Metamath Proof Explorer


Theorem eqab2

Description: Implication of a class abstraction. (Contributed by Peter Mazsa, 16-Apr-2019)

Ref Expression
Assertion eqab2 ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝜑 ) → ∀ 𝑥 ∈ 𝐴 𝜑 )

Proof

Step Hyp Ref Expression
1 biimp ⊢ ( ( 𝑥 ∈ 𝐴 ↔ 𝜑 ) → ( 𝑥 ∈ 𝐴 → 𝜑 ) )
2 1 alimi ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝜑 ) → ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) )
3 2 ralrid ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝜑 ) → ∀ 𝑥 ∈ 𝐴 𝜑 )