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Metamath Proof Explorer

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) Avoid ax-13 . (Revised by SN, 23-Nov-2022) Avoid ax-10 , ax-11 , ax-12 . (Revised by SN, 5-Oct-2024)

		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	1	ax-gen	\|- A. x ( x = A -> ( ph <-> ps ) )
3		elabgt	\|- ( ( A e. V /\ A. x ( x = A -> ( ph <-> ps ) ) ) -> ( A e. { x \| ph } <-> ps ) )
4	2 3	mpan2	\|- ( A e. V -> ( A e. { x \| ph } <-> ps ) )