Theorem nfd 1511

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 1507	. . 3 |- ( ph -> A. x ph )
3		nfd.2	. . 3 |- ( ph -> ( ps -> A. x ps ) )
4	2, 3	alrimih 1457	. 2 |- ( ph -> A. x ( ps -> A. x ps ) )
5		df-nf 1449	. 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 1341   F/wnf 1448

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 1435  ax-gen 1437  ax-4 1498

This theorem depends on definitions:  df-bi 116  df-nf 1449

This theorem is referenced by:  nfdh  1512  nfrimi  1513  nfnt  1644  cbv1h  1734  nfald  1748  a16nf  1854  dvelimALT  1998  dvelimfv  1999  nfsb4t  2002  hbeud  2036