Theorem ud1lem1 560

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

Assertion

Ref	Expression
ud1lem1	((a →1 b) →1 (b →1 a)) = (a ∪ (a⊥ ∩ b⊥ ))

Proof of Theorem ud1lem1

Step	Hyp	Ref	Expression
1		df-i1 44	. 2 ((a →1 b) →1 (b →1 a)) = ((a →1 b)⊥ ∪ ((a →1 b) ∩ (b →1 a)))
2		ud1lem0c 277	. . . 4 (a →1 b)⊥ = (a ∩ (a⊥ ∪ b⊥ ))
3		df-i1 44	. . . . 5 (a →1 b) = (a⊥ ∪ (a ∩ b))
4		df-i1 44	. . . . 5 (b →1 a) = (b⊥ ∪ (b ∩ a))
5	3, 4	2an 79	. . . 4 ((a →1 b) ∩ (b →1 a)) = ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (b ∩ a)))
6	2, 5	2or 72	. . 3 ((a →1 b)⊥ ∪ ((a →1 b) ∩ (b →1 a))) = ((a ∩ (a⊥ ∪ b⊥ )) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (b ∩ a))))
7		ancom 74	. . . . . . . 8 (b ∩ a) = (a ∩ b)
8	7	lor 70	. . . . . . 7 (b⊥ ∪ (b ∩ a)) = (b⊥ ∪ (a ∩ b))
9	8	lan 77	. . . . . 6 ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (b ∩ a))) = ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b)))
10		coman1 185	. . . . . . . . 9 (a ∩ b) C a
11	10	comcom2 183	. . . . . . . 8 (a ∩ b) C a⊥
12		coman2 186	. . . . . . . . 9 (a ∩ b) C b
13	12	comcom2 183	. . . . . . . 8 (a ∩ b) C b⊥
14	11, 13	fh3r 475	. . . . . . 7 ((a⊥ ∩ b⊥ ) ∪ (a ∩ b)) = ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b)))
15	14	ax-r1 35	. . . . . 6 ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (a ∩ b))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ b))
16	9, 15	ax-r2 36	. . . . 5 ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (b ∩ a))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ b))
17	16	lor 70	. . . 4 ((a ∩ (a⊥ ∪ b⊥ )) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (b ∩ a)))) = ((a ∩ (a⊥ ∪ b⊥ )) ∪ ((a⊥ ∩ b⊥ ) ∪ (a ∩ b)))
18		or12 80	. . . . 5 ((a ∩ (a⊥ ∪ b⊥ )) ∪ ((a⊥ ∩ b⊥ ) ∪ (a ∩ b))) = ((a⊥ ∩ b⊥ ) ∪ ((a ∩ (a⊥ ∪ b⊥ )) ∪ (a ∩ b)))
19	10	comcom 453	. . . . . . . 8 a C (a ∩ b)
20		comorr 184	. . . . . . . . . 10 a⊥ C (a⊥ ∪ b⊥ )
21	20	comcom2 183	. . . . . . . . 9 a⊥ C (a⊥ ∪ b⊥ )⊥
22	21	comcom5 458	. . . . . . . 8 a C (a⊥ ∪ b⊥ )
23	19, 22	fh4r 476	. . . . . . 7 ((a ∩ (a⊥ ∪ b⊥ )) ∪ (a ∩ b)) = ((a ∪ (a ∩ b)) ∩ ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)))
24	23	lor 70	. . . . . 6 ((a⊥ ∩ b⊥ ) ∪ ((a ∩ (a⊥ ∪ b⊥ )) ∪ (a ∩ b))) = ((a⊥ ∩ b⊥ ) ∪ ((a ∪ (a ∩ b)) ∩ ((a⊥ ∪ b⊥ ) ∪ (a ∩ b))))
25		orabs 120	. . . . . . . . . 10 (a ∪ (a ∩ b)) = a
26		df-a 40	. . . . . . . . . . . 12 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
27	26	lor 70	. . . . . . . . . . 11 ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)) = ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ )
28		df-t 41	. . . . . . . . . . . 12 1 = ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ )
29	28	ax-r1 35	. . . . . . . . . . 11 ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ ) = 1
30	27, 29	ax-r2 36	. . . . . . . . . 10 ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)) = 1
31	25, 30	2an 79	. . . . . . . . 9 ((a ∪ (a ∩ b)) ∩ ((a⊥ ∪ b⊥ ) ∪ (a ∩ b))) = (a ∩ 1)
32		an1 106	. . . . . . . . 9 (a ∩ 1) = a
33	31, 32	ax-r2 36	. . . . . . . 8 ((a ∪ (a ∩ b)) ∩ ((a⊥ ∪ b⊥ ) ∪ (a ∩ b))) = a
34	33	lor 70	. . . . . . 7 ((a⊥ ∩ b⊥ ) ∪ ((a ∪ (a ∩ b)) ∩ ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)))) = ((a⊥ ∩ b⊥ ) ∪ a)
35		ax-a2 31	. . . . . . 7 ((a⊥ ∩ b⊥ ) ∪ a) = (a ∪ (a⊥ ∩ b⊥ ))
36	34, 35	ax-r2 36	. . . . . 6 ((a⊥ ∩ b⊥ ) ∪ ((a ∪ (a ∩ b)) ∩ ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)))) = (a ∪ (a⊥ ∩ b⊥ ))
37	24, 36	ax-r2 36	. . . . 5 ((a⊥ ∩ b⊥ ) ∪ ((a ∩ (a⊥ ∪ b⊥ )) ∪ (a ∩ b))) = (a ∪ (a⊥ ∩ b⊥ ))
38	18, 37	ax-r2 36	. . . 4 ((a ∩ (a⊥ ∪ b⊥ )) ∪ ((a⊥ ∩ b⊥ ) ∪ (a ∩ b))) = (a ∪ (a⊥ ∩ b⊥ ))
39	17, 38	ax-r2 36	. . 3 ((a ∩ (a⊥ ∪ b⊥ )) ∪ ((a⊥ ∪ (a ∩ b)) ∩ (b⊥ ∪ (b ∩ a)))) = (a ∪ (a⊥ ∩ b⊥ ))
40	6, 39	ax-r2 36	. 2 ((a →1 b)⊥ ∪ ((a →1 b) ∩ (b →1 a))) = (a ∪ (a⊥ ∩ b⊥ ))
41	1, 40	ax-r2 36	1 ((a →1 b) →1 (b →1 a)) = (a ∪ (a⊥ ∩ b⊥ ))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12

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-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  ud1  595