Theorem u5lemana 609

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

Assertion

Ref	Expression
u5lemana	((a →5 b) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))

Proof of Theorem u5lemana

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⊥ )
3		comanr1 464	. . . . . 6 a C (a ∩ b)
4	3	comcom3 454	. . . . 5 a⊥ C (a ∩ b)
5		comanr1 464	. . . . 5 a⊥ C (a⊥ ∩ b)
6	4, 5	com2or 483	. . . 4 a⊥ C ((a ∩ b) ∪ (a⊥ ∩ b))
7		comanr1 464	. . . 4 a⊥ C (a⊥ ∩ b⊥ )
8	6, 7	fh1r 473	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a⊥ ) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ a⊥ ))
9	4, 5	fh1r 473	. . . . 5 (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a⊥ ) = (((a ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ a⊥ ))
10		ax-a2 31	. . . . . 6 (((a ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ a⊥ )) = (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a ∩ b) ∩ a⊥ ))
11		an32 83	. . . . . . . . 9 ((a⊥ ∩ b) ∩ a⊥ ) = ((a⊥ ∩ a⊥ ) ∩ b)
12		anidm 111	. . . . . . . . . 10 (a⊥ ∩ a⊥ ) = a⊥
13	12	ran 78	. . . . . . . . 9 ((a⊥ ∩ a⊥ ) ∩ b) = (a⊥ ∩ b)
14	11, 13	ax-r2 36	. . . . . . . 8 ((a⊥ ∩ b) ∩ a⊥ ) = (a⊥ ∩ b)
15		an32 83	. . . . . . . . 9 ((a ∩ b) ∩ a⊥ ) = ((a ∩ a⊥ ) ∩ b)
16		ancom 74	. . . . . . . . . 10 ((a ∩ a⊥ ) ∩ b) = (b ∩ (a ∩ a⊥ ))
17		dff 101	. . . . . . . . . . . . 13 0 = (a ∩ a⊥ )
18	17	lan 77	. . . . . . . . . . . 12 (b ∩ 0) = (b ∩ (a ∩ a⊥ ))
19	18	ax-r1 35	. . . . . . . . . . 11 (b ∩ (a ∩ a⊥ )) = (b ∩ 0)
20		an0 108	. . . . . . . . . . 11 (b ∩ 0) = 0
21	19, 20	ax-r2 36	. . . . . . . . . 10 (b ∩ (a ∩ a⊥ )) = 0
22	16, 21	ax-r2 36	. . . . . . . . 9 ((a ∩ a⊥ ) ∩ b) = 0
23	15, 22	ax-r2 36	. . . . . . . 8 ((a ∩ b) ∩ a⊥ ) = 0
24	14, 23	2or 72	. . . . . . 7 (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a ∩ b) ∩ a⊥ )) = ((a⊥ ∩ b) ∪ 0)
25		or0 102	. . . . . . 7 ((a⊥ ∩ b) ∪ 0) = (a⊥ ∩ b)
26	24, 25	ax-r2 36	. . . . . 6 (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a ∩ b) ∩ a⊥ )) = (a⊥ ∩ b)
27	10, 26	ax-r2 36	. . . . 5 (((a ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ a⊥ )) = (a⊥ ∩ b)
28	9, 27	ax-r2 36	. . . 4 (((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a⊥ ) = (a⊥ ∩ b)
29		an32 83	. . . . 5 ((a⊥ ∩ b⊥ ) ∩ a⊥ ) = ((a⊥ ∩ a⊥ ) ∩ b⊥ )
30	12	ran 78	. . . . 5 ((a⊥ ∩ a⊥ ) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
31	29, 30	ax-r2 36	. . . 4 ((a⊥ ∩ b⊥ ) ∩ a⊥ ) = (a⊥ ∩ b⊥ )
32	28, 31	2or 72	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ a⊥ )) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))
33	8, 32	ax-r2 36	. 2 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))
34	2, 33	ax-r2 36	1 ((a →5 b) ∩ a⊥ ) = ((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:  u5lemnoa  664