Theorem nfdh 1524

Description: Deduce that x is not free in ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)

Hypotheses

Ref	Expression
nfdh.1	|- ( ph -> A. x ph )
nfdh.2	|- ( ph -> ( ps -> A. x ps ) )

Assertion

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

Proof of Theorem nfdh

Step	Hyp	Ref	Expression
1		nfdh.1	. . 3 |- ( ph -> A. x ph )
2	1	nfi 1462	. 2 |- F/ x ph
3		nfdh.2	. 2 |- ( ph -> ( ps -> A. x ps ) )
4	2, 3	nfd 1523	1 |- ( ph -> F/ x ps )

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:  hbsbd  1982