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Metamath Proof Explorer

Theorem ss2abim

Description: Class abstractions in a subclass relationship. Reverse direction of ss2ab which requires fewer axioms. (Contributed by SN, 2-Feb-2026)

Ref Expression
Assertion ss2abim 
|- ( A. x ( ph -> ps ) -> { x | ph } C_ { x | ps } )

Proof

Step Hyp Ref Expression
1 spsbim 
 |-  ( A. x ( ph -> ps ) -> ( [ y / x ] ph -> [ y / x ] ps ) )
2 1 alrimiv 
 |-  ( A. x ( ph -> ps ) -> A. y ( [ y / x ] ph -> [ y / x ] ps ) )
3 df-ss 
 |-  ( { x | ph } C_ { x | ps } <-> A. y ( y e. { x | ph } -> y e. { x | ps } ) )
4 df-clab 
 |-  ( y e. { x | ph } <-> [ y / x ] ph )
5 df-clab 
 |-  ( y e. { x | ps } <-> [ y / x ] ps )
6 4 5 imbi12i 
 |-  ( ( y e. { x | ph } -> y e. { x | ps } ) <-> ( [ y / x ] ph -> [ y / x ] ps ) )
7 6 albii 
 |-  ( A. y ( y e. { x | ph } -> y e. { x | ps } ) <-> A. y ( [ y / x ] ph -> [ y / x ] ps ) )
8 3 7 bitr2i 
 |-  ( A. y ( [ y / x ] ph -> [ y / x ] ps ) <-> { x | ph } C_ { x | ps } )
9 2 8 sylib 
 |-  ( A. x ( ph -> ps ) -> { x | ph } C_ { x | ps } )