Theorem nbdi 468

Description: Negated biconditional (distributive form)

Assertion

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

Proof of Theorem nbdi

Step	Hyp	Ref	Expression
1		dfnb 87	. 2 (a ≡ b)⊥ = ((a ∪ b) ∩ (a⊥ ∪ b⊥ ))
2		comorr 176	. . . . 5 a C (a ∪ b)
3	2	comcom 435	. . . 4 (a ∪ b) C a
4	3	comcom2 175	. . 3 (a ∪ b) C a⊥
5		comorr 176	. . . . . 6 b C (b ∪ a)
6		ax-a2 30	. . . . . 6 (b ∪ a) = (a ∪ b)
7	5, 6	cbtr 174	. . . . 5 b C (a ∪ b)
8	7	comcom 435	. . . 4 (a ∪ b) C b
9	8	comcom2 175	. . 3 (a ∪ b) C b⊥
10	4, 9	fh1 451	. 2 ((a ∪ b) ∩ (a⊥ ∪ b⊥ )) = (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ b⊥ ))
11	1, 10	ax-r2 35	1 (a ≡ b)⊥ = (((a ∪ b) ∩ a⊥ ) ∪ ((a ∪ b) ∩ b⊥ ))

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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