Theorem ud5lem3c 593

Description: Lemma for unified disjunction. (Contributed by NM, 26-Nov-1997.)

Assertion

Ref	Expression
ud5lem3c	((a →5 b)⊥ ∩ (a ∪ (a⊥ ∩ b))⊥ ) = (((a ∪ b) ∩ (a ∪ b⊥ )) ∩ a⊥ )

Proof of Theorem ud5lem3c

Step	Hyp	Ref	Expression
1		ud5lem0c 281	. . 3 (a →5 b)⊥ = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b))
2		oran 87	. . . . 5 (a ∪ (a⊥ ∩ b)) = (a⊥ ∩ (a⊥ ∩ b)⊥ )⊥
3	2	con2 67	. . . 4 (a ∪ (a⊥ ∩ b))⊥ = (a⊥ ∩ (a⊥ ∩ b)⊥ )
4		anor2 89	. . . . . 6 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
5	4	con2 67	. . . . 5 (a⊥ ∩ b)⊥ = (a ∪ b⊥ )
6	5	lan 77	. . . 4 (a⊥ ∩ (a⊥ ∩ b)⊥ ) = (a⊥ ∩ (a ∪ b⊥ ))
7	3, 6	ax-r2 36	. . 3 (a ∪ (a⊥ ∩ b))⊥ = (a⊥ ∩ (a ∪ b⊥ ))
8	1, 7	2an 79	. 2 ((a →5 b)⊥ ∩ (a ∪ (a⊥ ∩ b))⊥ ) = ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b)) ∩ (a⊥ ∩ (a ∪ b⊥ )))
9		an32 83	. . 3 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b)) ∩ (a⊥ ∩ (a ∪ b⊥ ))) = ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a⊥ ∩ (a ∪ b⊥ ))) ∩ (a ∪ b))
10		an4 86	. . . . . . 7 (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a⊥ ∩ (a ∪ b⊥ ))) = (((a⊥ ∪ b⊥ ) ∩ a⊥ ) ∩ ((a ∪ b⊥ ) ∩ (a ∪ b⊥ )))
11		ancom 74	. . . . . . . . . 10 ((a⊥ ∪ b⊥ ) ∩ a⊥ ) = (a⊥ ∩ (a⊥ ∪ b⊥ ))
12		anabs 121	. . . . . . . . . 10 (a⊥ ∩ (a⊥ ∪ b⊥ )) = a⊥
13	11, 12	ax-r2 36	. . . . . . . . 9 ((a⊥ ∪ b⊥ ) ∩ a⊥ ) = a⊥
14		anidm 111	. . . . . . . . 9 ((a ∪ b⊥ ) ∩ (a ∪ b⊥ )) = (a ∪ b⊥ )
15	13, 14	2an 79	. . . . . . . 8 (((a⊥ ∪ b⊥ ) ∩ a⊥ ) ∩ ((a ∪ b⊥ ) ∩ (a ∪ b⊥ ))) = (a⊥ ∩ (a ∪ b⊥ ))
16		ancom 74	. . . . . . . 8 (a⊥ ∩ (a ∪ b⊥ )) = ((a ∪ b⊥ ) ∩ a⊥ )
17	15, 16	ax-r2 36	. . . . . . 7 (((a⊥ ∪ b⊥ ) ∩ a⊥ ) ∩ ((a ∪ b⊥ ) ∩ (a ∪ b⊥ ))) = ((a ∪ b⊥ ) ∩ a⊥ )
18	10, 17	ax-r2 36	. . . . . 6 (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a⊥ ∩ (a ∪ b⊥ ))) = ((a ∪ b⊥ ) ∩ a⊥ )
19	18	ran 78	. . . . 5 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a⊥ ∩ (a ∪ b⊥ ))) ∩ (a ∪ b)) = (((a ∪ b⊥ ) ∩ a⊥ ) ∩ (a ∪ b))
20		ancom 74	. . . . 5 (((a ∪ b⊥ ) ∩ a⊥ ) ∩ (a ∪ b)) = ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ a⊥ ))
21	19, 20	ax-r2 36	. . . 4 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a⊥ ∩ (a ∪ b⊥ ))) ∩ (a ∪ b)) = ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ a⊥ ))
22		anass 76	. . . . 5 (((a ∪ b) ∩ (a ∪ b⊥ )) ∩ a⊥ ) = ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ a⊥ ))
23	22	ax-r1 35	. . . 4 ((a ∪ b) ∩ ((a ∪ b⊥ ) ∩ a⊥ )) = (((a ∪ b) ∩ (a ∪ b⊥ )) ∩ a⊥ )
24	21, 23	ax-r2 36	. . 3 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a⊥ ∩ (a ∪ b⊥ ))) ∩ (a ∪ b)) = (((a ∪ b) ∩ (a ∪ b⊥ )) ∩ a⊥ )
25	9, 24	ax-r2 36	. 2 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b)) ∩ (a⊥ ∩ (a ∪ b⊥ ))) = (((a ∪ b) ∩ (a ∪ b⊥ )) ∩ a⊥ )
26	8, 25	ax-r2 36	1 ((a →5 b)⊥ ∩ (a ∪ (a⊥ ∩ b))⊥ ) = (((a ∪ b) ∩ (a ∪ b⊥ )) ∩ a⊥ )

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₅ wi5 16

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-t 41  df-f 42  df-i5 48

This theorem is used by:  ud5lem3  594