Theorem abssf

Description: Class abstraction in a subclass relationship. (Contributed by Glauco Siliprandi, 26-Jun-2021)

	Ref	Expression
Hypothesis	abssf.1	|- F/_ x A
Assertion	abssf	|- ( { x | ph } C_ A <-> A. x ( ph -> x e. A ) )

Proof

Step	Hyp	Ref	Expression
1		abssf.1	|- F/_ x A
2	1	abid2f	|- { x | x e. A } = A
3	2	sseq2i	|- ( { x | ph } C_ { x | x e. A } <-> { x | ph } C_ A )
4		ss2ab	|- ( { x | ph } C_ { x | x e. A } <-> A. x ( ph -> x e. A ) )
5	3 4	bitr3i	|- ( { x | ph } C_ A <-> A. x ( ph -> x e. A ) )