Theorem dfnb 95

Description: Negated biconditional. (Contributed by NM, 30-Aug-1997.)

Assertion

Ref	Expression
dfnb	(a ≡ b)⊥ = ((a ∪ b) ∩ (a⊥ ∪ b⊥ ))

Proof of Theorem dfnb

Step	Hyp	Ref	Expression
1		oran 87	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a ∩ b)⊥ ∩ (a⊥ ∩ b⊥ )⊥ )⊥
2	1	con2 67	. . 3 ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ = ((a ∩ b)⊥ ∩ (a⊥ ∩ b⊥ )⊥ )
3		ancom 74	. . 3 ((a ∩ b)⊥ ∩ (a⊥ ∩ b⊥ )⊥ ) = ((a⊥ ∩ b⊥ )⊥ ∩ (a ∩ b)⊥ )
4	2, 3	ax-r2 36	. 2 ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ = ((a⊥ ∩ b⊥ )⊥ ∩ (a ∩ b)⊥ )
5		dfb 94	. . 3 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
6	5	ax-r4 37	. 2 (a ≡ b)⊥ = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
7		oran 87	. . 3 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
8		df-a 40	. . . . 5 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
9	8	con2 67	. . . 4 (a ∩ b)⊥ = (a⊥ ∪ b⊥ )
10	9	ax-r1 35	. . 3 (a⊥ ∪ b⊥ ) = (a ∩ b)⊥
11	7, 10	2an 79	. 2 ((a ∪ b) ∩ (a⊥ ∪ b⊥ )) = ((a⊥ ∩ b⊥ )⊥ ∩ (a ∩ b)⊥ )
12	4, 6, 11	3tr1 63	1 (a ≡ b)⊥ = ((a ∪ b) ∩ (a⊥ ∪ b⊥ ))

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

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-b 39  df-a 40

This theorem is used by:  wnbdi  429  ska2  432  ska4  433  nbdi  486  test2  803