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 e. A <-> C e. B ) )

Proof

Step Hyp Ref Expression
1 id 
 |-  ( A = B -> A = B )
2 1 eleq2d 
 |-  ( A = B -> ( C e. A <-> C e. B ) )