# Equivalence of Definitions of Ordering

## Theorem

The following definitions of the concept of Ordering are equivalent:

### Definition 1

An ordering on $S$ is a relation $\RR$ on $S$ such that:

 $(1)$ $:$ $\RR$ is reflexive $\ds \forall a \in S:$ $\ds a \mathrel \RR a$ $(2)$ $:$ $\RR$ is transitive $\ds \forall a, b, c \in S:$ $\ds a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$ $(3)$ $:$ $\RR$ is antisymmetric $\ds \forall a \in S:$ $\ds a \mathrel \RR b \land b \mathrel \RR a \implies a = b$

### Definition 2

An ordering on $S$ is a relation $\RR$ on $S$ such that:

$(1): \quad \RR \circ \RR = \RR$
$(2): \quad \RR \cap \RR^{-1} = \Delta_S$

where:

$\circ$ denotes relation composition
$\RR^{-1}$ denotes the inverse of $\RR$
$\Delta_S$ denotes the diagonal relation on $S$.

## Proof 1

### Definition 1 implies Definition 2

Let $\RR$ be a relation on $S$ satisfying:

 $(1)$ $:$ $\RR$ is reflexive $\ds \forall a \in S:$ $\ds a \mathrel \RR a$ $(2)$ $:$ $\RR$ is transitive $\ds \forall a, b, c \in S:$ $\ds a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$ $(3)$ $:$ $\RR$ is antisymmetric $\ds \forall a \in S:$ $\ds a \mathrel \RR b \land b \mathrel \RR a \implies a = b$
$\RR \circ \RR = \RR$
$\RR \cap \RR^{-1} = \Delta_S$

Thus $\RR$ is an ordering by definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $\RR$ be a relation which fulfils the conditions:

$(1): \quad \RR \circ \RR = \RR$
$(2): \quad \RR \cap \RR^{-1} = \Delta_S$

By definition of set equality, it follows from $(1)$:

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

Thus, by definition, $\RR$ is transitive.

By Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation, it follows from $(2)$ that:

$\RR$ is reflexive
$\RR$ is antisymmetric.

Thus $\RR$ is an ordering by definition 1.

$\blacksquare$

## Proof 2

### Definition 1 implies Definition 2

Let $\RR$ be a relation on $S$ satisfying:

 $(1)$ $:$ $\RR$ is reflexive $\ds \forall a \in S:$ $\ds a \mathrel \RR a$ $(2)$ $:$ $\RR$ is transitive $\ds \forall a, b, c \in S:$ $\ds a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$ $(3)$ $:$ $\RR$ is antisymmetric $\ds \forall a \in S:$ $\ds a \mathrel \RR b \land b \mathrel \RR a \implies a = b$

#### Condition $(1)$

Let $\tuple {x, y} \in \RR \circ \RR$.

Then there exists a $z \in \RR$ such that:

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

By $\RR$ being transitive:

$\tuple {x, y} \in \RR$

Hence:

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

Now let $\tuple {x, y} \in \RR$.

By $\RR$ being reflexive:

$\tuple {y, y} \in \RR$

Hence by the definition of relation composition:

$\tuple {x, y} \in \RR \circ \RR$

Hence:

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

#### Condition $(2)$

Follows immediately from Relation is Antisymmetric iff Intersection with Inverse is Coreflexive and $\RR$ being reflexive.

Thus $\RR$ is an ordering by definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $\RR$ be a relation which fulfils the conditions:

$(1): \quad \RR \circ \RR = \RR$
$(2): \quad \RR \cap \RR^{-1} = \Delta_S$

#### Reflexivity

By Intersection is Subset the condition:

$\RR \cap \RR^{-1} = \Delta_S$

implies:

$\Delta_S \subseteq \RR$

Thus $\RR$ is reflexive by definition.

#### Antisymmetry

$\RR \cap \RR^{-1} = \Delta_S$

implies that $\RR$ is antisymmetric.

#### Transitivity

Let $\tuple {x, y}, \tuple {y, z} \in \RR$.

Then by the definition of relation composition:

$\tuple {x, z} \in \RR \circ \RR$

But by the condition:

$\RR \circ \RR = \RR$

It follows that:

$\tuple {x, z} \in \RR$

Hence $\RR$ is transitive.

Thus $\RR$ is an ordering by definition 1.

$\blacksquare$