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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 
|- ( ph -> R Er X )
ersym.2 
|- ( ph -> A R B )
Assertion ercl2 
|- ( ph -> B e. X )

Proof

Step Hyp Ref Expression
1 ersym.1 
 |-  ( ph -> R Er X )
2 ersym.2 
 |-  ( ph -> A R B )
3 1 2 ersym 
 |-  ( ph -> B R A )
4 1 3 ercl 
 |-  ( ph -> B e. X )