Metamath Proof Explorer


Theorem eleq2i

Description: Inference from equality to equivalence of membership. (Contributed by NM, 26-May-1993)

Ref Expression
Hypothesis eleq1i.1
|- A = B
Assertion eleq2i
|- ( C e. A <-> C e. B )

Proof

Step Hyp Ref Expression
1 eleq1i.1
 |-  A = B
2 eleq2
 |-  ( A = B -> ( C e. A <-> C e. B ) )
3 1 2 ax-mp
 |-  ( C e. A <-> C e. B )