Metamath Proof Explorer


Theorem ercl2

Description: Elementhood in the field of an equivalence relation. (Contributed by Mario Carneiro, 12-Aug-2015)

Ref Expression
Hypotheses ersym.1 ⊢ ( 𝜑 → 𝑅 Er 𝑋 )
ersym.2 ⊢ ( 𝜑 → 𝐴 𝑅 𝐵 )
Assertion ercl2 ( 𝜑 → 𝐵 ∈ 𝑋 )

Proof

Step Hyp Ref Expression
1 ersym.1 ⊢ ( 𝜑 → 𝑅 Er 𝑋 )
2 ersym.2 ⊢ ( 𝜑 → 𝐴 𝑅 𝐵 )
3 1 2 ersym ⊢ ( 𝜑 → 𝐵 𝑅 𝐴 )
4 1 3 ercl ⊢ ( 𝜑 → 𝐵 ∈ 𝑋 )