Operating on Transitive Relationships Compatible with Operation

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

Let $\prec$ be a transitive relation compatible with $\circ$.

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

If $x \preceq y$ and $z \preceq w$, then $x \circ y \preceq z \circ w$.

If $x \prec y$ and $z \preceq w$, then $x \circ y \prec z \circ w$.

If $x \preceq y$ and $z \prec w$, then $x \circ y \prec z \circ w$.

Proof
By the definition of compatibility,


 * $x \preceq y \implies x \circ z \preceq y \circ z$
 * $z \preceq w \implies y \circ z \preceq y \circ w$

By transitivity,


 * $(x \preceq y) \land (z \preceq w)\implies x \circ z \preceq y \circ w$

Again using the definition of compatibility,


 * $x \prec y \implies x \circ z \prec y \circ z$
 * $z \preceq w \implies y \circ z \preceq y \circ w$

By transitivity,
 * $(x \prec y) \land (z \preceq w)\implies x \circ z \prec y \circ w$.

A similar argument shows that
 * $(x \preceq y) \land (z \prec w)\implies x \circ z \prec y \circ w$.