Theorem nfnf 1565

Description: If x is not free in ph, it is not free in F/ y ph. (Contributed by Mario Carneiro, 11-Aug-2016.) (Proof shortened by Wolf Lammen, 30-Dec-2017.)

Hypothesis

Ref	Expression
nfal.1	|- F/ x ph

Assertion

Ref	Expression
nfnf	|- F/ x F/ y ph

Proof of Theorem nfnf

Step	Hyp	Ref	Expression
1		df-nf 1449	. 2 |- ( F/ y ph <-> A. y ( ph -> A. y ph ) )
2		nfal.1	. . . 4 |- F/ x ph
3	2	nfal 1564	. . . 4 |- F/ x A. y ph
4	2, 3	nfim 1560	. . 3 |- F/ x ( ph -> A. y ph )
5	4	nfal 1564	. 2 |- F/ x A. y ( ph -> A. y ph )
6	1, 5	nfxfr 1462	1 |- F/ x F/ y ph

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-7 1436  ax-gen 1437  ax-4 1498  ax-ial 1522  ax-i5r 1523

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

This theorem is referenced by:  nfnfc  2315