Theorem eqabrd

Description: Equality of a class variable and a class abstraction (deduction form of eqabb ). (Contributed by NM, 16-Nov-1995)

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

Proof

Step	Hyp	Ref	Expression
1		eqabrd.1	⊢ ( 𝜑 → 𝐴 = { 𝑥 ∣ 𝜓 } )
2	1	eleq2d	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ { 𝑥 ∣ 𝜓 } ) )
3		abid	⊢ ( 𝑥 ∈ { 𝑥 ∣ 𝜓 } ↔ 𝜓 )
4	2 3	bitrdi	⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝜓 ) )