Theorem nf5d 2118

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

Step	Hyp	Ref	Expression
1		nf5d.1	. . 3 |- F/ x ph
2		nf5d.2	. . 3 |- ( ph -> ( ps -> A. x ps ) )
3	1, 2	alrimi 2082	. 2 |- ( ph -> A. x ( ps -> A. x ps ) )
4		nf5-1 2023	. 2 |- ( A. x ( ps -> A. x ps ) -> F/ x ps )
5	3, 4	syl 17	1 |- ( ph -> F/ x ps )

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

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-ex 1705  df-nf 1710

This theorem is referenced by:  axc16nfOLD  2163  nfaldOLD  2166  dvelimhw  2173  cbv1h  2268  nfeqf  2301  axc16nfALT  2323  nfsb2  2360  distel  31709  bj-cbv1hv  32730  wl-ax11-lem3  33364