Theorem

Let A and B be objects. Let φ(a,b) be a proposition depending on a member a of A and a member b of B. Then we have {⟨a,b⟩∈A×B|φ(a,b)} is a binary relation on A and B.

Background

The following background is necessary to understand this theorem.

(.) Let A and x be objects. Then x∈A is a proposition.

(.) Let A, B and R be objects. Then "R is a binary relation on A and B" is a proposition.

(.) Let A and B be objects. Let φ(a,b) be a proposition depending on a member a of A and a member b of B. Then "{⟨a,b⟩∈A×B|φ(a,b)}" is an object.

Background for Proof

The following background is necessary for the proof.

(.) Let A and B be objects. Then "A⊆B" is a proposition.

(.) Let A and B be objects. Then "A×B" is an object.

(.) Let A, B and R be objects. Then R is a binary relation on A and B is the proposition given by R⊆A×B.

(.) Let A and B be objects. Let φ(a,b) be a proposition depending on a member a of A and a member b of B. Then we know {⟨a,b⟩∈A×B|φ(a,b)}⊆A×B.

Proof

Clearly, {⟨a,b⟩∈A×B|φ(a,b)}⊆A×B. Using this, we conclude {⟨a,b⟩∈A×B|φ(a,b)} is a binary relation on A and B.

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