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Theorem nfdv 1849
Description: Apply the definition of not-free in a context. (Contributed by Mario Carneiro, 11-Aug-2016.)
Hypothesis
Ref Expression
nfdv.1  |-  ( ph  ->  ( ps  ->  A. x ps ) )
Assertion
Ref Expression
nfdv  |-  ( ph  ->  F/ x ps )
Distinct variable group:    ph, x
Allowed substitution hint:    ps( x)

Proof of Theorem nfdv
StepHypRef Expression
1 nfdv.1 . . 3  |-  ( ph  ->  ( ps  ->  A. x ps ) )
21alrimiv 1846 . 2  |-  ( ph  ->  A. x ( ps 
->  A. x ps )
)
3 df-nf 1437 . 2  |-  ( F/ x ps  <->  A. x
( ps  ->  A. x ps ) )
42, 3sylibr 133 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1329   F/wnf 1436
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-5 1423  ax-gen 1425  ax-17 1506
This theorem depends on definitions:  df-bi 116  df-nf 1437
This theorem is referenced by: (None)
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