Reflexive Reduction of Ordering is Strict Ordering/Proof 1

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

Then $\mathcal R^\neq$, the reflexive reduction of $\mathcal R$, is a strict ordering on $S$.

Antireflexivity
By the definition of reflexive reduction:


 * $\mathcal R^\neq = \mathcal R \setminus \Delta_S$

where $\Delta_S$ denotes the diagonal relation on $S$.

By Set Difference Intersection Second Set is Empty Set:


 * $\left({\mathcal R \setminus \Delta_S}\right) \cap \Delta_S = \varnothing$

Thus $\mathcal R \setminus \Delta_S$ and $\Delta_S$ are disjoint.

Hence by Antireflexive Disjoint from Diagonal Relation, $\mathcal R^\neq$ is antireflexive.

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

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

We consider the two remaining cases.

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


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

and so:


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

Then by the antisymmetry of $\mathcal R$:


 * $x = y$

And:


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

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

Case 2: $x \neq z$
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:


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

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