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Theorem neldifsnd 4155
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 4154 . 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 1767    \ cdif 3473   {csn 4027
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1601  ax-4 1612  ax-5 1680  ax-6 1719  ax-7 1739  ax-10 1786  ax-11 1791  ax-12 1803  ax-13 1968  ax-ext 2445
This theorem depends on definitions:  df-bi 185  df-an 371  df-tru 1382  df-ex 1597  df-nf 1600  df-sb 1712  df-clab 2453  df-cleq 2459  df-clel 2462  df-nfc 2617  df-ne 2664  df-v 3115  df-dif 3479  df-sn 4028
This theorem is referenced by:  difsnb  4169  fsnunf2  6101  rpnnen2lem9  13820  ramub1lem1  14406  ramub1lem2  14407  acsfiindd  15667  gsummgp0  17069  islindf4  18680  gsummatr01lem3  18966  onint1  29767  prtlem80  30430  fsumsplitsndif  32040  mgpsumunsn  32246
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