Metamath Proof Explorer


Theorem bj-cbvaldv

Description: Version of cbvald with a disjoint variable condition, which does not require ax-13 . (Contributed by BJ, 16-Jun-2019) (Proof modification is discouraged.)

Ref Expression
Hypotheses bj-cbvaldv.1 y φ
bj-cbvaldv.2 φ y ψ
bj-cbvaldv.3 φ x = y ψ χ
Assertion bj-cbvaldv φ x ψ y χ

Proof

Step Hyp Ref Expression
1 bj-cbvaldv.1 y φ
2 bj-cbvaldv.2 φ y ψ
3 bj-cbvaldv.3 φ x = y ψ χ
4 nfv x φ
5 nfv x χ
6 5 a1i φ x χ
7 4 1 2 6 3 bj-cbv2v φ x ψ y χ