Theorem nfd 1516

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

Hypotheses

Ref	Expression
nfd.1	|- F/ x ph
nfd.2	|- ( ph -> ( ps -> A. x ps ) )

Assertion

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

Proof of Theorem nfd

Step	Hyp	Ref	Expression
1		nfd.1	. . . 4 |- F/ x ph
2	1	nfri 1512	. . 3 |- ( ph -> A. x ph )
3		nfd.2	. . 3 |- ( ph -> ( ps -> A. x ps ) )
4	2, 3	alrimih 1462	. 2 |- ( ph -> A. x ( ps -> A. x ps ) )
5		df-nf 1454	. 2 |- ( F/ x ps <-> A. x ( ps -> A. x ps ) )
6	4, 5	sylibr 133	1 |- ( ph -> F/ x ps )

Syntax hints:   -> wi 4   A.wal 1346   F/wnf 1453

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 1440  ax-gen 1442  ax-4 1503

This theorem depends on definitions:  df-bi 116  df-nf 1454

This theorem is referenced by:  nfdh  1517  nfrimi  1518  nfnt  1649  cbv1h  1739  nfald  1753  a16nf  1859  dvelimALT  2003  dvelimfv  2004  nfsb4t  2007  hbeud  2041