Metamath Proof Explorer


Theorem eqabdv

Description: Deduction from a wff to a class abstraction. (Contributed by NM, 9-Jul-1994) Avoid ax-11 . (Revised by Wolf Lammen, 6-May-2023)

Ref Expression
Hypothesis eqabdv.1 ⊢ φ → x ∈ A ↔ ψ
Assertion eqabdv ⊢ φ → A = x | ψ

Proof

Step Hyp Ref Expression
1 eqabdv.1 ⊢ φ → x ∈ A ↔ ψ
2 1 sbbidv ⊢ φ → y x x ∈ A ↔ y x ψ
3 clelsb1 ⊢ y x x ∈ A ↔ y ∈ A
4 3 bicomi ⊢ y ∈ A ↔ y x x ∈ A
5 df-clab ⊢ y ∈ x | ψ ↔ y x ψ
6 2 4 5 3bitr4g ⊢ φ → y ∈ A ↔ y ∈ x | ψ
7 6 eqrdv ⊢ φ → A = x | ψ