Theorem ud5lem2 590

Description: Lemma for unified disjunction. (Contributed by NM, 10-Apr-2012.)

Assertion

Ref	Expression
ud5lem2	((a ∪ b⊥ ) →5 a) = (a ∪ (a⊥ ∩ b))

Proof of Theorem ud5lem2

Step	Hyp	Ref	Expression
1		df-i5 48	. 2 ((a ∪ b⊥ ) →5 a) = ((((a ∪ b⊥ ) ∩ a) ∪ ((a ∪ b⊥ )⊥ ∩ a)) ∪ ((a ∪ b⊥ )⊥ ∩ a⊥ ))
2		ax-a3 32	. . 3 ((((a ∪ b⊥ ) ∩ a) ∪ ((a ∪ b⊥ )⊥ ∩ a)) ∪ ((a ∪ b⊥ )⊥ ∩ a⊥ )) = (((a ∪ b⊥ ) ∩ a) ∪ (((a ∪ b⊥ )⊥ ∩ a) ∪ ((a ∪ b⊥ )⊥ ∩ a⊥ )))
3		ancom 74	. . . . 5 ((a ∪ b⊥ ) ∩ a) = (a ∩ (a ∪ b⊥ ))
4		anabs 121	. . . . 5 (a ∩ (a ∪ b⊥ )) = a
5	3, 4	ax-r2 36	. . . 4 ((a ∪ b⊥ ) ∩ a) = a
6		ax-a2 31	. . . . 5 (((a ∪ b⊥ )⊥ ∩ a) ∪ ((a ∪ b⊥ )⊥ ∩ a⊥ )) = (((a ∪ b⊥ )⊥ ∩ a⊥ ) ∪ ((a ∪ b⊥ )⊥ ∩ a))
7		anor2 89	. . . . . . . . . 10 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
8	7	ax-r1 35	. . . . . . . . 9 (a ∪ b⊥ )⊥ = (a⊥ ∩ b)
9	8	ran 78	. . . . . . . 8 ((a ∪ b⊥ )⊥ ∩ a⊥ ) = ((a⊥ ∩ b) ∩ a⊥ )
10		an32 83	. . . . . . . . 9 ((a⊥ ∩ b) ∩ a⊥ ) = ((a⊥ ∩ a⊥ ) ∩ b)
11		anidm 111	. . . . . . . . . 10 (a⊥ ∩ a⊥ ) = a⊥
12	11	ran 78	. . . . . . . . 9 ((a⊥ ∩ a⊥ ) ∩ b) = (a⊥ ∩ b)
13	10, 12	ax-r2 36	. . . . . . . 8 ((a⊥ ∩ b) ∩ a⊥ ) = (a⊥ ∩ b)
14	9, 13	ax-r2 36	. . . . . . 7 ((a ∪ b⊥ )⊥ ∩ a⊥ ) = (a⊥ ∩ b)
15	8	ran 78	. . . . . . . 8 ((a ∪ b⊥ )⊥ ∩ a) = ((a⊥ ∩ b) ∩ a)
16		an32 83	. . . . . . . . 9 ((a⊥ ∩ b) ∩ a) = ((a⊥ ∩ a) ∩ b)
17		ancom 74	. . . . . . . . . 10 ((a⊥ ∩ a) ∩ b) = (b ∩ (a⊥ ∩ a))
18		ancom 74	. . . . . . . . . . . . 13 (a⊥ ∩ a) = (a ∩ a⊥ )
19		dff 101	. . . . . . . . . . . . . 14 0 = (a ∩ a⊥ )
20	19	ax-r1 35	. . . . . . . . . . . . 13 (a ∩ a⊥ ) = 0
21	18, 20	ax-r2 36	. . . . . . . . . . . 12 (a⊥ ∩ a) = 0
22	21	lan 77	. . . . . . . . . . 11 (b ∩ (a⊥ ∩ a)) = (b ∩ 0)
23		an0 108	. . . . . . . . . . 11 (b ∩ 0) = 0
24	22, 23	ax-r2 36	. . . . . . . . . 10 (b ∩ (a⊥ ∩ a)) = 0
25	17, 24	ax-r2 36	. . . . . . . . 9 ((a⊥ ∩ a) ∩ b) = 0
26	16, 25	ax-r2 36	. . . . . . . 8 ((a⊥ ∩ b) ∩ a) = 0
27	15, 26	ax-r2 36	. . . . . . 7 ((a ∪ b⊥ )⊥ ∩ a) = 0
28	14, 27	2or 72	. . . . . 6 (((a ∪ b⊥ )⊥ ∩ a⊥ ) ∪ ((a ∪ b⊥ )⊥ ∩ a)) = ((a⊥ ∩ b) ∪ 0)
29		or0 102	. . . . . 6 ((a⊥ ∩ b) ∪ 0) = (a⊥ ∩ b)
30	28, 29	ax-r2 36	. . . . 5 (((a ∪ b⊥ )⊥ ∩ a⊥ ) ∪ ((a ∪ b⊥ )⊥ ∩ a)) = (a⊥ ∩ b)
31	6, 30	ax-r2 36	. . . 4 (((a ∪ b⊥ )⊥ ∩ a) ∪ ((a ∪ b⊥ )⊥ ∩ a⊥ )) = (a⊥ ∩ b)
32	5, 31	2or 72	. . 3 (((a ∪ b⊥ ) ∩ a) ∪ (((a ∪ b⊥ )⊥ ∩ a) ∪ ((a ∪ b⊥ )⊥ ∩ a⊥ ))) = (a ∪ (a⊥ ∩ b))
33	2, 32	ax-r2 36	. 2 ((((a ∪ b⊥ ) ∩ a) ∪ ((a ∪ b⊥ )⊥ ∩ a)) ∪ ((a ∪ b⊥ )⊥ ∩ a⊥ )) = (a ∪ (a⊥ ∩ b))
34	1, 33	ax-r2 36	1 ((a ∪ b⊥ ) →5 a) = (a ∪ (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

This proof depends on definitions:  df-a 40  df-t 41  df-f 42  df-i5 48

This theorem is used by:  ud5  599