Metamath Proof Explorer

Theorem eleq1

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

Ref Expression
Assertion eleq1 
|- ( A = B -> ( A e. C <-> B e. C ) )

Proof

Step Hyp Ref Expression
1 id 
 |-  ( A = B -> A = B )
2 1 eleq1d 
 |-  ( A = B -> ( A e. C <-> B e. C ) )