text:
\section{Introduction}

A (noncommutative) \emph{semiring} $S$ is a set together with binary operations
\[+,*:S\times S\rightarrow S,\]respectively called \emph{addition} and \emph{multiplication}, and elements $0_S,1_S\in S$ such that $(S,+,0_S)$ is a commutative monoid, $(S,*,1_S)$ is a noncommutative monoid and such that the multiplication distributes over the addition \cite{golan2013semirings}. A (noncommutative) \emph{semialgebra} $S$ is a (noncommutative) semiring together with a (nonnegative) scalar multiplication
\[\mathbb{R}_+\times S\rightarrow S,(r,x)\mapsto r\cdot x\] that is a commutative monoid homomorphism from $(\mathbb{R}_+,+)$ or $(S,+)$ to $(S,+)$ in the first or the second argument respectively, and satisfies the following addition laws: \emph{i})\; $1\cdot x=x$,\, \emph{ii})\; $(r\cdot x)*(s\cdot y)=(rs)\cdot(x*y)$ \cite[Defintion 3.5]{algeP2}.

A semialgebra $S$ is \emph{zero-sum-free} if for any $a,b\in S$, $a+b=0_S$ implies $a=b=0_S$. It is \emph{zero-divisor-free} if for any $a,b\in S$, $a*b=0_S$ implies $a=0_S$ or $b=0_S$. In a zero-divisor-free semialgebra $S$, $r\cdot a=0_S$ implies $r=0$ or $a=0_S$ for any $r\in\mathbb{R}_+$ and $a\in S$: $r\cdot a=0_S$ $\Rightarrow$ $(r\cdot1_S)* a=0_S$ $\Rightarrow$ $r\cdot1_S=0_S$ or $a=0_S$. If $a\neq0_S$ and $r\neq0$, then $1_S=r^{-1}\cdot(r\cdot1_S)=r^{-1}\cdot0_S=0_S$. Hence $a=a*1_S=a*0_S=0_S$, which is a contradiction.

A \emph{preorder relation} (or, simply \emph{preorder}) $\leq$ on a set $X$ is a binary relation that is both reflexive and transitive (\cite{golan2013semirings}, pp. 119). Set $x,y\in X$, we sometimes write ``$x\geq y$" instead of ``$y\leq x$", and write ``$x<y$" or ``$y>x$" instead of the condition ``$x\leq y\text{ and }x\neq y$" throughout the paper. A preorder relation $\leq$ on $X$ is said to be \emph{trivial} if $x\leq y$ holds for any $x,y\in X$.

A \emph{preordered semiring} \cite[Definition 2.2]{fritz2021generalization} is a semiring with a preorder relation $\leq$ such that for all $a,x,y$ in the semiring,
\begin{equation}\label{preoc}
x\leq y\text{ implies }\left\{
\begin{array}{rcl}
a+x&\leq&a+y,\\
a*x&\leq&a*y,\\
x*a&\leq&y*a.
\end{array}
\right.
\end{equation}
Similarly, a \emph{preordered semialgebra} is a semialgebra with a preorder relation $\leq$ such that $x\leq y$ implies the inequalities in (\ref{preoc}). Note that it also implies the inequality
\begin{equation}\label{scalar}
r\cdot x\leq r\cdot y
\end{equation}
for any $r\in\mathbb{R}_+$, since $r\cdot x=(r\cdot1_S)*x\leq(r\cdot1_S)*y=r\cdot y$.

Let $E$ and $K$ be two semialgebras. A map $f$ from $E$ to $K$ is a \emph{semialgebra homomorphism} (or, simply \emph{homomorphism}) if $f(0_E)=0_K,f(1_E)=1_K$ and for any $x,y\in E$ and any $r\in\mathbb{R}_+$, the following equations hold
\[f(x+y)=f(x)+f(y),\;f(x*y)=f(x)*f(y)\text{ and }f(r\cdot x)=r\cdot f(x).\]If $\leq_{_E}$ and $\leq_{_K}$ are two preorder relations on  $E$ and $K$ respectively, then a map $f$ from $E$ to $K$ is \emph{monotone} (\emph{w.r.t.} $\leq_{_E}$ and $\leq_{_K}$) if for any $x,y\in E$, $x\leq_{_E} y$ implies $f(x)\leq_{_K} f(y)$.

The following definition of a \emph{power universal} element in a noncommutative preordered semialgebra is a natural generalization of   \cite[Definition 2.8]{fritz2021generalization}.
\begin{definition}\label{poweruni}
Let $S$ be a noncommutative preordered semialgebra with $1_S\geq0_S$. An element $u\in S$ with $u\geq1_S$ is \emph{power universal} $($\emph{w.r.t.} the preorder $\leq)$ if for every nonzero $x\in S$ there is a number  $k\in\mathbb{N}$ such that
\begin{equation}\label{PU}
u^k*x\geq1_S,\;\;x*u^k\geq1_S\;\text{ and }\;u^k\geq x.
\end{equation}
\end{definition}

 Recently, Strassen's  separation theorem \cite{strassen1988asymptotic,zuiddam2019asymptotic} for preordered semirings (called ``Strassen's Vergleichsstellensatz" \cite{algeP1}) has been generalized tremendously by Fritz and Vrana \cite{fritz2021generalization,vrana2021generalization,algeP1,algeP2} due to its varies applications to real algebraic geometry, probability theory, theoretical computer science and quantum information theory \cite{zuiddam2019asymptotic,fritz2021generalization,li2020quantum,perry2021semiring,bunth2021asymptotic,robere2022amortized,1stapp}. Strassen's Vergleichsstellensatz and its generalizations are analogs of those Positivstellensätze in real algebraic geometry focusing mainly on ordered rings and fields. Among them, Fritz's results \cite{fritz2021generalization,algeP1} recover the classical Positivstellensatz of Krivine-Kadison-Dubois, which suggests that the theory of semirings and semialgebras is able to provide us with new insights into real algebraic geometry.

While Strassen's Vergleichsstellensatz and its generalizations accept only the commutative setting for the semirings under consideration, we prove in this paper a noncommutative version of one of Fritz's generalized Vergleichsstellensätze in \cite{fritz2021generalization}. As in the commutative case, the main assumption on the preordered semialgebra $S$ considered in our theorem is:
\begin{assumption}\label{power}
The inequality $``\,1_S\geq0_S"$ holds in the preordered semialgebra $S$ and there is a {power universal} element $u\in S$.
\end{assumption}
\noindent{}The assumption of the existence of a power universal element is similar to the Archimedean condition in the traditional real algebraic geometry.
Moreover,  the semialgebra $S$ considered in this paper also   satisfies the following assumption:
\begin{assumption}\label{assfree}
The semialgebra $S$ is both {zero-sum-free} and {zero-divisor-free},  and the inequality $``\,1_S\neq0_S"$ holds.
\end{assumption}
Then our main result  can be stated as follows:
\begin{theorem}\label{main}
Let $(S,\leq)$ be a preordered semialgebra satisfying Assumptions \ref{power} and \ref{assfree}, with a power universal element $u\in S$. Then, for every nonzero $x,y\in S$, the following are equivalent:

\noindent$($a$)$ $f(x)\geq f(y)$ for every monotone semialgebra homomorphism $f:S\rightarrow\mathbb{R}_+$.

\noindent$($b$)$ For every real number $\epsilon>0$, there is a finite number $m\in\mathbb{N}$ such that
\begin{equation}
\overline{x}+\sum_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j\succcurlyeq\overline{y}.
\end{equation}

\noindent$($c$)$ For every  real number $r\in\mathbb{R}_+$ and every real number $\epsilon>0$, there is a polynomial $p\in\mathbb{Q}_+[X]$ such that $p(r)\leq\epsilon$ and
\[
\overline{x}+p(\overline{u})\succcurlyeq\overline{y}.
\]

\noindent$($d$)$ For every real number  $r\in\mathbb{R}_+$ and every real number $\epsilon>0$, there is a polynomial $p\in\mathbb{Q}_+[X]$ such that $p(r)\leq1+\epsilon$ and
\begin{equation}
p(\overline{u})*\overline{x}\succcurlyeq\overline{y}.
\end{equation}
\end{theorem}
\noindent{}The elements $\overline{x},\overline{y}$ and $\overline{u}$ above are the images of $x,y$ and $u$ with respect to the canonical map (Definition \ref{canonical}) from $S$ to the semialgebra of its fractions, and the preorder ``$\preccurlyeq$'' is a preorder on the semialgebra of the fractions which is directly derived from the original preorder ``$\leq$" on $S$ (Definition  \ref{derivedpreorder}).

The main technical difficulty to prove Theorem \ref{main} is to define properly the semialgebra of the fractions of a noncommutative semialgebra: In the commutative case, every fraction of  elements in a commutative semialgebra/semiring can be written in the simple form ``$\frac{a}{b}$" for two elements $a,b$ in the semialgebra/semiring. However, ``fractions" in the noncommutative case can be of more complicated forms, \emph{e.g.}, ``$d^{-1}*g*h^{-1}$" and ``$(d+g^{-1})^{-1}$", which can no longer be written in the form ``$\frac{a}{b}$". Our definition for the semialgebra of the fractions of a noncommutative semialgebra in Subsection \ref{subsec:frac} generalizes the definition in the literature and we also show  that it coincides with the usual definition in the commutative case (Proposition \ref{isom}). Now that theories on commutative semirings and semialgebras help us understand commutative real algebraic geometry better, it is reasonable to expect the same thing in the noncommutative case. Moreover, we hope that there could be applications of the theory of noncommutative semirings and semialgebras (including our new theorem) to   quantum information theory and other related  areas.

The main difference between Theorem \ref{main} and  Fritz's Vergleichsstellensatz (\cite[Theorem 2.12]{fritz2021generalization}) is that, in the commutative case, the corresponding inequalities in conditions (b) - (d) can be rewritten as inequalities in the original semialgebra $S$ (Proposition \ref{preorderisom}), while in Theorem \ref{main}, one may find it non-trivial to do the same thing.

In the next section, we define the semialgebra of the fractions of a noncommutative semialgebra that satisfies Assumption \ref{assfree}. This definition is the cornerstone of the main theorem of our paper. Section \ref{sec:derivedPreorder} indicates how a preorder relation on a semialgebra derives another preorder relation on the semialgebra of its fractions.  Section \ref{sec:extend} is devoted to interpreting how an $\mathbb{R}_+$-valued semialgebra homomorphism can be extended from a semialgebra to the semialgebra of its fractions. In Section \ref{sec:V}, we explain in detail how an $\mathbb{R}$-linear space can be constructed from a (noncommutative) semialgebra. This is mentioned in \cite{fritz2021generalization} but we include it for rigorousness and completeness. Finally, in  Section \ref{sec:mainTHM}, we present  the proof of Theorem \ref{main}.

\section{The Semialgebra of the Fractions of a Noncommutative Semialgebra}\label{sec:frac}
Throughout this paper, the letter ``$S$" stands for a noncommutative semialgebra that satisfies Assumption \ref{assfree}. Our definition for the semialgebra of the fractions of such a semialgebra $S$ consists of  three parts.
The first part is devoted to defining a set $W$ of formal rational expressions of  elements in $S$ containing formal operations while the second part aims to construct an equivalence relation $R$ on it. Finally, in the third part we define the semialgebra $F$ of the fractions of $S$ as $F=W/R$.

