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Theorem nfdh 1977
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 1682 . 2  |-  F/ x ph
3 nfdh.2 . 2  |-  ( ph  ->  ( ps  ->  A. x ps ) )
42, 3nfd 1976 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff setvar class
Syntax hints:    -> wi 4   A.wal 1450   F/wnf 1675
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1677  ax-4 1690  ax-5 1766  ax-6 1813  ax-7 1859  ax-12 1950
This theorem depends on definitions:  df-bi 190  df-ex 1672  df-nf 1676
This theorem is referenced by:  hbimd  2024  ax12indalem  32580  ax12inda2ALT  32581
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