Metamath Proof Explorer


Theorem bj-cbval2vv

Description: Version of cbval2vv 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-cbval2vv.1 ⊢ x = z ∧ y = w → φ ↔ ψ
Assertion bj-cbval2vv ⊢ ∀ x ∀ y φ ↔ ∀ z ∀ w ψ

Proof

Step Hyp Ref Expression
1 bj-cbval2vv.1 ⊢ x = z ∧ y = w → φ ↔ ψ
2 nfv ⊢ Ⅎ z φ
3 nfv ⊢ Ⅎ w φ
4 nfv ⊢ Ⅎ x ψ
5 nfv ⊢ Ⅎ y ψ
6 2 3 4 5 1 cbval2v ⊢ ∀ x ∀ y φ ↔ ∀ z ∀ w ψ