In
Proposition \ref{isom}, we  prove  that when the semialgebra $S$ is commutative, the semialgebra $F$ of its fractions defined in this section coincides with the corresponding concept in the commutative case, which is obtained naturally from Example 11.7 of \cite{golan2013semirings}.  Therefore,
the new definition of the semialgebra of the fractions indeed generalizes the corresponding concept in the commutative case. Furthermore, it also generalizes the definition in \cite[Proposition  11.5]{golan2013semirings}, where elements in the semiring of the fractions of a noncommutative semiring are only allowed to be of the form $a_1^{-1}*a_2$ (while our definition allows the inversion to appear in any position or even to be nested in an element, \emph{e.g.}, we have elements of the form $a^{-1}*b*c^{-1}+d*p^{-1}*q$ and $(a+b^{-1})^{-1}$, \emph{etc.}, which are apparently much more complicated).

In \cite[Chap.\,18]{golan2013semirings}, the author generalized the definition of the semiring of fractions in \cite[Chap.\,11]{golan2013semirings} via the concept of a \emph{Gabriel filter} of a semiring. Although our definition in this section generalizes the one in \cite[Chap.\,11]{golan2013semirings}, we do not know how our definition and the one given in \cite[Chap.\,18]{golan2013semirings} could possibly be related since they looks utterly different from each other.

Another topic which may related to the content in this section is the theory of \emph{skew fields} \cite{cohn1995skew}, which mainly studies the rings (or fields) of fractions of commutative and noncommutative rings. While there are many methods to construct skew fields of noncommutative rings, this section illustrates a particular way to construct a ``semi-skew-field" of a noncommutative semialgebra.

\subsection{Defining the set of formal rational expressions of the elements in $S$}\label{subsec:formalexpression}

Before defining the semialgebra of the fractions of a noncommutative semialgebra $S$, we need to define the set $W$ of all \emph{the formal rational expressions} of the elements in $S$ with the formal addition ``$\oplus$", the formal multiplication ``$\circledast$", the formal scalar multiplication ``$\odot$" and the formal inverse ``$\;^{-1}$". This is done in a recursive manner: First, set $W_0=S$ and $\mathcal{O}_0=\{0_S\}$, then set
\begin{equation}\label{W}
    W_{i+1}=W_i\cup(W_i\oplus W_i)\cup(W_i\circledast W_i)\cup(\mathbb{R}_{+}\odot W_i)\cup(W_i\backslash \mathcal{O}_i)^{-1},
\end{equation}
and
\begin{equation}\label{O}
    \mathcal{O}_{i+1}=\mathcal{O}_{i}\cup(\mathcal{O}_{i}\oplus\mathcal{O}_{i})\cup(\mathcal{O}_{i}\circledast W_i)\cup(W_i\circledast\mathcal{O}_{i})\cup(\{0_{\mathbb{R}}\}\odot W_i)\cup(\mathbb{R}_+\odot\mathcal{O}_{i}).
\end{equation}
Finally, we define $W=\cup_{i=0}^\infty W_i$ and $\mathcal{O}=\cup_{i=0}^\infty \mathcal{O}_i$. In the definition above, the symbol $0_\mathbb{R}$ denotes the real number $0$. Moreover, for any operation $\pi\in\{\oplus,\circledast,\odot\}$ and any two sets $A,B$, the set $A\pi B$ refers to
$\{(a\pi b)\;|\;a\in A,b\in B\}.$ Similarly, $A^{-1}$ stands for the set $\{(a)^{-1}|\;a\in A\}$. It is clear that each $\mathcal{O}_i$ is a subset of $W_i$ and thus $\mathcal{O}\subset W$. We will see later that the set $\mathcal{O}$ stands for those expressions that should be regarded as zero in the semialgebra of fractions that we are going to construct. This is why the formal inverse is only applied to those expressions in the set $W_i\backslash\mathcal{O}_i$ (rather than the whole set $W_i$) when each $W_{i+1}$ is constructed.

Note that the order in which the formal operations apply in the expressions in $W$ matters, \emph{e.g.}, two expressions like $(a\oplus b)\oplus c$ and $a\oplus (b\oplus c)$ are regarded as distinct elements in $W$.

The following observation is useful in the rest of the paper:

\begin{proposition}\label{basicrules}
For any $c,d\in W$ and $r\in\mathbb{R}_+$, we have

i) $c\oplus d\in\mathcal{O}$ iff $c\in\mathcal{O}$ and $d\in\mathcal{O}$;

ii) $c\circledast d\in\mathcal{O}$ iff $c\in\mathcal{O}$ or $d\in\mathcal{O}$;

iii) $r\odot c\in\mathcal{O}$ iff $r=0_\mathbb{R}$ or $c\in\mathcal{O}$.
\end{proposition}
\begin{proof}
The ``if" parts of the three conditions are straightforward from the equation (\ref{O}). For the ``only if" parts, we only explain the one of the second condition (the other two of the rest conditions can be proved similarly): as $c\circledast d\in\mathcal{O}=\cup_{i=0}^\infty\mathcal{O}_i$, there is a minimal $\ell\in\mathbb{N}$ such that $c\circledast d\in \mathcal{O}_\ell$. Clearly, $\ell>0$ and $c\circledast d\notin\mathcal{O}_{\ell-1}$ ($\ell>0$ since the only element $0_S\in\mathcal{O}_0$ contains no formal operations but there is a ``$\circledast$" in the expression $c\circledast d$). Note that the last formal operation in $c\circledast d$ is ``$\circledast$", we have $c\circledast d\in\mathcal{O}_{\ell-1}\circledast W_{\ell-1}$ or $c\circledast d\in W_{\ell-1}\circledast\mathcal{O}_{\ell-1}$ by the definition of $\mathcal{O}_\ell$ in (\ref{O}). Thus, $c\in\mathcal{O}_{\ell-1}$ or $d\in\mathcal{O}_{\ell-1}$.
\end{proof}

Since there is no formal operation in the element in $\mathcal{O}_0$ and the last operation of any expression in $\mathcal{O}_{\ell}$ ($\ell>0$) is ``$\oplus$", ``$\odot$" or ``$\circledast$" by equation (\ref{O}), which cannot be ``$\,^{-1}$", we have the following  proposition:
\begin{proposition}\label{invnotinO}
The last formal operation of any expression in $\mathcal{O}$ cannot be ``$\,^{-1}$", \emph{i.e.}, $a^{-1}\notin\mathcal{O}$ for any $a^{-1}\in W$.
\end{proposition}

\subsection{Constructing an equivalence relation on the set of formal rational expressions}
In this subsection we define an equivalence relation $R$ on the set $W$ so that the quotient $W/R$ becomes a semialgebra of the fractions of $S$ with every nonzero element invertible.

Set $Q=\{w\in W\,|\,w\text{ contains no formal inverse}\}$, then any $w\in Q$ can be evaluated in the semialgebra $S$: one takes off all the ``circles" of the symbols $\oplus,\circledast$ and $\odot$, respectively turning them into $+,*$ and $\cdot$ (the operations in $S$), whenever they occur in the expression $w$. In this way, the expression $w$ results in an element of $S$, denoted by $w_S$.
The equivalence relation $R$ is defined recursively as follows: The initial relations $R_0\subset R$,  used to  guarantee
that basic laws of operation hold in $W/R$, are given by $R_0=\cup_{i=0}^7A_i$, with
\begin{equation}\label{As}
\left.
\begin{array}{rcl}
    A_1&=&\{((a\oplus b)\oplus c,a\oplus (b\oplus c)),((a\circledast b)\circledast c,a\circledast (b\circledast c)),\\
        &&\textcolor{white}{h}(a\circledast(b\oplus c),(a\circledast b)\oplus(a\circledast c)),((b\oplus c)\circledast a,(b\circledast a)\oplus(c\circledast a)),\\
    &&\textcolor{white}{h}(a\oplus b,b\oplus a),(0_S\oplus a,a),(a\oplus0_S,a)\,|\,a,b,c\in W\},\\
    A_2&=&\{(1_S\circledast a,a),(a\circledast1_S,a),(0_S\circledast a,0_S),(a\circledast0_S,0_S)\,|\,a\in W\},\\
    A_3&=&\{(r\odot(a\oplus b),(r\odot a)\oplus( r\odot b)),((r+_{_{\mathbb{R}}}t)\odot a,(r\odot a)\oplus (t\odot a)),\\
    &&\textcolor{white}{h}(r\odot(a\circledast b),(r\odot a)\circledast b),(r\odot(a\circledast b),a\circledast (r\odot b)),\\
    &&\textcolor{white}{h}((rt)\odot a,r\odot(t\odot a)),(1_
    \mathbb{R}\odot a,a),(0_
    \mathbb{R}\odot a,0_S)\,|\,r,t\in\mathbb{R}_+,a,b\in W\},\\
   A_4&=&\{(a^{-1}\circledast a,1_S),(a\circledast a^{-1},1_S),((a\circledast b)^{-1},b^{-1}\circledast a^{-1}),(1_S^{-1},1_S),\\
    &&\textcolor{white}{h}(s\odot(a^{-1}),(\frac{1}{s}\odot a)^{-1})\,|\,a,b\in W\backslash\mathcal{O},s\in\mathbb{R}_{>0}\},\\
   A_5 &=&\{(a,b)\in Q\times Q\,|\,a_S=b_S\},\\
   A_6 &=&\{(a,a)\,|\,a\in W\},\\
   A_7 &=&\{(a,0_S)\,|\,a\in\mathcal{O}\}=\mathcal{O}\times\{0_S\}.\\
\end{array}
\right.
\end{equation}
Then, these relations are propagated through those formal operations of $W$, and also through the symmetry and transitivity of the relation $R$ that we are going to define:
\begin{equation}\label{Ri}
\left.
\begin{array}{rcl}
R_{i+1}&\hspace{-1mm}=&\hspace{-1mm}R_i\cup(R_i\oplus R_i)\cup(R_i\circledast R_i)\cup(\mathbb{R}_+\odot R_i)\cup(R_i\backslash((\mathcal{O}\times W)\cup(W\times\mathcal{O})))^{-1}\\
&&\hspace{-2mm}\cup\;\text{rev}(R_i)\cup\{(a,c)\in W\times W\,|\,\exists b\in W\text{ \emph{s.t.} }(a,b),(b,c)\in R_i\},
\end{array}
\right.
\end{equation}
where $\text{rev}(R_i)=\{(b,a)\,|\,(a,b)\in R_i\}$, and as usual,

i) for any $\pi\in\{\oplus,\circledast\}$ and any two sets $C,D\subset W\times W$, \[C\pi D=\{(a_1\pi a_2,b_1\pi b_2)\,|\,(a_1,b_1)\in C, (a_2,b_2)\in D\};\]

ii) for any $C\subset W\times W$, $\mathbb{R}_+\odot C=\{(r\odot a,r\odot b)\,|\,r\in\mathbb{R}_+,(a,b)\in C\}$;

iii) for any $C\subset (W\backslash\mathcal{O})\times(W\backslash\mathcal{O})$, $C^{-1}=\{(a^{-1},b^{-1})\,|\,(a,b)\in C\}$.

