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Theorem neldifsnd 3752
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 3751 . 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 1684    \ cdif 3149   {csn 3640
This theorem is referenced by:  difsneq  3757  acsfiindd  14280
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866  ax-ext 2264
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1310  df-ex 1529  df-nf 1532  df-sb 1630  df-clab 2270  df-cleq 2276  df-clel 2279  df-nfc 2408  df-ne 2448  df-v 2790  df-dif 3155  df-sn 3646
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