
30		or0 102	. . . . . . . 8 (0 ∪ 0) = 0
31	29, 30	ax-r2 36	. . . . . . 7 (((a ∩ b) ∩ b⊥ ) ∪ ((a⊥ ∩ b) ∩ b⊥ )) = 0
32	15, 31	ax-r2 36	. . . . . 6 (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ b⊥ ) = 0
33	14, 32	2or 72	. . . . 5 (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ 0)
34		or0 102	. . . . 5 ((a⊥ ∩ b⊥ ) ∪ 0) = (a⊥ ∩ b⊥ )
35	33, 34	ax-r2 36	. . . 4 (((a⊥ ∩ b⊥ ) ∩ b⊥ ) ∪ (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ b⊥ )) = (a⊥ ∩ b⊥ )
36	10, 35	ax-r2 36	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ b⊥ )) = (a⊥ ∩ b⊥ )
37	9, 36	ax-r2 36	. 2 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
38	2, 37	ax-r2 36	1 ((a →5 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →₅ wi5 16

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-r3 439

This theorem 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 referenced by:  u5lemnob  674  u5lembi  725