Metamath Proof Explorer


Theorem brdom2g

Description: Dominance relation. This variation of brdomg does not require the Axiom of Union. (Contributed by NM, 15-Jun-1998) Extract from a subproof of brdomg . (Revised by BTernaryTau, 29-Nov-2024)

Ref Expression
Assertion brdom2g ⊢ A ∈ V ∧ B ∈ W → A ≼ B ↔ ∃ f f : A ⟶ 1-1 B

Proof

Step Hyp Ref Expression
1 f1eq2 ⊢ x = A → f : x ⟶ 1-1 y ↔ f : A ⟶ 1-1 y
2 1 exbidv ⊢ x = A → ∃ f f : x ⟶ 1-1 y ↔ ∃ f f : A ⟶ 1-1 y
3 f1eq3 ⊢ y = B → f : A ⟶ 1-1 y ↔ f : A ⟶ 1-1 B
4 3 exbidv ⊢ y = B → ∃ f f : A ⟶ 1-1 y ↔ ∃ f f : A ⟶ 1-1 B
5 df-dom ⊢ ≼ = x y | ∃ f f : x ⟶ 1-1 y
6 2 4 5 brabg ⊢ A ∈ V ∧ B ∈ W → A ≼ B ↔ ∃ f f : A ⟶ 1-1 B