Theorem nfd 1416

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 1412	. . 3 |- ( ph -> A. x ph )
3		nfd.2	. . 3 |- ( ph -> ( ps -> A. x ps ) )
4	2, 3	alrimih 1358	. 2 |- ( ph -> A. x ( ps -> A. x ps ) )
5		df-nf 1350	. 2 |- ( F/ x ps <-> A. x ( ps -> A. x ps ) )
6	4, 5	sylibr 137	1 |- ( ph -> F/ x ps )

Syntax hints:   -> wi 4   A.wal 1241   F/wnf 1349

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-4 1400

This theorem depends on definitions:  df-bi 110  df-nf 1350

This theorem is referenced by:  nfdh  1417  nfrimi  1418  nfnt  1546  cbv1h  1633  nfald  1643  a16nf  1746  dvelimALT  1886  dvelimfv  1887  nfsb4t  1890  hbeud  1922