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Theorem nfd 1457
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 1453 . . 3  |-  ( ph  ->  A. x ph )
3 nfd.2 . . 3  |-  ( ph  ->  ( ps  ->  A. x ps ) )
42, 3alrimih 1399 . 2  |-  ( ph  ->  A. x ( ps 
->  A. x ps )
)
5 df-nf 1391 . 2  |-  ( F/ x ps  <->  A. x
( ps  ->  A. x ps ) )
64, 5sylibr 132 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1283   F/wnf 1390
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 1377  ax-gen 1379  ax-4 1441
This theorem depends on definitions:  df-bi 115  df-nf 1391
This theorem is referenced by:  nfdh  1458  nfrimi  1459  nfnt  1587  cbv1h  1674  nfald  1684  a16nf  1788  dvelimALT  1928  dvelimfv  1929  nfsb4t  1932  hbeud  1964
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