Theorem nfd 1456

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 1452	. . 3 |- ( ph -> A. x ph )
3		nfd.2	. . 3 |- ( ph -> ( ps -> A. x ps ) )
4	2, 3	alrimih 1398	. 2 |- ( ph -> A. x ( ps -> A. x ps ) )
5		df-nf 1390	. 2 |- ( F/ x ps <-> A. x ( ps -> A. x ps ) )
6	4, 5	sylibr 132	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:  nfdh  1457  nfrimi  1458  nfnt  1586  cbv1h  1673  nfald  1683  a16nf  1787  dvelimALT  1927  dvelimfv  1928  nfsb4t  1931  hbeud  1963