Metamath Proof Explorer


Theorem ssdf

Description: A sufficient condition for a subclass relationship. (Contributed by Glauco Siliprandi, 3-Jan-2021)

Ref Expression
Hypotheses ssdf.1
|- F/ x ph
ssdf.2
|- ( ( ph /\ x e. A ) -> x e. B )
Assertion ssdf
|- ( ph -> A C_ B )

Proof

Step Hyp Ref Expression
1 ssdf.1
 |-  F/ x ph
2 ssdf.2
 |-  ( ( ph /\ x e. A ) -> x e. B )
3 2 ex
 |-  ( ph -> ( x e. A -> x e. B ) )
4 1 3 ralrimi
 |-  ( ph -> A. x e. A x e. B )
5 dfss3
 |-  ( A C_ B <-> A. x e. A x e. B )
6 4 5 sylibr
 |-  ( ph -> A C_ B )