Metamath Proof Explorer


Theorem eqabf

Description: Equality of a class variable and a class abstraction. In this version, the fact that x is a nonfree variable in A is explicitly stated as a hypothesis. (Contributed by Thierry Arnoux, 11-May-2017) Avoid ax-13 . (Revised by Wolf Lammen, 13-May-2023)

Ref Expression
Hypothesis eqabf.0 ⊢ Ⅎ 𝑥 𝐴
Assertion eqabf ( 𝐴 = { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝜑 ) )

Proof

Step Hyp Ref Expression
1 eqabf.0 ⊢ Ⅎ 𝑥 𝐴
2 nfab1 ⊢ Ⅎ 𝑥 { 𝑥 ∣ 𝜑 }
3 1 2 cleqf ⊢ ( 𝐴 = { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ { 𝑥 ∣ 𝜑 } ) )
4 abid ⊢ ( 𝑥 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜑 )
5 4 bibi2i ⊢ ( ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ { 𝑥 ∣ 𝜑 } ) ↔ ( 𝑥 ∈ 𝐴 ↔ 𝜑 ) )
6 5 albii ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ { 𝑥 ∣ 𝜑 } ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝜑 ) )
7 3 6 bitri ⊢ ( 𝐴 = { 𝑥 ∣ 𝜑 } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝜑 ) )