Metamath Proof Explorer

Theorem sseld

Description: Membership deduction from subclass relationship. (Contributed by NM, 15-Nov-1995)

Ref Expression
Hypothesis sseld.1 
|- ( ph -> A C_ B )
Assertion sseld 
|- ( ph -> ( C e. A -> C e. B ) )

Proof

Step Hyp Ref Expression
1 sseld.1 
 |-  ( ph -> A C_ B )
2 ssel 
 |-  ( A C_ B -> ( C e. A -> C e. B ) )
3 1 2 syl 
 |-  ( ph -> ( C e. A -> C e. B ) )