Metamath Proof Explorer


Theorem nfs1

Description: If y is not free in ph , x is not free in [ y / x ] ph . Usage of this theorem is discouraged because it depends on ax-13 . Check out nfs1v for a version requiring less axioms. (Contributed by Mario Carneiro, 11-Aug-2016) (New usage is discouraged.)

Ref Expression
Hypothesis nfs1.1
|- F/ y ph
Assertion nfs1
|- F/ x [ y / x ] ph

Proof

Step Hyp Ref Expression
1 nfs1.1
 |-  F/ y ph
2 1 nf5ri
 |-  ( ph -> A. y ph )
3 2 hbsb3
 |-  ( [ y / x ] ph -> A. x [ y / x ] ph )
4 3 nf5i
 |-  F/ x [ y / x ] ph