Metamath Proof Explorer


Theorem eqrdv

Description: Deduce equality of classes from equivalence of membership. (Contributed by NM, 17-Mar-1996)

Ref Expression
Hypothesis eqrdv.1 ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
Assertion eqrdv ( 𝜑 → 𝐴 = 𝐵 )

Proof

Step Hyp Ref Expression
1 eqrdv.1 ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
2 1 alrimiv ⊢ ( 𝜑 → ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
3 dfcleq ⊢ ( 𝐴 = 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
4 2 3 sylibr ⊢ ( 𝜑 → 𝐴 = 𝐵 )