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Theorem nfd 1523
Description: Deduce that  x is not free in  ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)
Hypotheses
Ref Expression
nfd.1  |-  F/ x ph
nfd.2  |-  ( ph  ->  ( ps  ->  A. x ps ) )
Assertion
Ref Expression
nfd  |-  ( ph  ->  F/ x ps )

Proof of Theorem nfd
StepHypRef Expression
1 nfd.1 . . . 4  |-  F/ x ph
21nfri 1519 . . 3  |-  ( ph  ->  A. x ph )
3 nfd.2 . . 3  |-  ( ph  ->  ( ps  ->  A. x ps ) )
42, 3alrimih 1469 . 2  |-  ( ph  ->  A. x ( ps 
->  A. x ps )
)
5 df-nf 1461 . 2  |-  ( F/ x ps  <->  A. x
( ps  ->  A. x ps ) )
64, 5sylibr 134 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1351   F/wnf 1460
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1447  ax-gen 1449  ax-4 1510
This theorem depends on definitions:  df-bi 117  df-nf 1461
This theorem is referenced by:  nfdh  1524  nfrimi  1525  nfnt  1656  cbv1h  1746  nfald  1760  a16nf  1866  dvelimALT  2010  dvelimfv  2011  nfsb4t  2014  hbeud  2048
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