Theorem ud4lem1d 580

Description: Lemma for unified disjunction. (Contributed by NM, 25-Nov-1997.)

Assertion

Ref	Expression
ud4lem1d	(((a →4 b)⊥ ∪ (b →4 a)) ∩ (b →4 a)⊥ ) = (((a⊥ ∪ b⊥ ) ∩ (a⊥ ∪ b)) ∩ a)

Proof of Theorem ud4lem1d

Step	Hyp	Ref	Expression
1		ud4lem1c 579	. . 3 ((a →4 b)⊥ ∪ (b →4 a)) = (a ∪ b⊥ )
2		ud4lem0c 280	. . 3 (b →4 a)⊥ = (((b⊥ ∪ a⊥ ) ∩ (b ∪ a⊥ )) ∩ ((b ∩ a⊥ ) ∪ a))
3	1, 2	2an 79	. 2 (((a →4 b)⊥ ∪ (b →4 a)) ∩ (b →4 a)⊥ ) = ((a ∪ b⊥ ) ∩ (((b⊥ ∪ a⊥ ) ∩ (b ∪ a⊥ )) ∩ ((b ∩ a⊥ ) ∪ a)))
4		an12 81	. . 3 ((a ∪ b⊥ ) ∩ (((b⊥ ∪ a⊥ ) ∩ (b ∪ a⊥ )) ∩ ((b ∩ a⊥ ) ∪ a))) = (((b⊥ ∪ a⊥ ) ∩ (b ∪ a⊥ )) ∩ ((a ∪ b⊥ ) ∩ ((b ∩ a⊥ ) ∪ a)))
5		ax-a2 31	. . . . 5 (b⊥ ∪ a⊥ ) = (a⊥ ∪ b⊥ )
6		ax-a2 31	. . . . 5 (b ∪ a⊥ ) = (a⊥ ∪ b)
7	5, 6	2an 79	. . . 4 ((b⊥ ∪ a⊥ ) ∩ (b ∪ a⊥ )) = ((a⊥ ∪ b⊥ ) ∩ (a⊥ ∪ b))
8		comor2 462	. . . . . . . . 9 (a ∪ b⊥ ) C b⊥
9	8	comcom3 454	. . . . . . . 8 (a ∪ b⊥ )⊥ C b⊥
10	9	comcom5 458	. . . . . . 7 (a ∪ b⊥ ) C b
11		comor1 461	. . . . . . . 8 (a ∪ b⊥ ) C a
12	11	comcom2 183	. . . . . . 7 (a ∪ b⊥ ) C a⊥
13	10, 12	com2an 484	. . . . . 6 (a ∪ b⊥ ) C (b ∩ a⊥ )
14	13, 11	fh1 469	. . . . 5 ((a ∪ b⊥ ) ∩ ((b ∩ a⊥ ) ∪ a)) = (((a ∪ b⊥ ) ∩ (b ∩ a⊥ )) ∪ ((a ∪ b⊥ ) ∩ a))
15		ax-a2 31	. . . . . . . . 9 (a ∪ b⊥ ) = (b⊥ ∪ a)
16		anor1 88	. . . . . . . . 9 (b ∩ a⊥ ) = (b⊥ ∪ a)⊥
17	15, 16	2an 79	. . . . . . . 8 ((a ∪ b⊥ ) ∩ (b ∩ a⊥ )) = ((b⊥ ∪ a) ∩ (b⊥ ∪ a)⊥ )
18		dff 101	. . . . . . . . 9 0 = ((b⊥ ∪ a) ∩ (b⊥ ∪ a)⊥ )
19	18	ax-r1 35	. . . . . . . 8 ((b⊥ ∪ a) ∩ (b⊥ ∪ a)⊥ ) = 0
20	17, 19	ax-r2 36	. . . . . . 7 ((a ∪ b⊥ ) ∩ (b ∩ a⊥ )) = 0
21		ancom 74	. . . . . . . 8 ((a ∪ b⊥ ) ∩ a) = (a ∩ (a ∪ b⊥ ))
22		anabs 121	. . . . . . . 8 (a ∩ (a ∪ b⊥ )) = a
23	21, 22	ax-r2 36	. . . . . . 7 ((a ∪ b⊥ ) ∩ a) = a
24	20, 23	2or 72	. . . . . 6 (((a ∪ b⊥ ) ∩ (b ∩ a⊥ )) ∪ ((a ∪ b⊥ ) ∩ a)) = (0 ∪ a)
25		ax-a2 31	. . . . . . 7 (0 ∪ a) = (a ∪ 0)
26		or0 102	. . . . . . 7 (a ∪ 0) = a
27	25, 26	ax-r2 36	. . . . . 6 (0 ∪ a) = a
28	24, 27	ax-r2 36	. . . . 5 (((a ∪ b⊥ ) ∩ (b ∩ a⊥ )) ∪ ((a ∪ b⊥ ) ∩ a)) = a
29	14, 28	ax-r2 36	. . . 4 ((a ∪ b⊥ ) ∩ ((b ∩ a⊥ ) ∪ a)) = a
30	7, 29	2an 79	. . 3 (((b⊥ ∪ a⊥ ) ∩ (b ∪ a⊥ )) ∩ ((a ∪ b⊥ ) ∩ ((b ∩ a⊥ ) ∪ a))) = (((a⊥ ∪ b⊥ ) ∩ (a⊥ ∪ b)) ∩ a)
31	4, 30	ax-r2 36	. 2 ((a ∪ b⊥ ) ∩ (((b⊥ ∪ a⊥ ) ∩ (b ∪ a⊥ )) ∩ ((b ∩ a⊥ ) ∪ a))) = (((a⊥ ∪ b⊥ ) ∩ (a⊥ ∪ b)) ∩ a)
32	3, 31	ax-r2 36	1 (((a →4 b)⊥ ∪ (b →4 a)) ∩ (b →4 a)⊥ ) = (((a⊥ ∪ b⊥ ) ∩ (a⊥ ∪ b)) ∩ a)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →₄ wi4 15

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-i4 47  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  ud4lem1  581