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Theorem dfnb 95
 Description: Negated biconditional. (Contributed by NM, 30-Aug-1997.)
Assertion
Ref Expression
dfnb (ab) = ((ab) ∩ (ab ))

Proof of Theorem dfnb
StepHypRef Expression
1 oran 87 . . . 4 ((ab) ∪ (ab )) = ((ab) ∩ (ab ) )
21con2 67 . . 3 ((ab) ∪ (ab )) = ((ab) ∩ (ab ) )
3 ancom 74 . . 3 ((ab) ∩ (ab ) ) = ((ab ) ∩ (ab) )
42, 3ax-r2 36 . 2 ((ab) ∪ (ab )) = ((ab ) ∩ (ab) )
5 dfb 94 . . 3 (ab) = ((ab) ∪ (ab ))
65ax-r4 37 . 2 (ab) = ((ab) ∪ (ab ))
7 oran 87 . . 3 (ab) = (ab )
8 df-a 40 . . . . 5 (ab) = (ab )
98con2 67 . . . 4 (ab) = (ab )
109ax-r1 35 . . 3 (ab ) = (ab)
117, 102an 79 . 2 ((ab) ∩ (ab )) = ((ab ) ∩ (ab) )
124, 6, 113tr1 63 1 (ab) = ((ab) ∩ (ab ))
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7 This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 This theorem depends on definitions:  df-b 39  df-a 40 This theorem is referenced by:  wnbdi  429  ska2  432  ska4  433  nbdi  486  test2  803
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