\noindent{}Set $R=\cup^\infty_{i=0}R_i$, then we have
\begin{proposition}
The set $R\subset W\times W$ is an equivalence relation on the set $W$.
\end{proposition}
\begin{proof}
For any $a\in W$, $(a,a)\in R_0\subset R$. Assume that $(a,b)\in R$, then $(a,b)\in R_i$ for some $i$. Hence $(b,a)\in\text{rev}(R_i)\subset R_{i+1}\subset R$. Suppose that both $(a,b)$ and $(b,c)$ are in $R$, then $(a,b)\in R_i$ for some $i$ and $(b,c)\in R_j$ for some $j$. Set $k=\max\{i,j\}$, then $(a,b),(b,c)\in R_k$. Thus $(a,c)\in R_{k+1}\subset R$.
\end{proof}
\begin{definition}\label{canonical}
For any $w\in W$, we denote by $\overline{w}$ the $R$-equivalent class containing $w$. The map \[\overline{\textcolor{white}{a}}:S\rightarrow W/R,x\mapsto \overline{x}\] is called \emph{the canonical map}. A subset $B\subset W$ is called $R$-\emph{saturated} if for any $w\in W$, $w\in B$ implies $\overline{w}\subset B$.
\end{definition}

In this paper, sometimes we write ``$x\stackrel{\,\hspace{-0.3mm}_R}{\sim} y$" instead of ``$(x,y)\in R$" for any elements $x,y\in W$.

\subsection{Defining the semialgebra of the fractions of $S$}\label{subsec:frac}

We show in  Lemma \ref{Osat}  that the set $\mathcal{O}$ is $R$-saturated (its  proof  is quite tedious, we move it to Appendix A). Therefore,
the following proposition is straightforward:
\begin{proposition}\label{0class}
These two sets coincide: $\overline{0}_S=\mathcal{O}$.
\end{proposition}
\begin{proof}
Since $\mathcal{O}\times\{0_S\}\subset R_0\subset R$, $\overline{0}_S\supset\mathcal{O}$. Conversely, set $x\in\overline{0}_S$, then $(x,0_S)\in R$. Clearly, $(x,0_S)\in R_i$ for some $i\in\mathbb{N}$ and $0_S\in\mathcal{O}$. By Lemma \ref{Osat}, we have $x\in\mathcal{O}$.
\end{proof}

Set $F=W/R$ to be the set of equivalent classes in $W$,  we can define an addition, a multiplication (which derives a corresponding inverse) and a scalar multiplication in $F$ as follows, turning it into a semialgebra:
\[
\left.
\begin{array}{rcl}
\overline{a}+\overline{b}&=&\overline{a\oplus b}\\
\overline{a}*\overline{b}&=&\overline{a\circledast b}\\
r\cdot \overline{a}&=&\overline{r\odot a}\\
\overline{c}^{-1}&=&\overline{c^{-1}}
\end{array}
\right\},
\;\forall a,b\in W,\;\forall r\in\mathbb{R}_+,\;\forall c\in W\backslash\mathcal{O}.
\]
\begin{proposition}
The operations given above are well-defined and they turn $F$ into a semialgebra with any nonzero element invertible.
\end{proposition}
\begin{proof}
Let $a',b'\in W$ and suppose that $\overline{a'}=\overline{a}$ and $\overline{b'}=\overline{b}$, \emph{i.e.}, $(a',a),(b',b)\in R$. By equation (\ref{Ri}), we have
\[
\left\{
\begin{array}{rcl}
(a'\oplus b',a\oplus b)&\in&R_{\max\{m,n\}+1},\\
(a'\circledast b',a\circledast b)&\in&R_{\max\{m,n\}+1},\\
(r\odot a',r\odot a)&\in&R_{m+1},
\end{array}
\right.
\]
whenever $(a',a)\in R_m$ and $(b',b)\in R_n$ for some $m,n\in\mathbb{N}$. Therefore,  our definitions for  the addition, the multiplication and the scalar multiplication are all well-defined. Using the last two kinds of relations in $A_1$ and the first two kinds of relations in $A_2$, we know  that $\overline{0}_S$ and $\overline{1}_S$ are the identities of addition and multiplication respectively. The rest kinds of the relations in the set $A_1\cup A_2\cup A_3$ ensure that all the operation laws that should hold in a semialgebra are indeed valid in $F$ (while the relations in $A_5$ indicate that the polynomial identities in $S$ are all preserved in $F$).

Set $c'\in W$ and suppose that $\overline{c'}=\overline{c}$. Note that $\mathcal{O}$ is $R$-saturated, $c\notin\mathcal{O}$ implies $c'\notin\mathcal{O}$. Thus $(c')^{-1}$ is indeed a formal expression in $W$, and $\overline{(c')^{-1}}$ is well-defined. Suppose that $(c',c)\in R_n$ for some $n\in\mathbb{N}$, then $((c')^{-1},c^{-1})\in R_{n+1}$, which means $\overline{(c')^{-1}}=\overline{c^{-1}}$. Thus, the inverse is well-defined for any $c\notin\mathcal{O}$, \emph{i.e.}, for any $\overline{c}\neq \overline{0}_S$. Finally, the relations in the set $A_4$ guarantee  that the following equations hold for any $\overline{a},\overline{b}\neq\overline{0}_S$: $\overline{a}^{-1}*\overline{a}=\overline{1}_S=\overline{a}*\overline{a}^{-1}$, $(\overline{a}*\overline{b})^{-1}=\overline{b}^{-1}*\overline{a}^{-1}$.
\end{proof}
\begin{remark}
The semialgebra $F$ is also zero-sum-free and zero-divisor-free:
\begin{itemize}
    \item
$\overline{a}+\overline{b}=\overline{0}_S$ implies that $a\oplus b\in\mathcal{O}$. Hence $a,b\in\mathcal{O}$, that is, $\overline{a}=\overline{b}=\overline{0}_S$;
\item  $\overline{a}*\overline{b}=\overline{0}_S$ implies that $a\circledast b\in\mathcal{O}$. Hence $a\in\mathcal{O}$ or $b\in\mathcal{O}$, that is, $\overline{a}=\overline{0}_S$ or $\overline{b}=\overline{0}_S$;
\item  $r\cdot\overline{b}=\overline{0}_S$ implies that $r\odot b\in\mathcal{O}$. Hence $r=0_\mathbb{R}$ or $b\in\mathcal{O}$, that is, $r=0_\mathbb{R}$ or $\overline{b}=\overline{0}_S$.
\end{itemize}
\end{remark}
\section{Preordered Semialgebra and the Derived Preorder Relation on the Semialgebra of its Fractions}\label{sec:derivedPreorder}
In this section, we discuss how a preorder relation on a semialgebra $S$ derives a relative preorder relation on the semialgebra $F$ of the fractions of $S$.

\begin{definition}\label{derivedpreorder}
The \emph{derived preorder} ``$\preccurlyeq$" on the semialgebra $F$ of the fractions of $S$ generated by the preorder ``$\leq$" on $S$ is given by the following condition: for any $g,h\in F$, we write $g\preccurlyeq h$ iff the pair $(g,h)$ is in the set $L=\cup_{i=0}^\infty L_i$ with
\begin{equation}\label{Li}
\left\{
\begin{array}{lcl}
L_0&=&\{(\overline{x},\overline{y}),(\overline{0}_S,a),(a,a)\,|\,x,y\in S, x\leq y,a\in F\},\\
L_{i+1}&=&L_i\cup(\mathbb{R}_+\cdot L_i)\cup(L_i+L_i)\cup(F*L_i)\cup(L_i*F),\\
&&\cup\{(a,c)\,|\,\exists b\in F\text{ such that }(a,b),(b,c)\in L_i\},
\end{array}
\right.
\end{equation}
where \[\mathbb{R}_+\cdot L_i=\{(r\cdot b_1,r\cdot b_2)\,|\,(b_1,b_2)\in L_i,r\in\mathbb{R}_+\},\]\[L_i+L_i=\{(b_1+ c_1,b_2+c_2)\,|\,(b_1,b_2)\in L_i,(c_1,c_2)\in L_i\},\]\[F*L_i=\{(a* b_1,a* b_2)\,|\,(b_1,b_2)\in L_i,a\in F\},\] and\[L_i*F=\{(b_1* a,b_2* a)\,|\,(b_1,b_2)\in L_i,a\in F\}.\]
\end{definition}

\begin{proposition}
The binary relation $\preccurlyeq$ on $F$ makes $F$ a preordered semialgebra.
\end{proposition}
\begin{proof}
Since $(a,a)\in L_0$ for any $a\in F$, the relation $\preccurlyeq$ is reflexive. Suppose that $a\preccurlyeq b$ and $b\preccurlyeq c$ for $a,b,c\in F$, then $(a,b)\in L_n$ and $(b,c)\in L_m$ for some $n,m\in\mathbb{N}$. By the definition of $L_{i+1}$ in (\ref{Li}), we have $(a,c)\in L_{\max\{n,m\}+1}$. Hence $\preccurlyeq$ is transitive.

Assume that $x\preccurlyeq y$ for some $x,y\in F$, $a\in F$ and $r\in\mathbb{R}_+$. Then $(x,y)\in L_n$ for some $n\in\mathbb{N}$. Noting that $(a,a)\in L_0\subset L_n$, we have $(a+x,a+y),(a*x,a*y),(x*a,y*a),(r\cdot x,r\cdot y)\in L_{n+1}$ by equation (\ref{Li}).
\end{proof}
Recall that, for a commutative zero-divisor-free semiring $K$ with preorder relation ``$\leq$", the  derived preorder relation ``$\preccurlyeq$" on the semiring $K^\text{fr}$ of the fractions of $K$ (see Appendix \ref{coincide}) is defined as follows: for any $x,y\in K$ and any $a,b\in K\backslash\{0_K\}$,
\begin{equation}\label{compare}
\frac{x}{a}\preccurlyeq\frac{y}{b}\text{ \hspace{2.5mm}iff\hspace{3mm} }\exists t\in K\backslash\{0_K\},x*b*t\leq y*a*t \text{ (\cite{fritz2021generalization}, pp. 13)}.
\end{equation}
If $K$ is a commutative zero-sum-free semialgebra, then the derived preorder relation  ``$\preccurlyeq$" can naturally be defined as in (\ref{compare}), too. In Appendix \ref{fracorder} Proposition \ref{preorderisom}, we  indicate that, when $S$ is commutative, and if one takes $K=S$ in (\ref{compare}), then the derived preorder relation for $F$ given in Definition \ref{derivedpreorder} coincides with the definition in (\ref{compare}) for $S^{\text{fr}}$.
This explains the motivation of  Definition \ref{derivedpreorder}.

The following proposition is important for the proof of the main theorem:
\begin{proposition}\label{poweruniversal}
If $u\in S$ is power universal in $S$ $($with respect to $\leq)$, then $\overline{u}$ is power universal in $F$ $($with respect to $\preccurlyeq)$.
\end{proposition}
\begin{proof}
We may assume that $u\neq0_S$, since otherwise both the preorder relations $\leq$ on $S$ and $\preccurlyeq$ on $F$ would be trivial. In that case the property that we want to prove is trivially true. Since $(S\backslash\{0_S\})\cap\mathcal{O}=\emptyset$, $u\neq0_S$ clearly implies $\lambda\cdot\overline{u}\neq\overline{0}_S$ for any fixed real number and $\lambda>1$. In the following, we first prove that $\lambda\cdot\overline{u}$ is power universal in $F$, then indicate that $\overline{u}$ is power universal, too.

