Metamath Proof Explorer


Theorem ssexg

Description: A subclass of a set is a set. Exercise 3 of TakeutiZaring p. 22 (generalized). (Contributed by NM, 14-Aug-1994) (Proof shortened by BJ, 18-Jul-2026)

Ref Expression
Assertion ssexg
|- ( ( A C_ B /\ B e. C ) -> A e. _V )

Proof

Step Hyp Ref Expression
1 dfss2
 |-  ( A C_ B <-> ( A i^i B ) = A )
2 inex2g
 |-  ( B e. C -> ( A i^i B ) e. _V )
3 eleq1
 |-  ( ( A i^i B ) = A -> ( ( A i^i B ) e. _V <-> A e. _V ) )
4 3 biimpa
 |-  ( ( ( A i^i B ) = A /\ ( A i^i B ) e. _V ) -> A e. _V )
5 2 4 sylan2
 |-  ( ( ( A i^i B ) = A /\ B e. C ) -> A e. _V )
6 1 5 sylanb
 |-  ( ( A C_ B /\ B e. C ) -> A e. _V )