Metamath Proof Explorer


Theorem breng

Description: Equinumerosity relation. This variation of bren does not require the Axiom of Union. (Contributed by NM, 15-Jun-1998) Extract from a subproof of bren . (Revised by BTernaryTau, 23-Sep-2024)

Ref Expression
Assertion breng ⊢ A ∈ V ∧ B ∈ W → A ≈ B ↔ ∃ f f : A ⟶ 1-1 onto B

Proof

Step Hyp Ref Expression
1 f1oeq2 ⊢ x = A → f : x ⟶ 1-1 onto y ↔ f : A ⟶ 1-1 onto y
2 1 exbidv ⊢ x = A → ∃ f f : x ⟶ 1-1 onto y ↔ ∃ f f : A ⟶ 1-1 onto y
3 f1oeq3 ⊢ y = B → f : A ⟶ 1-1 onto y ↔ f : A ⟶ 1-1 onto B
4 3 exbidv ⊢ y = B → ∃ f f : A ⟶ 1-1 onto y ↔ ∃ f f : A ⟶ 1-1 onto B
5 df-en ⊢ ≈ = x y | ∃ f f : x ⟶ 1-1 onto y
6 2 4 5 brabg ⊢ A ∈ V ∧ B ∈ W → A ≈ B ↔ ∃ f f : A ⟶ 1-1 onto B