For any $x\in W\backslash\mathcal{O}$ (or equivalently, any nonzero $\overline{x}\in F$), we need to show that there is a number  $k\in\mathbb{N}$ such that
\begin{equation}\label{lpu}
(\lambda\cdot \overline{u})^k*\overline{x}\succcurlyeq1_S,\;\;\overline{x}*(\lambda\cdot \overline{u})^k\succcurlyeq1_S\;\text{ and }\;(\lambda\cdot \overline{u})^k\succcurlyeq \overline{x}.
\end{equation}
Suppose that $x\in W_i$ for $i\in\mathbb{N}$. We prove inductively on $i$:
\\

\noindent{}\emph{The $``i=0"$ case:}

When $i=0$, $x\in W_0=S$ and $x\neq0_S$. Since $u$ is power universal in $S$, there is an integer $k\in\mathbb{N}$ such that inequalities (\ref{PU}) hold. By the definition of $\preccurlyeq$ in equations (\ref{Li}), we also have
\[
\overline{u}^k*\overline{x}\succcurlyeq1_S,\;\;\overline{x}* \overline{u}^k\succcurlyeq1_S\;\text{ and }\;\overline{u}^k\succcurlyeq \overline{x}.
\] Since $\lambda>1$, $\lambda\cdot\overline{u}=\overline{u}+(\lambda-1)\cdot\overline{u}\succcurlyeq\overline{u}$. Thus $(\lambda\cdot\overline{u})^k\succcurlyeq\overline{u}^k$, and we obtain the inequalities in $(\ref{lpu})$.
\\

\noindent{}\emph{The induction step:}

Assume that for any $0\leq i\leq \ell$ ($\ell\in\mathbb{N}$) and any $x\in W_i\backslash\mathcal{O}$, there is an integer  $k\in\mathbb{N}$ such that the inequalities in (\ref{lpu}) hold. We prove that this is also valid for $i=\ell+1$:

Set $x\in W_{\ell+1}\backslash\mathcal{O}$. Then there are several cases corresponding to the components in the definition of $W_{\ell+1}$ in equation (\ref{W}):

1) $x\in W_\ell$: The conclusion follows directly from the inductive assumption.

2) $x\in W_\ell\oplus W_\ell$: Then $\overline{x}=\overline{b}+\overline{c}$ with $b,c\in W_\ell$. Since $\overline{x}\neq\overline{0}_S$, $\overline{b}\neq\overline{0}_S$ or $\overline{c}\neq\overline{0}_S$. If one of $\overline{b}$ and $\overline{c}$ equals $\overline{0}_S$, say, $\overline{b}=\overline{0}_S$, then $\overline{x}=\overline{c}$. According to the inductive assumption, for $c\in W_\ell$, there is an integer $k\in\mathbb{N}$ such that (\ref{lpu}) hold with ``$\overline{x}$" replaced by ``$\overline{c}$".
Therefore, inequalities in (\ref{lpu}) hold for $\overline{x}$ itself as well. Now, set $\overline{b}\neq\overline{0}_S$ and $\overline{c}\neq\overline{0}_S$, then we have $k_1,k_2\in\mathbb{N}$ such that
\begin{equation}\label{six}
\left\{
\begin{array}{rcl}
(\lambda\cdot \overline{u})^{k_1}*\overline{b}\succcurlyeq\overline{1}_S,&\overline{b}*(\lambda\cdot \overline{u})^{k_1}\succcurlyeq\overline{1}_S,&(\lambda\cdot \overline{u})^{k_1}\succcurlyeq\overline{b},\\
(\lambda\cdot \overline{u})^{k_2}*\overline{c}\succcurlyeq\overline{1}_S,&\overline{c}*(\lambda\cdot \overline{u})^{k_2}\succcurlyeq\overline{1}_S,&(\lambda\cdot \overline{u})^{k_2}\succcurlyeq\overline{c}.
\end{array}
\right.
\end{equation}
Set $\hat{k}=\max\{k_1,k_2\}$, and choose sufficiently large  $k\geq \hat{k}$ satisfying $\lambda^k\geq2\lambda^{\hat{k}}$, then
\[
(\lambda\cdot \overline{u})^k*\overline{x}\succcurlyeq(\lambda\cdot \overline{u})^{\hat{k}}*\overline{x}=(\lambda\cdot \overline{u})^{\hat{k}}*(\overline{b}+\overline{c})\succcurlyeq\overline{1}_S+\overline{1}_S\succcurlyeq\overline{1}_S.
\]
Similarly, we have $\overline{x}*(\lambda\cdot \overline{u})^k\succcurlyeq\overline{1}_S$. On the other hand,
\[
(\lambda\cdot \overline{u})^k=\lambda^k\cdot\overline{u}^k\succcurlyeq(2\lambda^{\hat{k}})\cdot\overline{u}^{\hat{k}}=2\cdot(\lambda\cdot\overline{u})^{\hat{k}}\succcurlyeq\overline{b}+\overline{c}=\overline{x}.
\]

3) $x\in W_\ell\circledast W_\ell$: Then $\overline{x}=\overline{b}*\overline{c}$ with $b,c\in W_\ell\backslash\mathcal{O}$ and the inequalities (\ref{six}) hold for some $k_1,k_2\in\mathbb{N}$. Set $k=k_1+k_2$. Multiplying the rightmost inequalities in each rows of (\ref{six}), we get $(\lambda\cdot \overline{u})^{k}\succcurlyeq\overline{b}*\overline{c}=\overline{x}$. Multiplying the leftmost inequality in the first row and the middle inequality in the second row of (\ref{six}), we obtain $(\lambda\cdot \overline{u})^{k_1}*\overline{b}*\overline{c}*(\lambda\cdot \overline{u})^{k_2}\succcurlyeq\overline{1}_S$. Then, by multiplying $(\lambda\cdot \overline{u})^{-k_1}$ and $(\lambda\cdot \overline{u})^{-k_2}$ from the left and the right sides respectively, one derives $\overline{x}\succcurlyeq(\lambda\cdot \overline{u})^{-k}$. From that, one obtains both $(\lambda\cdot \overline{u})^{k}*\overline{x}\succcurlyeq\overline{1}_S$ and $\overline{x}*(\lambda\cdot \overline{u})^{k}\succcurlyeq\overline{1}_S$ by simply multiplying $(\lambda\cdot \overline{u})^{k}$ from the left and the right sides respectively.

4) $x\in\mathbb{R}_+\odot W_\ell$: In this case, $x=r\cdot\overline{b}$ with $b\in W_\ell\backslash\mathcal{O}$ and $r$ is a positive real number. Hence there is an integer $k_1\in\mathbb{N}$ such that the inequalities in the first row or (\ref{six}) hold. Choose a  sufficiently large integer $m\in\mathbb{N}$ satisfying $\lambda^m\geq\max\{r,1/r\}$ and set $k=m+k_1$. Then we have
\[
\left.
\begin{array}{rcl}
(\lambda\cdot\overline{u})^k*\overline{x}&\succcurlyeq&(\lambda\cdot\overline{1}_S)^m*(\lambda\cdot\overline{u})^{k_1}*(r\cdot\overline{b})\\
&\succcurlyeq&(\lambda^mr)\cdot((\lambda\cdot\overline{u})^{k_1}*\overline{b})\\
&\succcurlyeq&1_\mathbb{R}\cdot\overline{1}_S\\
&=&\overline{1}_S,
\end{array}
\right.
\]
and similarly
\[
\left.
\begin{array}{rcl}
\overline{x}*(\lambda\cdot\overline{u})^k&\succcurlyeq&(r\cdot\overline{b})*(\lambda\cdot\overline{u})^{k_1}*(\lambda\cdot\overline{1}_S)^m\\
&\succcurlyeq&(\lambda^mr)\cdot(\overline{b}*(\lambda\cdot\overline{u})^{k_1})\\
&\succcurlyeq&1_\mathbb{R}\cdot\overline{1}_S\\
&=&\overline{1}_S.
\end{array}
\right.
\]
Moreover, we have $(\lambda\cdot\overline{u})^k=(\lambda\cdot\overline{u})^m*(\lambda\cdot\overline{u})^{k_1}\succcurlyeq(\lambda\cdot\overline{1}_S)^m*\overline{b}=\lambda^m\cdot\overline{b}\succcurlyeq r\cdot\overline{b}=\overline{x}$.

5) $x\in(W_\ell\backslash\mathcal{O}_\ell)^{-1}$: Then $\overline{x}=\overline{b}^{-1}$ with $b\in W_\ell\backslash\mathcal{O}_\ell$ satisfying the inequalities in the first row of (\ref{six}) for some integer $k_1\in\mathbb{N}$. From the rightmost inequality we obtain both $(\lambda\cdot\overline{u})^{k_1}*\overline{b}^{-1}\succcurlyeq\overline{1}_S$ and $\,\overline{b}^{-1}*(\lambda\cdot\overline{u})^{k_1}\succcurlyeq\overline{1}_S$ by multiplying $\overline{b}^{-1}$ from its right and left sides, respectively. From the leftmost one we obtain $(\lambda\cdot\overline{u})^{k_1}\succcurlyeq\overline{b}^{-1}$ by multiplying $\overline{b}^{-1}$ from its right side.
\\\vspace{-2mm}

By now we have shown that $\lambda\cdot\overline{u}$ is power universal in $F$ since the inequalities in (\ref{lpu}) hold. Now we show that $\overline{u}$ is also power universal: Apparently, $(\lambda\cdot1_S)^k\neq0_S$, hence there is $k_0\in\mathbb{N}$ such that $u^{k_0}\geq(\lambda\cdot1_S)^k$. Therefore $\overline{u}^{k_0}\succcurlyeq(\lambda\cdot\overline{1}_S)^k$ in $F$. Using the inequalities in (\ref{lpu}), we have $\overline{u}^{k_0+k}\succcurlyeq(\lambda\cdot\overline{1}_S)^k*(\lambda^{-k}\cdot\overline{x})=\overline{x}$, and $\overline{u}^{k_0+k}*\overline{x}\succcurlyeq(\lambda\cdot\overline{1}_S)^k*\overline{u}^k*\overline{x}=(\lambda\cdot\overline{u})^k*\overline{x}\succcurlyeq1_S$. Similarly, we also have $\overline{x}*\overline{u}^{k_0+k}\succcurlyeq1_S$. Hence $\overline{u}$ is indeed power universal.
\end{proof}
\section{Extending $\mathbb{R}_+$-Valued Homomorphisms from a Semialgebra to the Semialgebra of its Fractions}\label{sec:extend}
In this section, we explain how a monotone homomorphism from the preordered semialgebra $S$ to some other preordered semialgebra $T$ (which shares some good properties) can be extended to the semialgebra $F$ of the fractions of $S$. In particular, when $T=\mathbb{R}_+$, the extension is always guaranteed.

