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 eqss 
 |-  ( A = B <-> ( A C_ B /\ B C_ A ) )
2 1 simplbi 
 |-  ( A = B -> A C_ B )