Theorem nfdv 1865

Description: Apply the definition of not-free in a context. (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypothesis

Ref	Expression
nfdv.1	|- ( ph -> ( ps -> A. x ps ) )

Assertion

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

Distinct variable group:   ph, x

Allowed substitution hint:   ps( x)

Proof of Theorem nfdv

Step	Hyp	Ref	Expression
1		nfdv.1	. . 3 |- ( ph -> ( ps -> A. x ps ) )
2	1	alrimiv 1862	. 2 |- ( ph -> A. x ( ps -> A. x ps ) )
3		df-nf 1449	. 2 |- ( F/ x ps <-> A. x ( ps -> A. x ps ) )
4	2, 3	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-17 1514

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

This theorem is referenced by: (None)