Reflexive Reduction of Ordering is Strict Ordering/Proof 1

Theorem
Let $\mathcal R$ be an ordering on a set $S$.

Let $\mathcal R^\ne$ be the reflexive reduction of $\mathcal R$.

Then $\mathcal R^\ne$ is a strict ordering on $S$.

Antireflexivity
Follows from Reflexive Reduction is Antireflexive.

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

By antireflexivity $x \ne y$ and $y \ne z$.

We consider the two remaining cases.

Case 1: $x = z$
If $x = z$ then:


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

and so:


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

Then by the antisymmetry of $\mathcal R$:


 * $x = y$

And:


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

Which contradicts that $\mathcal R^\ne$ is antireflexive.

Case 2: $x \ne z$
By the transitivity of $\mathcal R$:


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

and by $x$ and $z$ being distinct:


 * $\left({x, z}\right) \notin \Delta_S$

It follows by the definition of reflexive reduction:


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

Hence $\mathcal R^\ne$ is transitive.