Metamath Proof Explorer


Theorem rfovf1od

Description: The value of the operator, ( A O B ) , which maps between relations and functions for relations between base sets, A and B , is a bijection. (Contributed by RP, 27-Apr-2021)

Ref Expression
Hypotheses rfovd.rf ⊢ O = a ∈ V , b ∈ V ⟼ r ∈ 𝒫 a × b ⟼ x ∈ a ⟼ y ∈ b | x r y
rfovd.a ⊢ φ → A ∈ V
rfovd.b ⊢ φ → B ∈ W
rfovcnvf1od.f ⊢ F = A O B
Assertion rfovf1od ⊢ φ → F : 𝒫 A × B ⟶ 1-1 onto 𝒫 B A

Proof

Step Hyp Ref Expression
1 rfovd.rf ⊢ O = a ∈ V , b ∈ V ⟼ r ∈ 𝒫 a × b ⟼ x ∈ a ⟼ y ∈ b | x r y
2 rfovd.a ⊢ φ → A ∈ V
3 rfovd.b ⊢ φ → B ∈ W
4 rfovcnvf1od.f ⊢ F = A O B
5 1 2 3 4 rfovcnvf1od ⊢ φ → F : 𝒫 A × B ⟶ 1-1 onto 𝒫 B A ∧ F -1 = f ∈ 𝒫 B A ⟼ x y | x ∈ A ∧ y ∈ f ⁡ x
6 5 simpld ⊢ φ → F : 𝒫 A × B ⟶ 1-1 onto 𝒫 B A