Theorem abss

Description: Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006)

	Ref	Expression
Assertion	abss	|- ( { x | ph } C_ A <-> A. x ( ph -> x e. A ) )

Proof

Step	Hyp	Ref	Expression
1		abid2	|- { x | x e. A } = A
2	1	sseq2i	|- ( { x | ph } C_ { x | x e. A } <-> { x | ph } C_ A )
3		ss2ab	|- ( { x | ph } C_ { x | x e. A } <-> A. x ( ph -> x e. A ) )
4	2 3	bitr3i	|- ( { x | ph } C_ A <-> A. x ( ph -> x e. A ) )