# Dual of Ordered Semigroup is Ordered Semigroup

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## Theorem

Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.

Then its dual $\struct {S, \circ, \succcurlyeq}$ is also an ordered semigroup.

## Proof

From Dual Ordering is Ordering, we have that $\struct {S, \succcurlyeq}$ is an ordered set.

We also note from the definition that $\struct {S, \circ}$ is a semigroup.

It remains to be demonstrated that $\succcurlyeq$ is compatible with $\circ$.

Recall that $\struct {S, \circ, \preccurlyeq}$ is an ordered semigroup.

Hence *a fortiori* $\preccurlyeq$ is compatible with $\circ$.

Let $x, y \in S$ be arbitrary such that $x \succcurlyeq y$.

We have:

\(\ds x\) | \(\succcurlyeq\) | \(\ds y\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(\preccurlyeq\) | \(\ds x\) | Definition of Dual Ordering | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {y \circ z}\) | \(\preccurlyeq\) | \(\ds \paren {x \circ z}\) | Definition of Relation Compatible with Operation | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ z}\) | \(\succcurlyeq\) | \(\ds \paren {y \circ z}\) |

and similarly:

\(\ds x\) | \(\succcurlyeq\) | \(\ds y\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(\preccurlyeq\) | \(\ds x\) | Definition of Dual Ordering | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {z \circ y}\) | \(\preccurlyeq\) | \(\ds \paren {z \circ x}\) | Definition of Relation Compatible with Operation | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {z \circ x}\) | \(\succcurlyeq\) | \(\ds \paren {z \circ y}\) |

Hence the result.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.3$