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Theorem nfdh 1460
Description: Deduce that  x is not free in  ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)
Hypotheses
Ref Expression
nfdh.1  |-  ( ph  ->  A. x ph )
nfdh.2  |-  ( ph  ->  ( ps  ->  A. x ps ) )
Assertion
Ref Expression
nfdh  |-  ( ph  ->  F/ x ps )

Proof of Theorem nfdh
StepHypRef Expression
1 nfdh.1 . . 3  |-  ( ph  ->  A. x ph )
21nfi 1394 . 2  |-  F/ x ph
3 nfdh.2 . 2  |-  ( ph  ->  ( ps  ->  A. x ps ) )
42, 3nfd 1459 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1285   F/wnf 1392
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1379  ax-gen 1381  ax-4 1443
This theorem depends on definitions:  df-bi 115  df-nf 1393
This theorem is referenced by:  hbsbd  1903
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