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 } )