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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