The following lemma due to \cite{fritz2021generalization} is straightforward:
\begin{lemma}\label{trivialnull}
Suppose that $(T,\leq_{_T})$ is a preordered semialgebra with nontrivial preorder relation $\leq_{_T}$ such that $1_T\geq_{_T}\hspace{-0.7mm}0_T$, and  $(E,\leq_{_E})$ is a preordered semialgebra with $1_E\geq_{_E}0_E$ and a power universal element $u$. Then for any monotone semialgebra homomorphism $f: E\rightarrow T$ and any nonzero $x\in E$, we have $f(x)\neq0_T$.
\end{lemma}
\begin{proof}
There is an integer $k\in\mathbb{N}$ satisfying
\begin{equation}\label{ukk}
u^k*x\geq_{_E}1_E.
\end{equation}
Applying the homomorphism $f$ to the inequality (\ref{ukk}), we get $f(u)^k*f(x)\geq_{_T}\hspace{-0.7mm}1_T$. Assume that $f(x)=0_T$, then we have $0_T\geq_{_T}\hspace{-0.7mm}1_T$. For any $y,z\in T$, we obtain
\begin{equation}\label{tri}
y=y*1_T\leq_{_T}\hspace{-0.7mm}y*0_T=z*0_T\leq_{_T}\hspace{-0.7mm}z*1_T=z,\end{equation}
which contradicts the assumption that $\leq_{_T}$ is nontrivial.
\end{proof}

Let $T$ be given  as in Lemma \ref{trivialnull} and suppose that $(S,\leq)$ is a preordered semialgebra satisfying Assumption \ref{power}. We assume further, throughout the rest of the paper, that $T$ is both zero-sum-free and zero-divisor-free, and that every nonzero element of $T$ is invertible (\emph{e.g.}, if $T=\mathbb{R}_+$, then all the assumptions on the semialgebra $T$ are satisfied).

For any monotone semialgebra homomorphism $f: S\rightarrow T$, the following proposition defines a map $f^W$ from the set of formal rational expressions $W$ (defined in Subsection \ref{subsec:formalexpression}) to the semialgebra $T$:
\begin{proposition}\label{fW}
The recursive assignment below for $f^W$ makes it a well-defined map from $W$ to $T$:

1) For any $x\in W_0=S$, set $f^W(x)=f(x)$;

2) Suppose that $x\in W_\ell$ for the  minimal integer  $\ell\in\mathbb{N}$ and $\ell>0$, then set
\begin{equation}\label{assgn}
f^W(x)=\left\{
\begin{array}{rcl}
f^W(a)+f^W(b),&if&x=a\oplus b\in W_{\ell-1}\oplus W_{\ell-1},\\f^W(a)*f^W(b),&if&x=a\circledast b\in W_{\ell-1}\circledast W_{\ell-1},\\
r\cdot f^W(a),&if&x=r\odot a\in \mathbb{R}_+\odot W_{\ell-1},\\
(f^W(a))^{-1},&if&x=a^{-1}\in (W_{\ell-1}\backslash\mathcal{O}_{\ell-1})^{-1}.
\end{array}
\right.
\end{equation}
\end{proposition}
\begin{proof}
It is sufficient to prove (inductively on the index $i$) that for any $i\in\mathbb{N}$ and any $x\in W_{i}$, $f^W(x)$ is a well-defined element in $T$ and $f^W(x)\neq0_T$ whenever $x\notin\mathcal{O}_i$ (thus the assignment in the last row of (\ref{assgn}) is well-defined due to the assumption that every nonzero element in $T$ is invertible and Lemma \ref{trivialnull}):

If $i=0$, $x\in W_0=S$, and $f^W(x)=f(x)$ is well-defined. When $x\notin\mathcal{O}_0=\{0_S\}$, we have $x\in S\backslash\{0_S\}$. Then Lemma \ref{trivialnull} indicates that $f^W(x)=f(x)\neq0_T$.

Suppose that for any $0\leq i<k$ ($k\in\mathbb{N}_{\geq 1}$) and any $x\in W_i$, $f^W(x)$ is a well-defined element in $T$ and $f^W(x)\neq0_T$ whenever $x\notin\mathcal{O}_i$. We then show that this is also valid for $i=k$ and any $x\in W_i$:

Let  $x\in W_k$, we may assume that $x\notin W_{k-1}$ (Since, if $x\in W_{k-1}$ then the inductive assumption implies the conclusions that we need). Hence $k\geq1$ is the minimal nature number so that $x\in W_k$. By equation (\ref{W}), there are several cases we need to check:

i) If $x=a\oplus b\in W_{k-1}\oplus W_{k-1}$, then $f^W(x)=f^W(a)+f^W(b)$ is well-defined since $f^W(a)$ and $f^W(b)$ are well-defined. When $x\notin\mathcal{O}_k$, we have either $a\notin\mathcal{O}_{k-1}$ or $b\notin\mathcal{O}_{k-1}$. Hence one of $f^W(a)$ and $f^W(b)$ is nonzero in $T$. We conclude that $f^W(x)=f^W(a)+f^W(b)$ is also nonzero due to the fact that $T$ is zero-sum-free.
\vspace{-3mm}

ii) If $x=a\circledast b\in W_{k-1}\circledast W_{k-1}$, then $f^W(x)=f^W(a)*f^W(b)$ is well-defined. When $x\notin\mathcal{O}_k$, we have $a,b\notin\mathcal{O}_{k-1}$. By the inductive assumption, both $f^W(a)$ and $f^W(b)$ are nonzero. Hence $f^W(x)=f^W(a)*f^W(b)$ is also nonzero (since $T$ is zero-divisor-free).
\vspace{0.5mm}

iii) If $x=r\odot a\in \mathbb{R}_+\odot W_{k-1}$, then $f^W(x)=r\cdot f^W(a)$ is well-defined. When $x\notin\mathcal{O}_k$, we have $r\neq0$ and $a\notin\mathcal{O}_{k-1}$. By the inductive assumption, $f^W(a)$ is nonzero in $T$. Hence $f^W(x)=r\cdot f^W(a)$ is also nonzero (using again the fact that $T$ is zero-divisor-free).
\vspace{0.5mm}

iv) If $x=a^{-1}\in (W_{k-1}\backslash\mathcal{O}_{k-1})^{-1}$ with $a\in W_{k-1}\backslash\mathcal{O}_{k-1}$, then $f^W(a)$ is well-defined and nonzero in $T$ by the inductive assumption. Since every nonzero element in $T$ is invertible, $f^W(x)=(f^W(a))^{-1}$ is also well-defined. Moreover, it is nonzero since $(f^W(a))^{-1}=0_T$ would imply \[1_T=f^W(a)*(f^W(a))^{-1}=f^W(a)*0_T=0_T,\] which indicates that $1_T\leq_{_T}0_T\leq_{_T}1_T$. And this would result in the triviality of $\leq_{_T}$ as shown in (\ref{tri}), contradicting one of the assumptions in Lemma \ref{trivialnull}.
\end{proof}
 Note that by formulae (\ref{assgn}) we have
\begin{equation}\label{prehomo}
\left\{
\begin{array}{rcl}
     f^W(a\oplus b)&=&f^W(a)+f^W(b), \\
     f^W(a\circledast b)&=&f^W(a)*f^W(b),\\
     f^W(r\odot a) &=&r\cdot f^W(a),\\
    f^W(c^{-1}) &=&(f^W(c))^{-1},
\end{array}
\right.
\end{equation}
for all $a,b\in W,c\in W\backslash\mathcal{O}$ and $r\in\mathbb{R}_+$. These equations are useful in the proof of the following proposition:

\begin{proposition}\label{f^F}
For any monotone semialgebra homomorphism $f: S\rightarrow T$, the map $f^F: F\rightarrow T,\, \overline{x}\mapsto f^W(x)\;(\forall x\in W)$ is well-defined.
\end{proposition}
\begin{proof}
For any $x,y\in W$ with $\overline{x}=\overline{y}$ (that is, $(x,y)\in R$), we will show that $f^W(x)=f^W(y)$. Since $R=\cup_{i=1}^\infty R_i$, it is sufficient to prove that for any $i\in\mathbb{N}$ and any $x,y\in W$ satisfying $(x,y)\in R_i$, we have $f^W(x)=f^W(y)$. The proof is inductive on the index $i$:

\noindent{}\emph{The $``i=0"$ case:}

Note that $R_0=\cup_{j=1}^7A_j$ in (\ref{As}), there are several subcases:

If $(x,y)\in \cup_{j=1}^4A_j\subset R_0$, $f^W(x)=f^W(y)$ follows from (\ref{prehomo}).

If $(x,y)\in A_5$, then $x,y\in Q$ and $x_S=y_S$. We claim that, for any $z\in Q$, it holds that $f^W(z)=f(z_S)$. Instead of giving a formal proof, we only provide a simple example to illustrate the claim above: if $z=s_1\oplus((r\odot s_2)\circledast s_3)\in Q$ for some elements $s_1,s_2,s_3\in S$ and a real number  $r\in\mathbb{R}_+$, then clearly we have

\[
\left.
\begin{array}{rcl}
f^W(z)&=&f(s_1)+((r\cdot f(s_2))*f(s_3))\\
&=&f(s_1+((r\cdot s_2)*s_3))\\
&=&f(z_S),
\end{array}
\right.
\]
where the first equality is due to the definition (\ref{assgn}) of $f^W$ and the identities in (\ref{prehomo}), while the second one is valid since $f$ is a semialgebra homomorphism from $S$ to $T$. Thus $f^W(x)=f(x_S)=f(y_S)=f^W(y)$.

If $(x,y)\in A_6$, then we have  $x=y$ and  $f^W(x)=f^W(y)$.

If $(x,y)\in A_7$, then we have $x\in\mathcal{O}$ and $y=0_S$. Since $f^W(0_S)=0_T$, we need to show that $f^W(x)=0_T$ for any $x\in\mathcal{O}$: Clearly, $f^W(x)=0_T$ for any $x\in\mathcal{O}_0=\{0_S\}$. Assume that for any $0\leq \hat{i}<k$ ($k\in\mathbb{N}_{\geq1}$) and any $x\in\mathcal{O}_{\hat{i}}$, we have $f^W(x)=0_T$. We then prove that this is also valid for $\hat{i}=k$. Set $x\in\mathcal{O}_k$ and $x\notin\mathcal{O}_{k-1}$. Then, by the equation (\ref{O}), we have $x=a\oplus b\in\mathcal{O}_{k-1}\oplus\mathcal{O}_{k-1}$, $x=a\circledast b\in(\mathcal{O}_{k-1}\circledast W_{k-1})\cup(W_{k-1}\circledast\mathcal{O}_{k-1})$ or $x=r\odot a\in(\mathbb{R}_+\odot\mathcal{O}_{k-1})\cup(\{0_\mathbb{R}\}\odot W_{k-1})$. From the equations in (\ref{prehomo}), we see that $f^W(x)=0_T$ holds in all cases.

Hence $f^W(x)=f^W(y)$ whenever $(x,y)\in R_0$.
\vspace{1mm}

\noindent{}\emph{The inductive case:}

Suppose that for any $0\leq i<\ell$ ($\ell\in\mathbb{N}_{\geq1}$) and any $(x,y)\in R_i$, we have $f^W(x)=f^W(y)$. We then prove that this is also valid for $i=\ell$.

