Theorem elabg

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of Quine p. 44. (Contributed by NM, 14-Apr-1995) Remove dependency on ax-13 . (Revised by Steven Nguyen, 23-Nov-2022)

	Ref	Expression
Hypothesis	elabg.1	|- ( x = A -> ( ph <-> ps ) )
Assertion	elabg	|- ( A e. V -> ( A e. { x | ph } <-> ps ) )

Proof

Step	Hyp	Ref	Expression
1		elabg.1	|- ( x = A -> ( ph <-> ps ) )
2		nfab1	|- F/_ x { x | ph }
3	2	nfel2	|- F/ x A e. { x | ph }
4		nfv	|- F/ x ps
5	3 4	nfbi	|- F/ x ( A e. { x | ph } <-> ps )
6		eleq1	|- ( x = A -> ( x e. { x | ph } <-> A e. { x | ph } ) )
7	6 1	bibi12d	|- ( x = A -> ( ( x e. { x | ph } <-> ph ) <-> ( A e. { x | ph } <-> ps ) ) )
8		abid	|- ( x e. { x | ph } <-> ph )
9	5 7 8	vtoclg1f	|- ( A e. V -> ( A e. { x | ph } <-> ps ) )