Theorem ud3lem1d 569

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

Assertion

Ref	Expression
ud3lem1d	((a →3 b) ∩ ((a →3 b)⊥ ∪ (b →3 a))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))

Proof of Theorem ud3lem1d

Step	Hyp	Ref	Expression
1		df-i3 46	. . 3 (a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))
2		ud3lem1c 568	. . 3 ((a →3 b)⊥ ∪ (b →3 a)) = (a ∪ b⊥ )
3	1, 2	2an 79	. 2 ((a →3 b) ∩ ((a →3 b)⊥ ∪ (b →3 a))) = ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∩ (a ∪ b⊥ ))
4		comor1 461	. . . . . . 7 (a ∪ b⊥ ) C a
5	4	comcom2 183	. . . . . 6 (a ∪ b⊥ ) C a⊥
6		comor2 462	. . . . . . 7 (a ∪ b⊥ ) C b⊥
7	6	comcom7 460	. . . . . 6 (a ∪ b⊥ ) C b
8	5, 7	com2an 484	. . . . 5 (a ∪ b⊥ ) C (a⊥ ∩ b)
9	5, 6	com2an 484	. . . . 5 (a ∪ b⊥ ) C (a⊥ ∩ b⊥ )
10	8, 9	com2or 483	. . . 4 (a ∪ b⊥ ) C ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))
11	5, 7	com2or 483	. . . . 5 (a ∪ b⊥ ) C (a⊥ ∪ b)
12	4, 11	com2an 484	. . . 4 (a ∪ b⊥ ) C (a ∩ (a⊥ ∪ b))
13	10, 12	fh1r 473	. . 3 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∩ (a ∪ b⊥ )) = ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ b⊥ )) ∪ ((a ∩ (a⊥ ∪ b)) ∩ (a ∪ b⊥ )))
14	8, 9	fh1r 473	. . . . 5 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ b⊥ )) = (((a⊥ ∩ b) ∩ (a ∪ b⊥ )) ∪ ((a⊥ ∩ b⊥ ) ∩ (a ∪ b⊥ )))
15		an32 83	. . . . . 6 ((a ∩ (a⊥ ∪ b)) ∩ (a ∪ b⊥ )) = ((a ∩ (a ∪ b⊥ )) ∩ (a⊥ ∪ b))
16		anabs 121	. . . . . . 7 (a ∩ (a ∪ b⊥ )) = a
17	16	ran 78	. . . . . 6 ((a ∩ (a ∪ b⊥ )) ∩ (a⊥ ∪ b)) = (a ∩ (a⊥ ∪ b))
18	15, 17	ax-r2 36	. . . . 5 ((a ∩ (a⊥ ∪ b)) ∩ (a ∪ b⊥ )) = (a ∩ (a⊥ ∪ b))
19	14, 18	2or 72	. . . 4 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ b⊥ )) ∪ ((a ∩ (a⊥ ∪ b)) ∩ (a ∪ b⊥ ))) = ((((a⊥ ∩ b) ∩ (a ∪ b⊥ )) ∪ ((a⊥ ∩ b⊥ ) ∩ (a ∪ b⊥ ))) ∪ (a ∩ (a⊥ ∪ b)))
20		ancom 74	. . . . . . . 8 ((a⊥ ∩ b) ∩ (a ∪ b⊥ )) = ((a ∪ b⊥ ) ∩ (a⊥ ∩ b))
21		anor2 89	. . . . . . . . . 10 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
22	21	lan 77	. . . . . . . . 9 ((a ∪ b⊥ ) ∩ (a⊥ ∩ b)) = ((a ∪ b⊥ ) ∩ (a ∪ b⊥ )⊥ )
23		dff 101	. . . . . . . . . 10 0 = ((a ∪ b⊥ ) ∩ (a ∪ b⊥ )⊥ )
24	23	ax-r1 35	. . . . . . . . 9 ((a ∪ b⊥ ) ∩ (a ∪ b⊥ )⊥ ) = 0
25	22, 24	ax-r2 36	. . . . . . . 8 ((a ∪ b⊥ ) ∩ (a⊥ ∩ b)) = 0
26	20, 25	ax-r2 36	. . . . . . 7 ((a⊥ ∩ b) ∩ (a ∪ b⊥ )) = 0
27		lear 161	. . . . . . . . 9 (a⊥ ∩ b⊥ ) ≤ b⊥
28		leor 159	. . . . . . . . 9 b⊥ ≤ (a ∪ b⊥ )
29	27, 28	letr 137	. . . . . . . 8 (a⊥ ∩ b⊥ ) ≤ (a ∪ b⊥ )
30	29	df2le2 136	. . . . . . 7 ((a⊥ ∩ b⊥ ) ∩ (a ∪ b⊥ )) = (a⊥ ∩ b⊥ )
31	26, 30	2or 72	. . . . . 6 (((a⊥ ∩ b) ∩ (a ∪ b⊥ )) ∪ ((a⊥ ∩ b⊥ ) ∩ (a ∪ b⊥ ))) = (0 ∪ (a⊥ ∩ b⊥ ))
32		or0r 103	. . . . . 6 (0 ∪ (a⊥ ∩ b⊥ )) = (a⊥ ∩ b⊥ )
33	31, 32	ax-r2 36	. . . . 5 (((a⊥ ∩ b) ∩ (a ∪ b⊥ )) ∪ ((a⊥ ∩ b⊥ ) ∩ (a ∪ b⊥ ))) = (a⊥ ∩ b⊥ )
34	33	ax-r5 38	. . . 4 ((((a⊥ ∩ b) ∩ (a ∪ b⊥ )) ∪ ((a⊥ ∩ b⊥ ) ∩ (a ∪ b⊥ ))) ∪ (a ∩ (a⊥ ∪ b))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))
35	19, 34	ax-r2 36	. . 3 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∩ (a ∪ b⊥ )) ∪ ((a ∩ (a⊥ ∪ b)) ∩ (a ∪ b⊥ ))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))
36	13, 35	ax-r2 36	. 2 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∩ (a ∪ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))
37	3, 36	ax-r2 36	1 ((a →3 b) ∩ ((a →3 b)⊥ ∪ (b →3 a))) = ((a⊥ ∩ b⊥ ) ∪ (a ∩ (a⊥ ∪ b)))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  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:  ud3lem1  570