Theorem nfdh 1457

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 1391	. 2 |- F/ x ph
3		nfdh.2	. 2 |- ( ph -> ( ps -> A. x ps ) )
4	2, 3	nfd 1456	1 |- ( ph -> F/ x ps )

Syntax hints:   -> wi 4   A.wal 1282   F/wnf 1389

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 1376  ax-gen 1378  ax-4 1440

This theorem depends on definitions:  df-bi 115  df-nf 1390

This theorem is referenced by:  hbsbd  1899