Theorem ud3lem1 570

Description: Lemma for unified disjunction. (Contributed by NM, 27-Nov-1997.)

Assertion

Ref	Expression
ud3lem1	((a →3 b) →3 (b →3 a)) = (a ∪ (a⊥ ∩ b⊥ ))

Proof of Theorem ud3lem1

Step	Hyp	Ref	Expression
1		df-i3 46	. 2 ((a →3 b) →3 (b →3 a)) = ((((a →3 b)⊥ ∩ (b →3 a)) ∪ ((a →3 b)⊥ ∩ (b →3 a)⊥ )) ∪ ((a →3 b) ∩ ((a →3 b)⊥ ∪ (b →3 a))))
2		ud3lem1a 566	. . . . . 6 ((a →3 b)⊥ ∩ (b →3 a)) = (a ∩ b⊥ )
3		ud3lem1b 567	. . . . . 6 ((a →3 b)⊥ ∩ (b →3 a)⊥ ) = 0
4	2, 3	2or 72	. . . . 5 (((a →3 b)⊥ ∩ (b →3 a)) ∪ ((a →3 b)⊥ ∩ (b →3 a)⊥ )) = ((a ∩ b⊥ ) ∪ 0)
5		or0 102	. . . . 5 ((a ∩ b⊥ ) ∪ 0) = (a ∩ b⊥ )
6	4, 5	ax-r2 36	. . . 4 (((a →3 b)⊥ ∩ (b →3 a)) ∪ ((a →3 b)⊥ ∩ (b →3 a)⊥ )) = (a ∩ b⊥ )
7		ud3lem1d 569	. . . 4 ((a →3 b) ∩ ((a →3 b)⊥ ∪ (b →3 a))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))
8	6, 7	2or 72	. . 3 ((((a →3 b)⊥ ∩ (b →3 a)) ∪ ((a →3 b)⊥ ∩ (b →3 a)⊥ )) ∪ ((a →3 b) ∩ ((a →3 b)⊥ ∪ (b →3 a)))) = ((a ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b))))
9		coman1 185	. . . . . . 7 (a ∩ b⊥ ) C a
10	9	comcom2 183	. . . . . . . 8 (a ∩ b⊥ ) C a⊥
11		coman2 186	. . . . . . . . 9 (a ∩ b⊥ ) C b⊥
12	11	comcom7 460	. . . . . . . 8 (a ∩ b⊥ ) C b
13	10, 12	com2or 483	. . . . . . 7 (a ∩ b⊥ ) C (a⊥ ∪ b)
14	9, 13	fh3 471	. . . . . 6 ((a ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b))) = (((a ∩ b⊥ ) ∪ a) ∩ ((a ∩ b⊥ ) ∪ (a⊥ ∪ b)))
15		ax-a2 31	. . . . . . . . 9 ((a ∩ b⊥ ) ∪ a) = (a ∪ (a ∩ b⊥ ))
16		orabs 120	. . . . . . . . 9 (a ∪ (a ∩ b⊥ )) = a
17	15, 16	ax-r2 36	. . . . . . . 8 ((a ∩ b⊥ ) ∪ a) = a
18		ax-a2 31	. . . . . . . . 9 ((a ∩ b⊥ ) ∪ (a⊥ ∪ b)) = ((a⊥ ∪ b) ∪ (a ∩ b⊥ ))
19		anor1 88	. . . . . . . . . . 11 (a ∩ b⊥ ) = (a⊥ ∪ b)⊥
20	19	lor 70	. . . . . . . . . 10 ((a⊥ ∪ b) ∪ (a ∩ b⊥ )) = ((a⊥ ∪ b) ∪ (a⊥ ∪ b)⊥ )
21		df-t 41	. . . . . . . . . . 11 1 = ((a⊥ ∪ b) ∪ (a⊥ ∪ b)⊥ )
22	21	ax-r1 35	. . . . . . . . . 10 ((a⊥ ∪ b) ∪ (a⊥ ∪ b)⊥ ) = 1
23	20, 22	ax-r2 36	. . . . . . . . 9 ((a⊥ ∪ b) ∪ (a ∩ b⊥ )) = 1
24	18, 23	ax-r2 36	. . . . . . . 8 ((a ∩ b⊥ ) ∪ (a⊥ ∪ b)) = 1
25	17, 24	2an 79	. . . . . . 7 (((a ∩ b⊥ ) ∪ a) ∩ ((a ∩ b⊥ ) ∪ (a⊥ ∪ b))) = (a ∩ 1)
26		an1 106	. . . . . . 7 (a ∩ 1) = a
27	25, 26	ax-r2 36	. . . . . 6 (((a ∩ b⊥ ) ∪ a) ∩ ((a ∩ b⊥ ) ∪ (a⊥ ∪ b))) = a
28	14, 27	ax-r2 36	. . . . 5 ((a ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b))) = a
29	28	lor 70	. . . 4 ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))) = ((a⊥ ∩ b⊥ ) ∪ a)
30		or12 80	. . . 4 ((a ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))) = ((a⊥ ∩ b⊥ ) ∪ ((a ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b))))
31		ax-a2 31	. . . 4 (a ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ a)
32	29, 30, 31	3tr1 63	. . 3 ((a ∩ b⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))) = (a ∪ (a⊥ ∩ b⊥ ))
33	8, 32	ax-r2 36	. 2 ((((a →3 b)⊥ ∩ (b →3 a)) ∪ ((a →3 b)⊥ ∩ (b →3 a)⊥ )) ∪ ((a →3 b) ∩ ((a →3 b)⊥ ∪ (b →3 a)))) = (a ∪ (a⊥ ∩ b⊥ ))
34	1, 33	ax-r2 36	1 ((a →3 b) →3 (b →3 a)) = (a ∪ (a⊥ ∩ b⊥ ))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8  0wf 9   →₃ wi3 14

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-i3 46  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  ud3  597  u3lem11a  787