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 ⊢ Ⅎ y φ
bj-cbvaldv.2 ⊢ φ → Ⅎ 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 ⊢ Ⅎ y φ
2 bj-cbvaldv.2 ⊢ φ → Ⅎ y ψ
3 bj-cbvaldv.3 ⊢ φ → x = y → ψ ↔ χ
4 nfv ⊢ Ⅎ x φ
5 nfv ⊢ Ⅎ x χ
6 5 a1i ⊢ φ → Ⅎ x χ
7 4 1 2 6 3 bj-cbv2v ⊢ φ → ∀ x ψ ↔ ∀ y χ