# Subsemigroup of Ordered Semigroup is Ordered

## Theorem

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

Let $\struct {T, \circ_T}$ be a subsemigroup of $\struct {S, \circ}$.

Then the ordered structure $\struct {T, \circ_T, \preceq_T}$ is also an ordered semigroup.

In the above:

$\circ_T$ denotes the operation induced on $T$ by $\circ$
$\preceq_T$ denotes the restriction of $\preceq$ to $T \times T$.

## Proof

It is necessary to ascertain that $\struct {T, \circ {\restriction_T} }$ fulfils the ordered semigroup axioms:

An ordered semigroup is an algebraic system $\struct {S, \circ, \preceq}$ which satisfies the following properties:

 $(\text {OS} 0)$ $:$ Closure $\ds \forall a, b \in S:$ $\ds a \circ b \in S$ $(\text {OS} 1)$ $:$ Associativity $\ds \forall a, b, c \in S:$ $\ds a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ $(\text {OS} 2)$ $:$ Compatibility of $\preceq$ with $\circ$ $\ds \forall a, b, c \in S:$ $\ds a \preceq b \implies \paren {a \circ c} \preceq \paren {b \circ c}$ where $\preceq$ is an ordering $\ds a \preceq b \implies \paren {c \circ a} \preceq \paren {c \circ b}$

In this context, we see that $\text {OS} 0$ and $\text {OS} 1$ are fulfilled a fortiori by dint of $\struct {T, \circ {\restriction_T} }$ being a subsemigroup of $\struct {S, \circ}$.

We have that $\struct {S, \circ, \preceq}$ is an ordered semigroup.

From Restriction of Ordering is Ordering, we have that $\preceq_T$ is an ordering.

Hence:

 $\ds \forall a, b \in S: \,$ $\ds a \preceq b$ $\implies$ $\ds \paren {a \circ c} \preceq \paren {b \circ c}$ Ordered Semigroup Axiom $\text {OS} 2$ on $\struct {S, \circ, \preceq}$ $\ds \leadsto \ \$ $\ds \forall a, b \in T: \,$ $\ds a \preceq_T b$ $\implies$ $\ds \paren {a \circ_T c} \preceq_T \paren {b \circ_T c}$

and:

 $\ds \forall a, b \in S: \,$ $\ds a \preceq b$ $\implies$ $\ds \paren {c \circ a} \preceq \paren {c \circ b}$ Ordered Semigroup Axiom $\text {OS} 2$ on $\struct {S, \circ, \preceq}$ $\ds \leadsto \ \$ $\ds \forall a, b \in T: \,$ $\ds a \preceq_T b$ $\implies$ $\ds \paren {c \circ_T a} \preceq_T \paren {c \circ_T b}$

Hence $\preceq_T$ fulfils Ordered Semigroup Axiom $\text {OS} 2$ on $\struct {T, \circ_T, \preceq_T}$.

$\blacksquare$