# Equivalence of Definitions of Transitive Relation

## Theorem

The following definitions of the concept of Transitive Relation are equivalent:

### Definition 1

$\RR$ is transitive if and only if:

$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$

that is:

$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \in \RR$

### Definition 2

$\RR$ is transitive if and only if:

$\RR \circ \RR \subseteq \RR$

where $\circ$ denotes composite relation.

## Proof

### Definition 1 implies Definition 2

Let $\RR$ be a relation which fulfills the condition:

$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$

Then:

 $\ds \tuple {x, z}$ $\in$ $\ds \RR \circ \RR$ $\ds \leadsto \ \$ $\ds \exists y \in \RR: \,$ $\ds \tuple {x, y}$ $\in$ $\ds \RR$ Definition of Composition of Relations $\, \ds \land \,$ $\ds \tuple {y, z}$ $\in$ $\ds \RR$ $\ds \leadsto \ \$ $\ds \tuple {x, z}$ $\in$ $\ds \RR$ Definition 1 of Transitive Relation $\ds \leadsto \ \$ $\ds \RR \circ \RR$ $\subseteq$ $\ds \RR$ Definition of Subset

Thus $\RR$ is transitive by definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $\RR$ be a relation that fulfills the condition:

$\RR \circ \RR \subseteq \RR$

Aiming for a contradiction, suppose $\RR$ does not fulfill the condition:

$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$

Then:

 $\ds \exists \paren {\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR}: \,$ $\ds \tuple {x, z}$ $\notin$ $\ds \RR$ Definition of Non-Transitive Relation $\ds \leadsto \ \$ $\ds \exists \tuple {x, z} \in \RR \circ \RR: \,$ $\ds \tuple {x, z}$ $\notin$ $\ds \RR$ Definition of Composition of Relations $\ds \leadsto \ \$ $\ds \RR \circ \RR$ $\nsubseteq$ $\ds \RR$ Definition of Subset

This contradicts our statement that $\RR \circ \RR \subseteq \RR$.

Hence by Proof by Contradiction $\RR$ does fulfills the condition:

$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$

Thus $\RR$ is transitive by definition 1.

$\blacksquare$