Metamath Proof Explorer


Theorem dfss5

Description: Alternate definition of subclass relationship: a class A is a subclass of another class B iff each element of A is equal to an element of B . (Contributed by AV, 13-Nov-2020)

Ref Expression
Assertion dfss5
|- ( A C_ B <-> A. x e. A E. y e. B x = y )

Proof

Step Hyp Ref Expression
1 dfss3
 |-  ( A C_ B <-> A. x e. A x e. B )
2 clel5
 |-  ( x e. B <-> E. y e. B x = y )
3 2 ralbii
 |-  ( A. x e. A x e. B <-> A. x e. A E. y e. B x = y )
4 1 3 bitri
 |-  ( A C_ B <-> A. x e. A E. y e. B x = y )