Operating on Ordered Group Inequalities

Theorem
Let $\left({G, \circ, \le}\right)$ be an ordered group.

Let $<$ be the reflexive reduction of $\le$.

Let $x, y, z, w \in \left({G, \circ, \le}\right)$.

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

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

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

Proof
Because $G$ is a group and $\le$ is compatible with it, so is $<$.

Because $\le$ is transitive and antisymmetric, $<$ is transitive by Reflexive Reduction of Transitive Antisymmetric Relation is Transitive.

Thus the theorem holds by User:Dfeuer/Operating on Transitive Compatible Relationships.