Metamath Proof Explorer


Theorem nfsbv

Description: If z is not free in ph , then it is not free in [ y / x ] ph when z is disjoint from both x and y . Version of nfsb with an additional disjoint variable condition on x , z but not requiring ax-13 . (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 𝑧 𝜑
Assertion nfsbv 𝑧 [ 𝑦 / 𝑥 ] 𝜑

Proof

Step Hyp Ref Expression
1 nfsbv.nf 𝑧 𝜑
2 1 nf5ri ( 𝜑 → ∀ 𝑧 𝜑 )
3 2 hbsbw ( [ 𝑦 / 𝑥 ] 𝜑 → ∀ 𝑧 [ 𝑦 / 𝑥 ] 𝜑 )
4 3 nf5i 𝑧 [ 𝑦 / 𝑥 ] 𝜑