Metamath Proof Explorer


Theorem ssfid

Description: A subset of a finite set is finite, deduction version of ssfi . (Contributed by Glauco Siliprandi, 21-Nov-2020)

Ref Expression
Hypotheses ssfid.1
|- ( ph -> A e. Fin )
ssfid.2
|- ( ph -> B C_ A )
Assertion ssfid
|- ( ph -> B e. Fin )

Proof

Step Hyp Ref Expression
1 ssfid.1
 |-  ( ph -> A e. Fin )
2 ssfid.2
 |-  ( ph -> B C_ A )
3 ssfi
 |-  ( ( A e. Fin /\ B C_ A ) -> B e. Fin )
4 1 2 3 syl2anc
 |-  ( ph -> B e. Fin )