Theorem nfd 1716

Description: Deduce that x is not free in ps in a context. (Contributed by Wolf Lammen, 16-Sep-2021.)

Hypothesis

Ref	Expression
nfd.1	|- ( ph -> ( E. x ps -> A. x ps ) )

Assertion

Ref	Expression
nfd	|- ( ph -> F/ x ps )

Proof of Theorem nfd

Step	Hyp	Ref	Expression
1		nfd.1	. 2 |- ( ph -> ( E. x ps -> A. x ps ) )
2		df-nf 1710	. 2 |- ( F/ x ps <-> ( E. x ps -> A. x ps ) )
3	1, 2	sylibr 224	1 |- ( ph -> F/ x ps )

Syntax hints:   -> wi 4   A.wal 1481   E.wex 1704   F/wnf 1708

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 197  df-nf 1710

This theorem is referenced by:  nfimdOLDOLD  1824  nf5-1  2023  axc16nf  2137  nfald  2165