Theorem elabf

Description: Membership in a class abstraction, using implicit substitution. (Contributed by NM, 1-Aug-1994) (Revised by Mario Carneiro, 12-Oct-2016)

	Ref	Expression
Hypotheses	elabf.1	|- F/ x ps
	elabf.2	|- A e. _V
	elabf.3	|- ( x = A -> ( ph <-> ps ) )
Assertion	elabf	|- ( A e. { x | ph } <-> ps )

Proof

Step	Hyp	Ref	Expression
1		elabf.1	|- F/ x ps
2		elabf.2	|- A e. _V
3		elabf.3	|- ( x = A -> ( ph <-> ps ) )
4		nfcv	|- F/_ x A
5	4 1 3	elabgf	|- ( A e. _V -> ( A e. { x | ph } <-> ps ) )
6	2 5	ax-mp	|- ( A e. { x | ph } <-> ps )