Theorem ud4lem3b 584

Description: Lemma for unified disjunction. (Contributed by NM, 23-Nov-1997.)

Assertion

Ref	Expression
ud4lem3b	((a →4 b)⊥ ∪ (a ∪ b)) = (a ∪ b)

Proof of Theorem ud4lem3b

Step	Hyp	Ref	Expression
1		ud4lem0c 280	. . 3 (a →4 b)⊥ = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b))
2	1	ax-r5 38	. 2 ((a →4 b)⊥ ∪ (a ∪ b)) = ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∪ (a ∪ b))
3		comor1 461	. . . . . . 7 (a ∪ b) C a
4	3	comcom2 183	. . . . . 6 (a ∪ b) C a⊥
5		comor2 462	. . . . . . 7 (a ∪ b) C b
6	5	comcom2 183	. . . . . 6 (a ∪ b) C b⊥
7	4, 6	com2or 483	. . . . 5 (a ∪ b) C (a⊥ ∪ b⊥ )
8	3, 6	com2or 483	. . . . 5 (a ∪ b) C (a ∪ b⊥ )
9	7, 8	com2an 484	. . . 4 (a ∪ b) C ((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ ))
10	3, 6	com2an 484	. . . . 5 (a ∪ b) C (a ∩ b⊥ )
11	10, 5	com2or 483	. . . 4 (a ∪ b) C ((a ∩ b⊥ ) ∪ b)
12	9, 11	fh3r 475	. . 3 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∪ (a ∪ b)) = ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∪ (a ∪ b)) ∩ (((a ∩ b⊥ ) ∪ b) ∪ (a ∪ b)))
13	7, 8	fh3r 475	. . . . . . 7 (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∪ (a ∪ b)) = (((a⊥ ∪ b⊥ ) ∪ (a ∪ b)) ∩ ((a ∪ b⊥ ) ∪ (a ∪ b)))
14		ax-a2 31	. . . . . . . . 9 ((a⊥ ∪ b⊥ ) ∪ (a ∪ b)) = ((a ∪ b) ∪ (a⊥ ∪ b⊥ ))
15		or4 84	. . . . . . . . . 10 ((a ∪ b) ∪ (a⊥ ∪ b⊥ )) = ((a ∪ a⊥ ) ∪ (b ∪ b⊥ ))
16		df-t 41	. . . . . . . . . . . . 13 1 = (b ∪ b⊥ )
17	16	lor 70	. . . . . . . . . . . 12 ((a ∪ a⊥ ) ∪ 1) = ((a ∪ a⊥ ) ∪ (b ∪ b⊥ ))
18	17	ax-r1 35	. . . . . . . . . . 11 ((a ∪ a⊥ ) ∪ (b ∪ b⊥ )) = ((a ∪ a⊥ ) ∪ 1)
19		or1 104	. . . . . . . . . . 11 ((a ∪ a⊥ ) ∪ 1) = 1
20	18, 19	ax-r2 36	. . . . . . . . . 10 ((a ∪ a⊥ ) ∪ (b ∪ b⊥ )) = 1
21	15, 20	ax-r2 36	. . . . . . . . 9 ((a ∪ b) ∪ (a⊥ ∪ b⊥ )) = 1
22	14, 21	ax-r2 36	. . . . . . . 8 ((a⊥ ∪ b⊥ ) ∪ (a ∪ b)) = 1
23		ax-a2 31	. . . . . . . . 9 ((a ∪ b⊥ ) ∪ (a ∪ b)) = ((a ∪ b) ∪ (a ∪ b⊥ ))
24		or4 84	. . . . . . . . . 10 ((a ∪ b) ∪ (a ∪ b⊥ )) = ((a ∪ a) ∪ (b ∪ b⊥ ))
25	16	lor 70	. . . . . . . . . . . 12 ((a ∪ a) ∪ 1) = ((a ∪ a) ∪ (b ∪ b⊥ ))
26	25	ax-r1 35	. . . . . . . . . . 11 ((a ∪ a) ∪ (b ∪ b⊥ )) = ((a ∪ a) ∪ 1)
27		or1 104	. . . . . . . . . . 11 ((a ∪ a) ∪ 1) = 1
28	26, 27	ax-r2 36	. . . . . . . . . 10 ((a ∪ a) ∪ (b ∪ b⊥ )) = 1
29	24, 28	ax-r2 36	. . . . . . . . 9 ((a ∪ b) ∪ (a ∪ b⊥ )) = 1
30	23, 29	ax-r2 36	. . . . . . . 8 ((a ∪ b⊥ ) ∪ (a ∪ b)) = 1
31	22, 30	2an 79	. . . . . . 7 (((a⊥ ∪ b⊥ ) ∪ (a ∪ b)) ∩ ((a ∪ b⊥ ) ∪ (a ∪ b))) = (1 ∩ 1)
32	13, 31	ax-r2 36	. . . . . 6 (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∪ (a ∪ b)) = (1 ∩ 1)
33		an1 106	. . . . . 6 (1 ∩ 1) = 1
34	32, 33	ax-r2 36	. . . . 5 (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∪ (a ∪ b)) = 1
35		lea 160	. . . . . . 7 (a ∩ b⊥ ) ≤ a
36	35	leror 152	. . . . . 6 ((a ∩ b⊥ ) ∪ b) ≤ (a ∪ b)
37	36	df-le2 131	. . . . 5 (((a ∩ b⊥ ) ∪ b) ∪ (a ∪ b)) = (a ∪ b)
38	34, 37	2an 79	. . . 4 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∪ (a ∪ b)) ∩ (((a ∩ b⊥ ) ∪ b) ∪ (a ∪ b))) = (1 ∩ (a ∪ b))
39		ancom 74	. . . . 5 (1 ∩ (a ∪ b)) = ((a ∪ b) ∩ 1)
40		an1 106	. . . . 5 ((a ∪ b) ∩ 1) = (a ∪ b)
41	39, 40	ax-r2 36	. . . 4 (1 ∩ (a ∪ b)) = (a ∪ b)
42	38, 41	ax-r2 36	. . 3 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∪ (a ∪ b)) ∩ (((a ∩ b⊥ ) ∪ b) ∪ (a ∪ b))) = (a ∪ b)
43	12, 42	ax-r2 36	. 2 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∪ (a ∪ b)) = (a ∪ b)
44	2, 43	ax-r2 36	1 ((a →4 b)⊥ ∪ (a ∪ b)) = (a ∪ b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →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:  ud4lem3  585