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Theorem nfd2 2002
Description: Deduce that  x is not free in  ps in a context. (Contributed by Wolf Lammen, 16-Sep-2021.)
Hypothesis
Ref Expression
nfd2.1  |-  ( ph  ->  ( E. x ps 
->  A. x ps )
)
Assertion
Ref Expression
nfd2  |-  ( ph  ->  F/ x ps )

Proof of Theorem nfd2
StepHypRef Expression
1 nfd2.1 . 2  |-  ( ph  ->  ( E. x ps 
->  A. x ps )
)
2 nf2 1648 . 2  |-  ( F/ x ps  <->  ( E. x ps  ->  A. x ps ) )
31, 2sylibr 133 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1333   F/wnf 1440   E.wex 1472
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-gen 1429  ax-ie2 1474  ax-4 1490  ax-ial 1514
This theorem depends on definitions:  df-bi 116  df-nf 1441
This theorem is referenced by:  nf5-1  2004
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