Metamath Proof Explorer


Theorem ss2ab1

Description: Class abstractions in a subclass relationship, closed form. One direction of ss2ab using fewer axioms. (Contributed by SN, 22-Dec-2024)

Ref Expression
Assertion ss2ab1
|- ( A. x ( ph -> ps ) -> { x | ph } C_ { x | ps } )

Proof

Step Hyp Ref Expression
1 spsbim
 |-  ( A. x ( ph -> ps ) -> ( [ t / x ] ph -> [ t / x ] ps ) )
2 df-clab
 |-  ( t e. { x | ph } <-> [ t / x ] ph )
3 df-clab
 |-  ( t e. { x | ps } <-> [ t / x ] ps )
4 1 2 3 3imtr4g
 |-  ( A. x ( ph -> ps ) -> ( t e. { x | ph } -> t e. { x | ps } ) )
5 4 ssrdv
 |-  ( A. x ( ph -> ps ) -> { x | ph } C_ { x | ps } )