Metamath Proof Explorer


Theorem ssdf2

Description: A sufficient condition for a subclass relationship. (Contributed by Glauco Siliprandi, 2-Jan-2022)

Ref Expression
Hypotheses ssdf2.p
|- F/ x ph
ssdf2.a
|- F/_ x A
ssdf2.b
|- F/_ x B
ssdf2.x
|- ( ( ph /\ x e. A ) -> x e. B )
Assertion ssdf2
|- ( ph -> A C_ B )

Proof

Step Hyp Ref Expression
1 ssdf2.p
 |-  F/ x ph
2 ssdf2.a
 |-  F/_ x A
3 ssdf2.b
 |-  F/_ x B
4 ssdf2.x
 |-  ( ( ph /\ x e. A ) -> x e. B )
5 4 ex
 |-  ( ph -> ( x e. A -> x e. B ) )
6 1 2 3 5 ssrd
 |-  ( ph -> A C_ B )