Theorem nbdi 486

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

Assertion

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

Proof of Theorem nbdi

Step	Hyp	Ref	Expression
1		dfnb 95	. 2 (a ≡ b)⊥ = ((a ∪ b) ∩ (a⊥ ∪ b⊥ ))
2		comorr 184	. . . . 5 a C (a ∪ b)
3	2	comcom 453	. . . 4 (a ∪ b) C a
4	3	comcom2 183	. . 3 (a ∪ b) C a⊥
5		comorr 184	. . . . . 6 b C (b ∪ a)
6		ax-a2 31	. . . . . 6 (b ∪ a) = (a ∪ b)
7	5, 6	cbtr 182	. . . . 5 b C (a ∪ b)
8	7	comcom 453	. . . 4 (a ∪ b) C b
9	8	comcom2 183	. . 3 (a ∪ b) C b⊥
10	4, 9	fh1 469	. 2 ((a ∪ b) ∩ (a⊥ ∪ b⊥ )) = (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ b⊥ ))
11	1, 10	ax-r2 36	1 (a ≡ b)⊥ = (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ 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-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-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)