Theorem nfd 1534

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 1530	. . 3 |- ( ph -> A. x ph )
3		nfd.2	. . 3 |- ( ph -> ( ps -> A. x ps ) )
4	2, 3	alrimih 1480	. 2 |- ( ph -> A. x ( ps -> A. x ps ) )
5		df-nf 1472	. 2 |- ( F/ x ps <-> A. x ( ps -> A. x ps ) )
6	4, 5	sylibr 134	1 |- ( ph -> F/ x ps )

Syntax hints:   -> wi 4   A.wal 1362   F/wnf 1471

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-gen 1460  ax-4 1521

This theorem depends on definitions:  df-bi 117  df-nf 1472

This theorem is referenced by:  nfdh  1535  nfrimi  1536  nfnt  1667  cbv1h  1757  nfald  1771  a16nf  1877  dvelimALT  2026  dvelimfv  2027  nfsb4t  2030  hbeud  2064