Metamath Proof Explorer


Theorem nfsbv

Description: If z is not free in ph , it is not free in [ y / x ] ph when z is distinct from x and y . Version of nfsb requiring more disjoint variables, but fewer axioms. (Contributed by Mario Carneiro, 11-Aug-2016) (Revised by Wolf Lammen, 7-Feb-2023) Remove disjoint variable condition on x , y . (Revised by Steven Nguyen, 13-Aug-2023) (Proof shortened by Wolf Lammen, 25-Oct-2024)

Ref Expression
Hypothesis nfsbv.nf
|- F/ z ph
Assertion nfsbv
|- F/ z [ y / x ] ph

Proof

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