Metamath Proof Explorer


Theorem ssabdv

Description: Deduction of abstraction subclass from implication. (Contributed by SN, 22-Dec-2024)

Ref Expression
Hypothesis ssabdv.1
|- ( ph -> ( x e. A -> ps ) )
Assertion ssabdv
|- ( ph -> A C_ { x | ps } )

Proof

Step Hyp Ref Expression
1 ssabdv.1
 |-  ( ph -> ( x e. A -> ps ) )
2 abid1
 |-  A = { x | x e. A }
3 1 ss2abdv
 |-  ( ph -> { x | x e. A } C_ { x | ps } )
4 2 3 eqsstrid
 |-  ( ph -> A C_ { x | ps } )