Operating on Ordered Group Inequalities

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

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

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

Then the following implications hold: If $x < y$ and $z < w$, then $x \circ z < y \circ w$.

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

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

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

Proof
Because $\left({G, \circ, \le}\right)$ is a group and $\le$ is compatible with $\circ$, $<$ is compatible with $\circ$ by Reflexive Reduction of Relation Compatible with Group Operation is Compatible.

By the definition of an ordering, $\le$ is transitive and antisymmetric.

Therefore by Reflexive Reduction of Transitive Antisymmetric Relation is Transitive, $<$ is transitive.

Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation.