Metamath Proof Explorer


Theorem bj-cbvexdvav

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

Ref Expression
Hypothesis bj-cbvaldvav.1 ⊢ φ ∧ x = y → ψ ↔ χ
Assertion bj-cbvexdvav ⊢ φ → ∃ x ψ ↔ ∃ y χ

Proof

Step Hyp Ref Expression
1 bj-cbvaldvav.1 ⊢ φ ∧ x = y → ψ ↔ χ
2 nfv ⊢ Ⅎ y φ
3 nfvd ⊢ φ → Ⅎ y ψ
4 1 ex ⊢ φ → x = y → ψ ↔ χ
5 2 3 4 bj-cbvexdv ⊢ φ → ∃ x ψ ↔ ∃ y χ