Metamath Proof Explorer

Theorem ssid

Description: Any class is a subclass of itself. Exercise 10 of TakeutiZaring p. 18. (Contributed by NM, 21-Jun-1993) (Proof shortened by Andrew Salmon, 14-Jun-2011)

Ref Expression
Assertion ssid
|- A C_ A


Step Hyp Ref Expression
1 id
 |-  ( x e. A -> x e. A )
2 1 ssriv
 |-  A C_ A