Equivalence of Definitions of Ordering/Proof 2

Theorem
The following definitions of ordering are equivalent:

Definition 1 implies Definition 2
Let $\mathcal R$ be a relation on $S$ satisfying:

Condition $(1)$
Let $\left({x, y}\right) \in \mathcal R \circ \mathcal R$.

Then there exists a $z \in \mathcal R$ such that:


 * $\left({x, z}\right), \left({z, y}\right) \in \mathcal R$

By $\mathcal R$ being transitive:


 * $\left({x, y}\right) \in \mathcal R$

Hence:


 * $\mathcal R \circ \mathcal R \subseteq \mathcal R$

Now let $\left({x, y}\right) \in \mathcal R$.

By $\mathcal R$ being reflexive:


 * $\left({y, y}\right) \in \mathcal R$

Hence by the definition of relation composition:


 * $\left({x, y}\right) \in \mathcal R \circ \mathcal R$

Hence:


 * $\mathcal R \subseteq \mathcal R \circ \mathcal R$

Condition $(2)$
Follows immediately from Relation is Antisymmetric iff Intersection with Inverse is Coreflexive and $\mathcal R$ being reflexive.

Thus $\mathcal R$ is an ordering by definition 2.

Definition 2 implies Definition 1
Let $\mathcal R$ be a relation which fulfils the conditions:
 * $(1): \quad \mathcal R \circ \mathcal R = \mathcal R$
 * $(2): \quad \mathcal R \cap \mathcal R^{-1} = \Delta_S$

Reflexivity
By Intersection is Subset the condition:


 * $\mathcal R \cap \mathcal R^{-1} = \Delta_S$

implies:


 * $\Delta_S \subseteq \mathcal R$

Thus $\mathcal R$ is reflexive by definition

Antisymmetry
By Relation is Antisymmetric iff Intersection with Inverse is Coreflexive the condition:


 * $\mathcal R \cap \mathcal R^{-1} = \Delta_S$

implies that $\mathcal R$ is antisymmetric

Transitivity
Let $\left({x, y}\right), \left({y, z}\right) \in \mathcal R$.

Then by the definition of relation composition:


 * $\left({x, z}\right) \in \mathcal R \circ \mathcal R$

But by the condition:


 * $\mathcal R \circ \mathcal R = \mathcal R$

It follows that:


 * $\left({x, z}\right) \in \mathcal R$

Hence $\mathcal R$ is transitive.

Thus $\mathcal R$ is an ordering by definition 1.