Metamath Proof Explorer


Theorem alexbii

Description: Biconditional form of aleximi . (Contributed by BJ, 16-Nov-2020)

Ref Expression
Hypothesis alexbii.1 ⊢ φ → ψ ↔ χ
Assertion alexbii ⊢ ∀ x φ → ∃ x ψ ↔ ∃ x χ

Proof

Step Hyp Ref Expression
1 alexbii.1 ⊢ φ → ψ ↔ χ
2 1 biimpd ⊢ φ → ψ → χ
3 2 aleximi ⊢ ∀ x φ → ∃ x ψ → ∃ x χ
4 1 biimprd ⊢ φ → χ → ψ
5 4 aleximi ⊢ ∀ x φ → ∃ x χ → ∃ x ψ
6 3 5 impbid ⊢ ∀ x φ → ∃ x ψ ↔ ∃ x χ