Metamath Proof Explorer


Theorem ss2ab

Description: Class abstractions in a subclass relationship. (Contributed by NM, 3-Jul-1994)

Ref Expression
Assertion ss2ab
|- ( { x | ph } C_ { x | ps } <-> A. x ( ph -> ps ) )

Proof

Step Hyp Ref Expression
1 nfab1
 |-  F/_ x { x | ph }
2 nfab1
 |-  F/_ x { x | ps }
3 1 2 dfss2f
 |-  ( { x | ph } C_ { x | ps } <-> A. x ( x e. { x | ph } -> x e. { x | ps } ) )
4 abid
 |-  ( x e. { x | ph } <-> ph )
5 abid
 |-  ( x e. { x | ps } <-> ps )
6 4 5 imbi12i
 |-  ( ( x e. { x | ph } -> x e. { x | ps } ) <-> ( ph -> ps ) )
7 6 albii
 |-  ( A. x ( x e. { x | ph } -> x e. { x | ps } ) <-> A. x ( ph -> ps ) )
8 3 7 bitri
 |-  ( { x | ph } C_ { x | ps } <-> A. x ( ph -> ps ) )