Theorem abssdv

Description: Deduction of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006) (Proof shortened by SN, 22-Dec-2024)

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

Proof

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