Metamath Proof Explorer


Theorem eleq2

Description: Equality implies equivalence of membership. (Contributed by NM, 26-May-1993) (Proof shortened by Wolf Lammen, 20-Nov-2019)

Ref Expression
Assertion eleq2 ( 𝐴 = 𝐵 → ( 𝐶 ∈ 𝐴 ↔ 𝐶 ∈ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 id ⊢ ( 𝐴 = 𝐵 → 𝐴 = 𝐵 )
2 1 eleq2d ⊢ ( 𝐴 = 𝐵 → ( 𝐶 ∈ 𝐴 ↔ 𝐶 ∈ 𝐵 ) )