Metamath Proof Explorer


Theorem sseldd

Description: Membership inference from subclass relationship. (Contributed by NM, 14-Dec-2004)

Ref Expression
Hypotheses sseld.1
|- ( ph -> A C_ B )
sseldd.2
|- ( ph -> C e. A )
Assertion sseldd
|- ( ph -> C e. B )

Proof

Step Hyp Ref Expression
1 sseld.1
 |-  ( ph -> A C_ B )
2 sseldd.2
 |-  ( ph -> C e. A )
3 1 sseld
 |-  ( ph -> ( C e. A -> C e. B ) )
4 2 3 mpd
 |-  ( ph -> C e. B )