Metamath Proof Explorer


Theorem nfs1f

Description: If x is not free in ph , it is not free in [ y / x ] ph . (Contributed by Mario Carneiro, 11-Aug-2016)

Ref Expression
Hypothesis nfs1f.1
|- F/ x ph
Assertion nfs1f
|- F/ x [ y / x ] ph

Proof

Step Hyp Ref Expression
1 nfs1f.1
 |-  F/ x ph
2 1 sbf
 |-  ( [ y / x ] ph <-> ph )
3 2 1 nfxfr
 |-  F/ x [ y / x ] ph