Theorem nfnf 1600

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 1484	. 2 |- ( F/ y ph <-> A. y ( ph -> A. y ph ) )
2		nfal.1	. . . 4 |- F/ x ph
3	2	nfal 1599	. . . 4 |- F/ x A. y ph
4	2, 3	nfim 1595	. . 3 |- F/ x ( ph -> A. y ph )
5	4	nfal 1599	. 2 |- F/ x A. y ( ph -> A. y ph )
6	1, 5	nfxfr 1497	1 |- F/ x F/ y ph

Syntax hints:   -> wi 4   A.wal 1371   F/wnf 1483

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 1470  ax-7 1471  ax-gen 1472  ax-4 1533  ax-ial 1557  ax-i5r 1558

This theorem depends on definitions:  df-bi 117  df-nf 1484

This theorem is referenced by:  nfnfc  2355