Metamath Proof Explorer


Theorem equsalhw

Description: Version of equsalh with a disjoint variable condition, which does not require ax-13 . (Contributed by NM, 29-Nov-2015) (Proof shortened by Wolf Lammen, 8-Jul-2022)

Ref Expression
Hypotheses equsalhw.1 ⊢ ψ → ∀ x ψ
equsalhw.2 ⊢ x = y → φ ↔ ψ
Assertion equsalhw ⊢ ∀ x x = y → φ ↔ ψ

Proof

Step Hyp Ref Expression
1 equsalhw.1 ⊢ ψ → ∀ x ψ
2 equsalhw.2 ⊢ x = y → φ ↔ ψ
3 1 nf5i ⊢ Ⅎ x ψ
4 3 2 equsalv ⊢ ∀ x x = y → φ ↔ ψ