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Theorem neldifsnd 4015
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 4014 . 2  |-  -.  A  e.  ( B  \  { A } )
21a1i 11 1  |-  ( ph  ->  -.  A  e.  ( B  \  { A } ) )
Colors of variables: wff setvar class
Syntax hints:   -. wn 3    -> wi 4    e. wcel 1756    \ cdif 3337   {csn 3889
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1591  ax-4 1602  ax-5 1670  ax-6 1708  ax-7 1728  ax-10 1775  ax-11 1780  ax-12 1792  ax-13 1943  ax-ext 2423
This theorem depends on definitions:  df-bi 185  df-an 371  df-tru 1372  df-ex 1587  df-nf 1590  df-sb 1701  df-clab 2430  df-cleq 2436  df-clel 2439  df-nfc 2577  df-ne 2620  df-v 2986  df-dif 3343  df-sn 3890
This theorem is referenced by:  difsnb  4027  fsnunf2  5929  rpnnen2lem9  13517  ramub1lem1  14099  ramub1lem2  14100  acsfiindd  15359  gsummgp0  16711  islindf4  18279  gsummatr01lem3  18475  onint1  28307  prtlem80  29015  fsumsplitsndif  30250  mgpsumunsn  30771
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