Theorem ud5lem0c 281

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

Assertion

Ref	Expression
ud5lem0c	(a →5 b)⊥ = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b))

Proof of Theorem ud5lem0c

Step	Hyp	Ref	Expression
1		df-i5 48	. . 3 (a →5 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))
2		oran 87	. . . 4 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = (((a ∩ b) ∪ (a⊥ ∩ b))⊥ ∩ (a⊥ ∩ b⊥ )⊥ )⊥
3		oran 87	. . . . . . . 8 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a ∩ b)⊥ ∩ (a⊥ ∩ b)⊥ )⊥
4		df-a 40	. . . . . . . . . . 11 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
5	4	con2 67	. . . . . . . . . 10 (a ∩ b)⊥ = (a⊥ ∪ b⊥ )
6		anor2 89	. . . . . . . . . . 11 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
7	6	con2 67	. . . . . . . . . 10 (a⊥ ∩ b)⊥ = (a ∪ b⊥ )
8	5, 7	2an 79	. . . . . . . . 9 ((a ∩ b)⊥ ∩ (a⊥ ∩ b)⊥ ) = ((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ ))
9	8	ax-r4 37	. . . . . . . 8 ((a ∩ b)⊥ ∩ (a⊥ ∩ b)⊥ )⊥ = ((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ ))⊥
10	3, 9	ax-r2 36	. . . . . . 7 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ ))⊥
11	10	con2 67	. . . . . 6 ((a ∩ b) ∪ (a⊥ ∩ b))⊥ = ((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ ))
12		oran 87	. . . . . . 7 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
13	12	ax-r1 35	. . . . . 6 (a⊥ ∩ b⊥ )⊥ = (a ∪ b)
14	11, 13	2an 79	. . . . 5 (((a ∩ b) ∪ (a⊥ ∩ b))⊥ ∩ (a⊥ ∩ b⊥ )⊥ ) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b))
15	14	ax-r4 37	. . . 4 (((a ∩ b) ∪ (a⊥ ∩ b))⊥ ∩ (a⊥ ∩ b⊥ )⊥ )⊥ = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b))⊥
16	2, 15	ax-r2 36	. . 3 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b))⊥
17	1, 16	ax-r2 36	. 2 (a →5 b) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b))⊥
18	17	con2 67	1 (a →5 b)⊥ = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (a ∪ b))

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-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-i5 48

This theorem is used by:  ud5lem1b  587  ud5lem1c  588  ud5lem3b  592  ud5lem3c  593