Assume without loss of generality that $(x,y)\in R_\ell\backslash R_{\ell-1}$. There are several cases left:

i) If $(x,y)=(x_1,y_1)\oplus(x_2,y_2)\in R_{\ell-1}\oplus R_{\ell-1}$, then $x=x_1\oplus x_2$, $y=y_1\oplus y_2$ and $(x_1,y_1),(x_2,y_2)\in R_{\ell-1}$. Thus $f^W(x)=f^W(x_1)+f^W(x_2)=f^W(y_1)+f^W(y_2)=f^W(y)$.

ii) Similarly, we can deal with the cases where $x\in(R_{\ell-1}\circledast R_{\ell-1})\cup(\mathbb{R}_+\odot R_{\ell-1})$.

iii) If $(x,y)=(x_1^{-1},y_1^{-1})\in(R_{\ell-1}\backslash((\mathcal{O}\times W)\cup(W\times \mathcal{O})))^{-1}$, then $x_1,y_1\notin\mathcal{O},x=x_1^{-1},y=y_1^{-1}$ and $(x_1,y_1)\in R_{\ell-1}$. Since $x_1,y_1\notin\mathcal{O}$, neither of $f^W(x_1)$ and $f^W(y_1)$ is zero according to the proof of Proposition \ref{fW}. Thus both $(f^W(x_1))^{-1}$ and $(f^W(y_1))^{-1}$ are well-defined and they are equal to each other since $f^W(x_1)=f^W(y_1)$ by the inductive assumption.

iv) If $(x,y)\in\text{rev}(R)_{\ell-1}$, then $(y,x)\in R_{\ell-1}$. The inductive assumption indicates that $f^W(y)=f^W(x)$.

v) If there is $z\in W$ such that $(x,z),(z,y)\in R_{\ell-1}$, then $f^W(x)=f^W(z)$ and $f^W(z)=f^W(y)$ again by the inductive assumption.

The above discussion show that $f^F$ is a well-defined map from $F$ to $T$.
\end{proof}

\begin{proposition}\label{f^F2}
The map $f^F: F\rightarrow T$ is a monotone semialgebra homomorphism.
\end{proposition}
\begin{proof}
The fact that it is a semialgebra homomorphism  directly follows from the identities in (\ref{prehomo}): \emph{e.g.}, \[f^F(\overline{x}+\overline{y})=f^F(\overline{x\oplus y})=f^W(x\oplus y)=f^W(x)+f^W(y)=f^F(\overline{x})+f^F(\overline{y}).\] The rest three identities corresponding to the last three rows in (\ref{prehomo}) can be derived similarly.

We now prove that $f^F$ is monotone.

Remember that the preorder relation $\preccurlyeq$ on $F$ is given by the set $L\subset F\times F$ defined in (\ref{Li}). We need to show that $f^F(g)\leq_{_{T}}\hspace{-1mm}f^F(h)$ for any $(g,h)\in L$:

If $(g,h)\in L_0$, then $g=\overline{0}_S$, $g=h$ or $g=\overline{x},h=\overline{y}$ for some $x,y\in S$ such that $x\leq y$. The first two sub-cases are trivial while the last one follows from \[f^F(\overline{x})=f^W(x)=f(x)\leq_{_{T}}\hspace{-1mm}f(y)=f^W(y)=f^F(\overline{y}).\]

Assume that for any $0\leq i<k$ ($k\in\mathbb{N}_{\geq1}$) and any $(g,h)\in L_{i}$, we have $f^F(g)\leq_{_{T}}\hspace{-1mm}f^F(h)$. We then prove the case where $i=k$.

We may assume that $(g,h)\notin L_{k-1}$, thus
\begin{equation}\label{cases}
\left.
\begin{array}{rcl}
(g,h)&\in &L_{k}\backslash L_{k-1}\\
&=&\Big((\mathbb{R}_+\cdot L_{k-1})\cup(L_{k-1}+L_{k-1})\cup(F*L_{k-1})\cup(L_{k-1}*F)\\
&&\cup\{(g,h)\,|\,\exists d\in F\text{ so that }(g,d),(d,h)\in L_{k-1}\}\Big)\Big\backslash L_{k-1}.
\end{array}
\right.
\end{equation}
Those cases corresponding to the components in the formula (\ref{cases}) can be dealt with separately, through using the transitivity and the compatible properties (\ref{preoc}) of a preorder relation on a semialgebra. To illustrate, consider the case where $(g,h)=(g_1,h_1)+(g_2,h_2)\in L_{k-1}+L_{k-1}$. We have $g=g_1+g_2$, $h=h_1+h_2$ and
\[\left.\begin{array}{rll}
f^F(g)=f^F(g_1+g_2)&=&\hspace{-2mm}f^F(g_1)+f^F(g_2)\\
&\leq_{_T}&\hspace{-2mm}f^F(h_1)+f^F(g_2)\\
&\leq_{_T}&\hspace{-2mm}f^F(h_1)+f^F(h_2)=f^F(h_1+h_2)=f^F(h).
\end{array}\right.\]
The two $\leq_{_T}\hspace{-0.4mm}$'s above are due to the inductive assumption and the first rule in (\ref{preoc}). The other cases can be handled similarly, hence we omit the details.
\end{proof}
\begin{remark}
For any monotone semialgebra homomorphism $g: F\rightarrow T$ such that $g(\overline{x})=f^F(\overline{x})$ for all $x\in S$ $($\emph{i.e.}, it extends the monotone semialgebra homomorphism $f)$, we can prove inductively that $g=f^F$. This indicates the universal property of the semialgebra $F$ of the fractions of $S$.
\end{remark}
\begin{remark}\label{realvaluedextension}
Set $T=\mathbb{R}_+$, then Propositions \ref{f^F} and \ref{f^F2} indicate that any monotone semialgebra homomorphism $f: S\rightarrow \mathbb{R}_+$ can be extended to a monotone semialgebra homomorphism mapping from $F$ to $\mathbb{R}_+$.
\end{remark}
\section{From A Semialgebra to A Linear Space}\label{sec:V}
In this section, we define a preordered $\mathbb{R}$-linear space based on the semialgebra $F$. This definition is already given by \cite{fritz2021generalization}, while we rewrite it in detail for the completeness of the present paper.

Define $V=F-F$ to be the Grothendieck group \cite[Chap.\,II]{karoubi1978k} of the commutative monoid $(F,+)$, with each element in $V$ of the form $a-b$ for some $a,b\in F$, and with $a-b=a'-b'$ in $V$ iff $\exists c\in F$ \emph{s.t.} $a+b'+c=a'+b+c$ in $F$. Then, $V$ becomes an $\mathbb{R}$-linear space once we define a scalar multiplication on it as below:
\[
r\cdot(a-b)=\left\{
\begin{array}{rclcl}
r\cdot a\hspace{-4.2mm}&\hspace{-1mm}-&\hspace{-1.4mm}r\cdot b,&\text{if\hspace{-1.7mm}}&r\in\mathbb{R}_{\geq0},\\
(-r)\cdot b\hspace{-4.2mm}&\hspace{-1mm}-&\hspace{-1.4mm}(-r)\cdot a,&\text{if\hspace{-1.7mm}}&r\in\mathbb{R}_{<0}.
\end{array}
\right.
\]
It's straightforward to verify that this operation is well-defined. One also observes the following identities:
\begin{equation}\label{idn}
\left.
\begin{array}{rcl}
1_\mathbb{R}\cdot(a-b)&=&a-b,\\
(rs)\cdot(a-b)&=&r\cdot(s\cdot(a-b)),\\
r\cdot((a-b)+(c-d))&=&r\cdot(a-b)+r\cdot(c-d),\\
(r+s)\cdot(a-b)&=&r\cdot(a-b)+s\cdot(a-b)
\end{array}
\right.
\end{equation}
with $r,s\in\mathbb{R}$ and $a,b,c,d\in F$. All of the above identities can be verified straightforwardly. While most of them can be checked concisely, it is not the case for the last one: One may need to consider six cases corresponding to the different sign(s) of the real numbers $r,s$ and $r+s$. To illustrate, we deal with the case where $r\geq0,s<0$ and $r+s<0$ but leave the rest of them to the readers.

Now that $r+s<0$,
\[(r+s)\cdot(a-b)=(-r-s)\cdot b-(-r-s)\cdot a\]
by the definition. On the other hand, we have
\[
\left.
\begin{array}{rcl}\vspace{1.6mm}
&&\textcolor{white}{(}r\cdot(a-b)+s\cdot(a-b)\\\vspace{1.6mm}
&=&(r\cdot a-r\cdot b)+\big((-s)\cdot b-(-s)\cdot a\big)\\
&=&\big(r\cdot a+(-s)\cdot b\big)-\big(r\cdot b+(-s)\cdot a\big).
\end{array}
\right.
\]
Then it is sufficient to show that
\[
\left.
\begin{array}{lrcll}\vspace{1.6mm}
    &(-r-s)\cdot b-(-r-s)\cdot a&=&\big(r\cdot a+(-s)\cdot b\big)-\big(r\cdot b+(-s)\cdot a\big)&\text{(in $V$)}  \\\vspace{1.6mm}
    \Leftarrow&(-r-s)\cdot b+r\cdot b+(-s)\cdot a&=&(-r-s)\cdot a+\big(r\cdot a+(-s)\cdot b\big)&\text{(in $F$)}\\\vspace{1.6mm}
    \Leftarrow&(-s)\cdot b+(-s)\cdot a&=&(-s)\cdot b+(-s)\cdot a.&\text{(in $F$)}
\end{array}
\right.
\]The last equality is trivially true. Hence the last identity in (\ref{idn}) is verified for the case of $r\geq0,s<0$ and $r+s<0$.

The identities in (\ref{idn}) suggest that $V$ is indeed an $\mathbb{R}$-linear space. In the sequel we define a preorder relation on $V$ based on the preorder relation $\preccurlyeq$ on $F$.

For any $a,b,c,d\in F$ (or equivalently, for any $a-b,c-d\in V$), we write $a-b\unlhd c-d$ iff there is a $g\in F$ so that $a+d+g\preccurlyeq b+c+g$ in $F$. It is straightforward to prove that the relation $\unlhd$ is well-defined and it is also a preorder. Furthermore, it is not hard to show that
\begin{equation}\label{Ccone}
a-b\unlhd c-d\text{\hspace{1.6mm} implies }
\left\{
\begin{array}{rcl}
     (a-b)+(g-h)&\unlhd&(c+d)+(g-h), \\
     r\cdot(a-b)&\unlhd& r\cdot(c-d),
\end{array}
\right.
\end{equation}
for any $a,b,c,d,g,h\in F$ and any $a\in\mathbb{R}_+$. These inequalities are similar to the ones given in (\ref{preoc}).

To conclude, the $\mathbb{R}$-linear space $V$ is preordered with the relation $\unlhd$ which is compatible with the addition and the scalar multiplication defined in itself.

