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Theorem neldifsnd 3765
Description:  A is not in  ( B  \  { A } ). Deduction form. (Contributed by David Moews, 1-May-2017.)
Assertion
Ref Expression
neldifsnd  |-  ( ph  ->  -.  A  e.  ( B  \  { A } ) )

Proof of Theorem neldifsnd
StepHypRef Expression
1 neldifsn 3764 . 2  |-  -.  A  e.  ( B  \  { A } )
21a1i 10 1  |-  ( ph  ->  -.  A  e.  ( B  \  { A } ) )
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    e. wcel 1696    \ cdif 3162   {csn 3653
This theorem is referenced by:  difsnb  3773  acsfiindd  14296
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547  ax-17 1606  ax-9 1644  ax-8 1661  ax-6 1715  ax-7 1720  ax-11 1727  ax-12 1878  ax-ext 2277
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639  df-clab 2283  df-cleq 2289  df-clel 2292  df-nfc 2421  df-ne 2461  df-v 2803  df-dif 3168  df-sn 3659
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