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