\section{The Main Theorem and the Proof}\label{sec:mainTHM}
This section is devoted to the proof of the main theorem. We state it again for convenience:

\noindent\textbf{Theorem \ref{main}}\emph{
Let $(S,\leq)$ be a preordered semialgebra satisfying Assumptions \ref{power} and \ref{assfree}, with a power universal element $u\in S$. Suppose that $(F,\preccurlyeq)$ is the preordered semialgebra of the fractions of $S$ defined in Subsection \ref{subsec:frac} with ``$\preccurlyeq$" in Def. \ref{derivedpreorder} and  $\overline{{}\textcolor{white}{a}}: S\rightarrow F, x\mapsto \overline{x}$ the canonical map in Def. \ref{canonical}. Then, for every nonzero $x,y\in S$, the following are equivalent:}

\noindent\emph{
$($a$)$ $f(x)\geq f(y)$ for every monotone semialgebra homomorphism $f:S\rightarrow\mathbb{R}_+$.}

\noindent\emph{
$($b$)$ For every real number $\epsilon>0$, there is $m\in\mathbb{N}$ such that
\begin{equation}\label{binq}
\overline{x}+\sum_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j\succcurlyeq\overline{y}.
\end{equation}}

\noindent\emph{
$($c$)$ For every $r\in\mathbb{R}_+$ and every real number $\epsilon>0$, there is a polynomial $p\in\mathbb{Q}_+[X]$ such that $p(r)\leq\epsilon$ and
\begin{equation}
\overline{x}+p(\overline{u})\succcurlyeq\overline{y}.
\end{equation}}

\noindent\emph{
$($d$)$ For every $r\in\mathbb{R}_+$ and every real number $\epsilon>0$, there is a polynomial $p\in\mathbb{Q}_+[X]$ such that $p(r)\leq1+\epsilon$ and
\begin{equation}\label{dd}
p(\overline{u})*\overline{x}\succcurlyeq\overline{y}.
\end{equation}
}

\begin{proof}
The proof of the implication ``$($b$)\Rightarrow($c$)$" is almost the same as  its counterpart in the proof of Theorem 3.1 in \cite{fritz2021generalization}.

The proof of ``$($c$)\Rightarrow($d$)$" is also similar to the corresponding part of the proof of Theorem 3.1 in \cite{fritz2021generalization}. But since we are dealing with noncommutative semialgebras, there are some tiny differences. So we rewrite the proof of ``$($c$)\Rightarrow($d$)$" for clearness:

Since $x\in S$ and $x\neq0_S$, $x\notin\mathcal{O}$, we have $\overline{x}\neq\overline{0}_S$. By Proposition  \ref{poweruniversal}, $\overline{u}$ is power universal in $F$. Hence there is an integer $k\in\mathbb{N}$ so that $\overline{u}^k*\overline{x}\succcurlyeq\overline{1}_S$. For any $r\in\mathbb{R}_+$ and any real number $\epsilon>0$, set
\[
\epsilon'=\left\{
\begin{array}{rcl}
\frac{\epsilon}{r^k},&\text{if}&r>0,\\
1,&\text{if}&r=0.
\end{array}
\right.
\]
Applying (c) to the numbers $r$ and $\epsilon'$, we obtain $p_0\in\mathbb{Q}_+[X]$ such that $p_0(r)\leq\epsilon'$ and $p_0(\overline{u})+\overline{x}\succcurlyeq\overline{y}$. Set $p=1+p_0X^k$, then \[p(r)=1+p_0(r)r^k\leq1+\epsilon'r^k\leq1+\epsilon\] and
\[p(\overline{u})*\overline{x}=\overline{x}+p_0(\overline{u})*\overline{u}^k*\overline{x}\succcurlyeq \overline{x}+p_0(\overline{u})\succcurlyeq\overline{y}.\]
Hence (d) is deduced.

``(d)$\Rightarrow$(a)": Let $f: S\rightarrow\mathbb{R}_+$  be a monotone semialgebra homomorphism, we need to show that $f(x)\geq f(y)$. By Remark \ref{realvaluedextension}, the map \[f^F: F\rightarrow \mathbb{R}_+,\overline{a}\mapsto f^W(a) \,\;\;(\forall a\in W)\] is a well-defined monotone semialgebra homomorphism. Setting $r=f^F(\overline{u})$ and applying $f^F$ to inequality (\ref{dd}), we obtain
\[p(r)f^F(\overline{x})\geq f^F(\overline{y}),\] where $p(r)\leq1+\epsilon$, $f^F(\overline{x})=f^W(x)=f(x)$ and $f^F(\overline{y})=f(y)$. Thus, we have $(1+\epsilon)f(x)\geq f(y)$ for any real number $\epsilon>0$. Taking $\epsilon\rightarrow0$, we obtain $f(x)\geq f(y)$.

``(a)$\Rightarrow$(b)": This part of the proof is almost the same as  the proof of Theorem 3.1 in \cite{fritz2021generalization}. This is mainly due to the well-defined semialgebra of the fractions of a noncommutative semialgebra given in previous  sections. We only sketch  the proof of this part below.

Assume that $F$ is \emph{order-cancellative}, i.e.  $g+w\succcurlyeq h+w$ implies $g\succcurlyeq h$ ($\forall g,h,w\in F$). Let $V$ be the linear space defined in Section \ref{sec:V}. Define \[C=\{v\in V\,|\,v\unrhd0_{_V}\}=\{g-h\in V\,|\,g,h\in F,g\succcurlyeq h\},\]
then $C$ is a cone in  $V$ by (\ref{Ccone}). For any real number $\epsilon>0$, set
\[\mathcal{N}_\epsilon=\{a\in F\,|\,\exists m\in\mathbb{N}\text{ \emph{s.t.} }a\preccurlyeq\sum^m_{j=0}\epsilon^{j+1}\cdot\overline{u}^j\},\]
and define $\mathcal{N}_\epsilon-\mathcal{N}_\epsilon=\{a-b\in V\,|\,a,b\in\mathcal{N}_\epsilon\}$. Then, as in \cite{fritz2021generalization}, using the set $\mathcal{N}_\epsilon-\mathcal{N}_\epsilon$ as $\epsilon$ varies a basis of neighborhoods of $0_{_V}$ \hspace{-1mm}makes $V$ a locally convex topological vector space. Denoting by $\overline{C}$ the closure of $C$ in that topology,  we have
\begin{equation}\label{inClosure}
\overline{x}-\overline{y}\in\overline{C}\;\,\Leftrightarrow \text{ the condition (b) holds,}
\end{equation}
where the proof of the direction ``$\Rightarrow$" relies on the assumption that $\preccurlyeq$ is order-cancellative. Denote by $V^*$ the dual space of $V$, \emph{i.e.}, the topological vector space consisting of all continues linear (real-valued) functions on $V$ and equipped with the weak-* topology. Define
\[C^*=\big\{\ell\in V^*\,|\,\ell(C)\subset[0,\infty)\big\},\]then $C^*$ is a closed convex cone in $V^*$ such that
\begin{equation}\label{KM}
C^*=\overline{\text{conv}}(\text{Er}(C^*)),
\end{equation}
with Er$(\cdot)$ the set of all the extreme rays of a cone and $\overline{\text{conv}}(\cdot)$ the closure of the convex hull of a set.
In the following, we assume that (b) does not hold and deduce the negation of the condition (a).

By observation (\ref{inClosure}), ``(b) does not hold" means $\overline{x}-\overline{y}\notin\overline{C}$. By the geometric Hahn-Banach theorem \cite{rudin1991functional}, there is an $\ell\in C^*$ so that $\ell(\overline{x}-\overline{y})<0$. Combining with the property in (\ref{KM}), one further claims that there is an  $f\in\text{Er}(C^*)\subset C^*$ satisfying $f(\overline{x}-\overline{y})<0$. Then we have  $f\big(\hspace{0.4mm}\overline{1}_S-\overline{0}_S\hspace{-0.4mm}\big)>0$ and the function
\begin{equation}\label{fhat}
\hat{f}: S\rightarrow\mathbb{R}_+,c\mapsto \frac{1}{f\big(\hspace{0.4mm}\overline{1}_S-\overline{0}_S\hspace{-0.4mm}\big)}f(\overline{c}-\overline{0}_S)
\end{equation}
is a monotone semialgebra homomorphism such that $\hat{f}(x)<\hat{f}(y)$, contradicting  to the condition (a).

If $F$ is not order-cancellative, the method used in \cite{fritz2021generalization} works as well in the present case.
\end{proof}

In the following, we verify the observation (\ref{inClosure}), the equation (\ref{KM}), the claim that ``there is an  $f\in\text{Er}(C^*)$ such that $f(\overline{x}-\overline{y})<0$" and the claim that ``the function $\hat{f}$ in (\ref{fhat}) is a monotone homomorphism" in Observations \ref{closure}--\ref{prooffhat}. These claims have already been  obtained by Fritz in \cite{fritz2021generalization}. We rewrite the proof of the observation (\ref{inClosure}) since our proof is not exactly the same as the one given by Fritz, while re-prove the equation (\ref{KM}) and the second last claim because (at least part of) the details of their proofs are omitted in the original one in \cite{fritz2021generalization}. Finally, the proof of the last claim is re-provided since it is a key step of both Fritz's proof and ours.

Note that the proof of the equation (\ref{KM}) below (Observation \ref{KMobs}) is provided by Fritz through private communications.
\begin{observation}\label{closure}
$\overline{x}-\overline{y}\in\overline{C}\;\,\Leftrightarrow$ \text{ the condition \emph{(b)} holds}.
\end{observation}
\begin{proof}

The condition $\overline{x}-\overline{y}\in\overline{C}$ implies that
\[\forall \epsilon>0, \,\exists v\in C\text{\emph{ s.t. }}v\in(\overline{x}-\overline{y})+(\mathcal{N}_\epsilon-\mathcal{N}_\epsilon).\]
This means that there are $c,d\in\mathcal{N}_\epsilon$ such that $(\overline{x}-\overline{y})+(c-d)=v\unrhd0_{_V}$. By the assumption that $F$ is order-cancellative, this implies $\overline{x}+c\succcurlyeq\overline{y}+d\succcurlyeq\overline{y}$. Since $c\in\mathcal{N}_\epsilon$, then  (\ref{binq}) holds.

Now that the inequality (\ref{binq}) holds, we have
\[\big(\overline{x}+\sum_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j\big)-\overline{0}_S\unrhd\overline{y}-\overline{0}_S.\]Adding $\overline{0}_S-\overline{y}$ to the both sides, one obtains
\[\big(\overline{x}+\sum_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j\big)-\overline{y}\unrhd\overline{y}-\overline{y}=0_{_V},\]
which is equivalent to
\begin{equation}\label{indicate}
(\overline{x}-\overline{y})+\big(\sum_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)\in C
\end{equation}
with $\sum_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\in\mathcal{N}_\epsilon-\mathcal{N}_\epsilon$. By  inequality (\ref{indicate}), we know that there is a point in $C$ which is also in the neighborhood $(\overline{x}-\overline{y})+(\mathcal{N}_\epsilon-\mathcal{N}_\epsilon)$ for any $\epsilon>0$. Thus $\overline{x}-\overline{y}\in\overline{C}$.
\end{proof}
\begin{observation}\label{KMobs}
$C^*=\overline{\text{\emph{conv}}}(\text{\emph{Er}}(C^*))$.
\end{observation}
\begin{proof}
Proposition 21 in \cite{choquet1962cones} indicates that any \emph{well-capped} \cite{asimow1969extremal} closed convex cone in a locally convex topological vector space is the closed convex hull of its extreme rays. To prove (\ref{KM}), it is sufficient to show that $C^*$ is well-capped (since $V^*$ is already locally convex). In \cite{fritz2021generalization}, Fritz defined the sets
\[C_\epsilon^*=\big\{\ell\in C^*\,|\,\ell(\mathcal{N}_\epsilon-\overline{0}_S\hspace{-0.6mm})\text{ is bounded}\big\}\]
and
\[
D_\epsilon=\big\{\ell\in C_\epsilon^*\,|\,\ell(\mathcal{N}_\epsilon-\overline{0}_S\hspace{-0.6mm})\subset[0,1]\big\},
\]
which satisfy the equation \begin{equation}\label{wellcapped}
C^*=\bigcup_{\epsilon>0}C_\epsilon^*=\bigcup_{\substack{\epsilon>0\\n\in\mathbb{N}}}n D_\epsilon
\end{equation}
and claimed that $D_\epsilon$ was a \emph{cap} \cite{asimow1969extremal} of $C^*$ (hence $C^*$ is well-capped by equation (\ref{wellcapped})). However, the proof of the condition ``the set $C^*\backslash D_\epsilon$ is convex", which is required in the definition of a cap, is omitted in \cite{fritz2021generalization}. The following proof of the condition is provided by Fritz through private communications:

