Theorem nfdv 1639

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.xps))

Assertion

Ref	Expression
nfdv	|- (ph -> F/xps)

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.xps))
2	1	alrimiv 1631	. 2 |- (ph -> A.x(ps -> A.xps))
3		df-nf 1545	. 2 |- ( F/xps <-> A.x(ps -> A.xps))
4	2, 3	sylibr 203	1 |- (ph -> F/xps)

Syntax hints:   -> wi 4  A.wal 1540   F/wnf 1544

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616

This theorem depends on definitions:  df-bi 177  df-nf 1545

This theorem is referenced by: (None)