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Theorem nf5d 2005
Description: Deduce that  x is not free in  ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)
Hypotheses
Ref Expression
nf5d.1  |-  F/ x ph
nf5d.2  |-  ( ph  ->  ( ps  ->  A. x ps ) )
Assertion
Ref Expression
nf5d  |-  ( ph  ->  F/ x ps )

Proof of Theorem nf5d
StepHypRef Expression
1 nf5d.1 . . 3  |-  F/ x ph
2 nf5d.2 . . 3  |-  ( ph  ->  ( ps  ->  A. x ps ) )
31, 2alrimi 1502 . 2  |-  ( ph  ->  A. x ( ps 
->  A. x ps )
)
4 nf5-1 2004 . 2  |-  ( A. x ( ps  ->  A. x ps )  ->  F/ x ps )
53, 4syl 14 1  |-  ( ph  ->  F/ x ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4   A.wal 1333   F/wnf 1440
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1427  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-4 1490  ax-ial 1514  ax-i5r 1515
This theorem depends on definitions:  df-bi 116  df-nf 1441
This theorem is referenced by:  nfabdw  2318
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