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=BAB

Proof

Step Hyp Ref Expression
1 id A=BA=B
2 1 eqimssd A=BAB