Metamath Proof Explorer


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