Operating on Ordered Group Inequalities

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

Let $\prec$ be the reflexive reduction of $\preceq$.

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

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

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

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

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

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

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

Therefore by Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering, $\prec$ is transitive.

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

Also see

 * Properties of Ordered Group