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Theorem nfdv 1620
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 1617 . 2  |-  ( ph  ->  A. x ( ps 
->  A. x ps )
)
3 df-nf 1532 . 2  |-  ( F/ x ps  <->  A. x
( ps  ->  A. x ps ) )
42, 3sylibr 203 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1527   F/wnf 1531
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
This theorem depends on definitions:  df-bi 177  df-nf 1532
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