Operating on Transitive Relationships Compatible with Operation

Theorem
Let $\left({S, \circ}\right)$ be a magma.

Let $\mathcal R$ be a transitive relation compatible with $\circ$.

Let $\mathcal R^=$ be the reflexive closure of $\mathcal R$

Let $x, y, z, w \in S$.

Then the following implications hold:

$(1)$ If $x \mathrel{\mathcal R^=} y$ and $z \mathrel{\mathcal R^=} w$, then $x \circ y \mathrel{\mathcal R^=} z \circ w$.

$(2)$ If $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R^=} w$, then $x \circ y \mathrel{\mathcal R} z \circ w$.

$(3)$ If $x \mathrel{\mathcal R^=} y$ and $z \mathrel{\mathcal R} w$, then $x \circ y \mathrel{\mathcal R} z \circ w$.

Proof
By the definition of compatibility,


 * $x \mathrel{\mathcal R^=} y \implies x \circ z \mathrel{\mathcal R^=} y \circ z$
 * $z \mathrel{\mathcal R^=} w \implies y \circ z \mathrel{\mathcal R^=} y \circ w$

By transitivity,


 * $(x \mathrel{\mathcal R^=} y) \land (z \mathrel{\mathcal R^=} w)\implies x \circ z \mathrel{\mathcal R^=} y \circ w$

Again using the definition of compatibility,


 * $x \mathrel{\mathcal R} y \implies x \circ z \mathrel{\mathcal R} y \circ z$
 * $z \mathrel{\mathcal R^=} w \implies y \circ z \mathrel{\mathcal R^=} y \circ w$

By transitivity,
 * $(x \mathrel{\mathcal R} y) \land (z \mathrel{\mathcal R^=} w)\implies x \circ z \mathrel{\mathcal R} y \circ w$.

A similar argument shows that
 * $(x \mathrel{\mathcal R^=} y) \land (z \mathrel{\mathcal R} w)\implies x \circ z \mathrel{\mathcal R} y \circ w$.