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 )