Operating on Ordered Group Inequalities

Theorem
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.

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

Let $x, y, z, w \in \struct {G, \circ, \preceq}$.

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 $z \preccurlyeq w$, then $x \circ z \prec y \circ w$.

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

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

Proof
Because:
 * $\struct {G, \circ, \preccurlyeq}$ is a group
 * $\preccurlyeq$ is compatible with $\circ$

it follows from Reflexive Reduction of Relation Compatible with Group Operation is Compatible that $\prec$ is compatible with $\circ$.

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

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

The result follows by Operating on Transitive Relationships Compatible with Operation.

Also see

 * Properties of Ordered Group