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