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 B

Proof

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