Reflexive Reduction of Relation Compatible with Cancellable Operation is Compatible

Theorem
Let $\struct {S, \circ}$ be an algebraic structure such that $\circ$ is a cancellable operation.

Let $\RR$ be a relation on $S$ which is compatible with $\circ$.

Let $\RR^\ne$ be the reflexive reduction of $\RR$.

Then $\RR^\ne$ is compatible with $\circ$.

Proof
$\RR^\ne$ is not compatible with $\circ$.

Let $x, y \in S$ such that:
 * $x \mathrel \RR y$

but:
 * $x \mathrel {\RR^\ne} y$

Then by definition of reflexive reduction:
 * $x \ne y$

Then as $\RR^\ne$ is not compatible with $\circ$:
 * $\exists z \in S: \lnot \paren {z \circ x \mathrel {\RR^\ne} z \circ y}$

But as $\RR$ is compatible with $\circ$:
 * $z \circ x \mathrel \RR z \circ y$

That means:
 * $z \circ x = z \circ y$

As $\circ$ is cancellable this leads to:
 * $x = y$

which is a contradiction.

Hence $\RR^\ne$ must be compatible with $\circ$.