Reflexive Reduction of Ordering is Strict Ordering/Proof 1

Antireflexivity
Follows from Reflexive Reduction is Antireflexive.

Transitivity
Suppose $\tuple {x, y}, \tuple {y, z} \in \RR^\ne$.

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

We consider the two remaining cases.

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


 * $\tuple {x, y}, \tuple {y, x} \in \RR^\ne$

and so:


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

Then by the antisymmetry of $\RR$:
 * $x = y$

and:
 * $\tuple {x, x} \in \RR^\ne$

which contradicts that $\RR^\ne$ is antireflexive.

Case 2: $x \ne z$
By the transitivity of $\RR$:
 * $\tuple {x, z} \in \RR$

and by $x$ and $z$ being distinct:
 * $\tuple {x, z} \notin \Delta_S$

It follows by the definition of reflexive reduction:
 * $\tuple {x, z} \in \RR^\ne$

Hence $\RR^\ne$ is transitive.