Theorem eqabcdv

Description: Deduction from a wff to a class abstraction. (Contributed by NM, 9-Jul-1994) (Proof shortened by Wolf Lammen, 16-Nov-2019)

	Ref	Expression
Hypothesis	eqabcdv.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝑥 ∈ 𝐴 ) )
Assertion	eqabcdv	⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } = 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		eqabcdv.1	⊢ ( 𝜑 → ( 𝜓 ↔ 𝑥 ∈ 𝐴 ) )
2	1	bicomd	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝜓 ) )
3	2	eqabdv	⊢ ( 𝜑 → 𝐴 = { 𝑥 ∣ 𝜓 } )
4	3	eqcomd	⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } = 𝐴 )