Metamath Proof Explorer


Theorem f1oen4g

Description: The domain and range of a one-to-one, onto set function are equinumerous. This variation of f1oeng does not require the Axiom of Replacement nor the Axiom of Power Sets nor the Axiom of Union. (Contributed by BTernaryTau, 7-Dec-2024)

Ref Expression
Assertion f1oen4g ⊢ F ∈ V ∧ A ∈ W ∧ B ∈ X ∧ F : A ⟶ 1-1 onto B → A ≈ B

Proof

Step Hyp Ref Expression
1 f1oeq1 ⊢ f = F → f : A ⟶ 1-1 onto B ↔ F : A ⟶ 1-1 onto B
2 1 spcegv ⊢ F ∈ V → F : A ⟶ 1-1 onto B → ∃ f f : A ⟶ 1-1 onto B
3 2 imp ⊢ F ∈ V ∧ F : A ⟶ 1-1 onto B → ∃ f f : A ⟶ 1-1 onto B
4 3 3ad2antl1 ⊢ F ∈ V ∧ A ∈ W ∧ B ∈ X ∧ F : A ⟶ 1-1 onto B → ∃ f f : A ⟶ 1-1 onto B
5 breng ⊢ A ∈ W ∧ B ∈ X → A ≈ B ↔ ∃ f f : A ⟶ 1-1 onto B
6 5 3adant1 ⊢ F ∈ V ∧ A ∈ W ∧ B ∈ X → A ≈ B ↔ ∃ f f : A ⟶ 1-1 onto B
7 6 adantr ⊢ F ∈ V ∧ A ∈ W ∧ B ∈ X ∧ F : A ⟶ 1-1 onto B → A ≈ B ↔ ∃ f f : A ⟶ 1-1 onto B
8 4 7 mpbird ⊢ F ∈ V ∧ A ∈ W ∧ B ∈ X ∧ F : A ⟶ 1-1 onto B → A ≈ B