Metamath Proof Explorer

Theorem eqimss

Description: Equality implies inclusion. (Contributed by NM, 21-Jun-1993) (Proof shortened by Andrew Salmon, 21-Jun-2011)

Ref Expression
Assertion eqimss 
|- ( A = B -> A C_ B )

Proof

Step Hyp Ref Expression
1 id 
 |-  ( A = B -> A = B )
2 1 eqimssd 
 |-  ( A = B -> A C_ B )