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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 ∣ 𝜑 } = 𝐴
	Assertion	eqabcri	⊢ ( 𝜑 ↔ 𝑥 ∈ 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		eqabcri.1	⊢ { 𝑥 ∣ 𝜑 } = 𝐴
2	1	eqcomi	⊢ 𝐴 = { 𝑥 ∣ 𝜑 }
3	2	eqabri	⊢ ( 𝑥 ∈ 𝐴 ↔ 𝜑 )
4	3	bicomi	⊢ ( 𝜑 ↔ 𝑥 ∈ 𝐴 )