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Metamath Proof Explorer

Theorem eqabcri

Description: Equality of a class variable and a class abstraction (inference form). (Contributed by NM, 31-Jul-1994) (Proof shortened by Wolf Lammen, 15-Nov-2019)

		Ref	Expression
	Hypothesis	eqabcri.1	\|- { x \| ph } = A
	Assertion	eqabcri	\|- ( ph <-> x e. A )

Proof

Step	Hyp	Ref	Expression
1		eqabcri.1	\|- { x \| ph } = A
2	1	eqcomi	\|- A = { x \| ph }
3	2	eqabri	\|- ( x e. A <-> ph )
4	3	bicomi	\|- ( ph <-> x e. A )