Ordered semigroup with an absorbing element According to Wikipedia, a partial order $\le$ on a semigroup $S(\bullet)$ is compatible with the semigroup operation if:

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*$a \le b \implies a \bullet c \le b \bullet c$ and $c \bullet a \le c \bullet b$ for any elements $a, b, c$ of $S$.

(https://en.wikipedia.org/wiki/Ordered_semigroup)
Let's take a simple and generic semigroup $\mathbb Z(\cdot)$ and check if the definition works on it:

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*$0 \le 1$, but $-1 \cdot 0 \le -1 \cdot 1$ is not true.

What is the point of giving such a definition that does not work for the simpliest and most common structures?
But it is even worse. Let's now consider $\mathbb Z(+, \cdot)$. All of a sudden we find that the same operation on the same set becomes compatible with the same order!
In other words, whether or not an order on a semigroup is compatible with the operation depends on the existence of another operation.
I started thinking on where could be the problem, and found that nobody consider the absorbing element as a "structuring" element of a semigroup:
https://en.wikipedia.org/wiki/Absorbing_element.
Meanwhile, an absorbing element, if exists, is unique in any magma and, therefore, semigroup.
And there is a fundamental property of an absorbing element to "remain on the same place" in an ordered semigroup or magma.
If we include it into the semigroup signature $S(\bullet, 0)$, we could give a different definition of an ordered semigroup:
A partial order $\le$ on a semigroup with an absorbing element $S(\bullet, 0)$ is compatible with the semigroup operation if:

*

*Every element of $S$ is comparable with $0$;

*$a \le b \implies a \bullet c \le b \bullet c$ and $c \bullet a \le c \bullet b$ for any elements $a, b$ and any element $c, 0 \le c$;

*$a \le b \implies b \bullet d \le a \bullet d$ and $d \bullet b \le d \bullet a$ for any elements $a, b$ and any element $d, d \le 0$.

Any semigroup without an absorbing element can be embedded into a semigroup with an absorbing element by simply adding $0$ into it.
Thus, we can define that a semigroup $S$ with or without an absorbing element is ordered if it can be embedded into an ordered semigroup with an absorbing element $S_0$ in a way that the order of $S$ is a subset of the order of $S_0$.
Another way to formulate the same idea is:
A semigroup $S$ is ordered if every element $s$ of it falls into one of the two categories:

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*$a \le b \implies s \bullet a \le s \bullet b$ and $a \bullet s \le b \bullet s$ for any elements $a, b$ of $S$;

*$a \le b \implies s \bullet b \le s \bullet a$ and $b \bullet s \le a \bullet s$ for any elements $a, b$ of $S$.

This way we could apply the definition to multiplication on rings without modifications.
Would it be a correct definition?
Can it be applied to all semigroups and magmas without changing the existing models?
Are there other definitions of ordered semigroups and magmas with an absorbing element?
Does it make sense to introduce a separate class of semigroups or magmas with an absorbing element?
 A: Consider an ordered semigroup $(S,\cdot,\le)$, where $S$ happens to have an absorbing element $0$. If $\cdot$ and $\le$ also make it an ordered semigroup with absorbing element, then we demand more than for a mere odered semigroup:

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*We additionally (quite explicitly) demand that $0$ is comparable with every element

*If $a\le b$ and  $d\le 0$, then $a\cdot d\le b\cdot d$ because we have an ordreed semigroup, and $b\cdot d\le a\cdot d$ because of your second postulate. Hence $a\cdot d=b\cdot d$ in that case. Likewise $b\cdot a=d\cdot b$.

While one can write down your definition, it is at least confusingly contradicting what we expect from just "accidentally" having an absorbing element in an ordered semigroup. The main question about definitions however is: Are they helpful? Do there exist interesting theorems e.g. of the form "If $X$ obeys the definition, then $X$ has other interesting properties"?
A: A semigroup with an absorbing element is usually called a semigroup with zero in the literature. A simple example of ordered semigroup with zero is $S = \{a, b, 0\}$ with $a^2 = a$, $b^2 = b$ and $ab = ba = 0$, ordered by $a < 0 < b$.
Now, I don't see any problem with the definition of an ordered semigroup, with or without zero. It is a perfectly sound definition and it works very well in practice.