Note that $\sum\limits_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j\in\mathcal{N}_\epsilon$ for any $m\in\mathbb{N}$, we have
\[
D_\epsilon=\Big\{\ell\in C^*_\epsilon\,\big|\,\forall m\in\mathbb{N}, \ell\big(\sum\limits_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)\leq1\Big\}.
\]
Then
\[C^*_\epsilon\backslash D_\epsilon=\Big\{\ell\in C^*_\epsilon\,\big|\,\exists m\in\mathbb{N}, \ell\big(\sum\limits_{j=0}^m\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)>1\Big\}\]
is convex. Set $f_i\in C^*_\epsilon\backslash D_\epsilon$ with $m_i$ such that $f_i\big(\sum\limits_{j=0}^{m_i}\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)>1$, $i=1,2$. For any $\mu_1,\mu_2=1-\mu_1\in(0,1)$ and $\hat{m}=\max\{m_1,m_2\}$, we have
\begin{equation}\label{g1convex}
\left.
\begin{array}{rcl}
&&(\mu_1f_1+\mu_2f_2)\big(\sum\limits_{j=0}^{\hat{m}}\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)\\
&=&\mu_1f_1\big(\sum\limits_{j=0}^{\hat{m}}\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)+\mu_2f_2\big(\sum\limits_{j=0}^{\hat{m}}\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)\\
&\geq&\mu_1f_1\big(\sum\limits_{j=0}^{m_1}\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)+\mu_2f_2\big(\sum\limits_{j=0}^{m_2}\epsilon^{j+1}\cdot\overline{u}^j-\overline{0}_S\big)\\
&>&1.
\end{array}
\right.
\end{equation}
Thus $\mu_1f_1+\mu_2f_2\in C_\epsilon^*\backslash D_\epsilon$.

The set $C^*\backslash D_\epsilon$ is also convex.  Set $f_1,f_2\in C^*\backslash D_\epsilon$ and $\mu_1,\mu_2=1-\mu_1\in(0,1)$. If one of $f_1$ and $f_2$ is not in $C^*_\epsilon$ (\emph{i.e.}, is not bounded over the set $\mathcal{N}_\epsilon-\overline{0}_S$), then neither is the functional $\mu_1f_1+\mu_2f_2$. Thus, $\mu_1f_1+\mu_2f_2\in C^*\backslash C^*_\epsilon\subset C^*\backslash D_\epsilon$. On the contrary, if both $f_1$ and $f_2$ are in $C^*_\epsilon$, then they are in $C^*_\epsilon\backslash D_\epsilon$. Now (\ref{g1convex}) implies $\mu_1f_1+\mu_2f_2\in C^*_\epsilon\backslash D_\epsilon\subset C^*\backslash D_\epsilon$.
\end{proof}
\begin{observation}\label{extray}
$\exists f\in\text{\emph{Er}}(C^*)\text{ \emph{s.t.} }f(\overline{x}-\overline{y})<0$.
\end{observation}
\begin{proof}
This part was also omitted in \cite{fritz2021generalization}, hence we give a proof here. Since there is an $\ell\in C^*$ so that $\ell(\overline{x}-\overline{y})<0$, the intersection $\mathcal{H}\cap C^*$ is not empty, where
\[\mathcal{H}=\{\phi\in V^*\,|\,\phi(\overline{x}-\overline{y})<0\}\]
is an open half-space in $V^*$. It is sufficient to show that there is $f\in\text{Er}(C^*)$ so that $f\in\mathcal{H}$. Assume on the contrary that none of the extreme rays of $C^*$ is in $\mathcal{H}$. By the equation (\ref{KM}), for any $\phi\in C^*$ and any neighborhood $O\ni\phi$, there is $\phi'\in O\cap\text{conv}(\text{Er}(C^*))$. Thus $\phi'=\sum\limits_{i=1}^s\mu_i\phi_i$ for some $s\in\mathbb{N}$, some $\phi_i\in\text{Er}(C^*)$ and some real numbers $\mu_i\in[0,1]$ with $\sum\limits_{i=1}^s\mu_i=1$. Since $\phi_i\notin\mathcal{H}$ by the assumption, $\phi_i(\overline{x}-\overline{y})\geq0$ for any $1\leq i\leq s$. Hence $\phi'(\overline{x}-\overline{y})\geq0$, i.e., $\phi'\notin\mathcal{H}$. This means that $O\cap(V^*\backslash\mathcal{H})\neq\emptyset$ for any neighborhood $O$ of $\phi$. Since $V^*\backslash\mathcal{H}$ is clearly closed, $\phi\in V^*\backslash\mathcal{H}$. Noting that $\phi$ is any element in $C^*$, we conclude that $C^*\cap\mathcal{H}=\emptyset$, which is a contradiction.
\end{proof}
\begin{observation}\label{prooffhat}
Let $f\in\text{\emph{Er}}(C^*)$ be any extreme ray of $C^*$, then $f(\overline{1}_S-\overline{0}_S)>0$ and $\hat{f}$ in (\ref{fhat}) is a monotone semialgebra homomorphism.
\end{observation}
\begin{proof}
We may assume, without loos of generality, that for any nonzero $a\in F$, $\overline{1}_S+a\neq\overline{0}_S$. Otherwise, there would be a nonzero $a\in F$ so that $\overline{0}_S=\overline{1}_S+a\succcurlyeq\overline{1}_S$. Note that $\overline{1}_S\succcurlyeq\overline{0}_S$, one concludes that the preorder $\preccurlyeq$ is trivial, as is shown in (\ref{tri}). Now the conditions (a)--(d) are all trivially true, so is the conclusion of the theorem.

Set $f_F(w)=f(w-\overline{0}_S)$ for any $w\in F$, then we have
\[f_F\big((\overline{1}_S+a)^{-1}*w\big)+f_F\big(a*(\overline{1}_S+a)^{-1}*w\big)=f_F(w)\]
by the linearity of $f$. Let
\[
\left\{
\begin{array}{rcl}
g(w)&=&f_F\big((\overline{1}_S+a)^{-1}*w\big)\\
h(w)&=&f_F\big(a*(\overline{1}_S+a)^{-1}*w\big)
\end{array}
\right.
\]
then $g,h$ are both linear and monotone on $F$ since $f_F$ is. Set $g^V(b-d)=g(b)-g(d)$ and $h^V(b-d)=h(b)-h(d)$ for any $b,d\in F$ (or equivalently, for any $b-d\in V$), then both $g^V$ and $h^V$ are well-defined on $V$. Moreover, a routine check indicates that they are monotone and linear maps from $V$ to $\mathbb{R}$ and \begin{equation}\label{decom}g^V+h^V=f.\end{equation}

For any nonzero $t\in F$, the operator
\begin{equation}\label{leftmul}
t_*: b-d\mapsto t*b-t*d,\,V\rightarrow V
\end{equation}
is well-defined and continuous (Lemma \ref{t*cont}). Noting that $g^V=f\circ\big((\overline{1}_S+a)^{-1}\big)_*$ and $h^V=f\circ\big(a*(\overline{1}_S+a)^{-1}\big)_*$, one observes that both $g^V$ and $h^V$ are continuous since $f$ is. To conclude, we have $g^V,h^V\in C^*$.

By the equation (\ref{decom}) and the fact that $f$ is an extreme ray, we conclude that there are nonnegative real numbers $s_1$ and $s_2$ such that $g^V=s_1f$ and $h^V=s_2f$. As a ray of $C^*$, $f\neq0$. There is some $b_0-d_0\in V$ such that $f(b_0-d_0)\neq0$. Thus \[g^V\big((\overline{1}_S+a)*b_0-(\overline{1}_S+a)*d_0\big)=f(b_0-d_0)\neq0\] and \[f^V\big((\overline{1}_S+a)*a^{-1}*b_0-(\overline{1}_S+a)*a^{-1}*d_0\big)=f(b_0-d_0)\neq0,\] which means $g^V\neq0$ and $h^V\neq0$. Hence $s_1,s_2\in\mathbb{R}_{>0}$. Therefore, we have $h^V=\gamma g^V$ for the positive number $\gamma=s_2/s_1$ and $h^V(w-\overline{0}_S)=\gamma g^V(w-\overline{0}_S)$ for any $w\in F$, \emph{i.e.},
\begin{equation}\label{fidentity}
f_F\big(a*(\overline{1}_S+a)^{-1}*w\big)=\gamma f_F\big((\overline{1}_S+a)^{-1}*w\big).
\end{equation}
Replacing $(\overline{1}_S+a)^{-1}*w$ by $w$ in (\ref{fidentity}) one obtains
\begin{equation}\label{fidentitySIM}
f_F(a*w)=\gamma f_F(w)
\end{equation}
for any $w\in F$ (with $\gamma$ in (\ref{fidentitySIM}) depending on $a$). Taking $w=\overline{1}_S$ in (\ref{fidentitySIM}), we have
\begin{equation}\label{gamma}
f_F(a)=\gamma f_F(\overline{1}_S).
\end{equation}
Now $0\neq f(b_0-d_0)=f_F(b_0)-f_F(d_0)$ indicates that $f_F(b_0)\neq0$ or $f_F(d_0)\neq0$. We may assume $f_F(b_0)\neq0$, then clearly $b_0\neq\overline{0}_S$. Taking $a=b_0$ in (\ref{gamma}), we have $f_F(b_0)=\gamma f_F(\overline{1}_S)$. Since $f_F(b_0)\neq0$, we obtain $f_F(\overline{1}_S)\neq0$. In fact, $f_F(\overline{1}_S)>0$ since $f\in C^*$. From (\ref{gamma}) we see that $\gamma=f_F(a)/f_F(\overline{1}_S)$. Substituting that back to (\ref{fidentitySIM}), we have
\begin{equation}\label{nearhomo}
f_F(a*w)=\frac{f_F(a)}{f_F(\overline{1}_S)}f_F(w)
\end{equation}
for any $w\in F$ and any nonzero $a\in F$. Indeed, when $a=\overline{0}_S$, the identity (\ref{nearhomo}) also holds: both of its two sides equal $0$. Define $\hat{f}(c)=\frac{1}{f_F(\overline{1}_S)}f_F(\overline{c})$ for any $c\in S$, then the identity (\ref{nearhomo}) indicates that $\hat{f}: S\rightarrow\mathbb{R}_+$ preserves the multiplications. The facts that it also preserves the additions and the scalar multiplications, and that it is monotone are straightforward. Thus, $\hat{f}$ is a monotone semialgebra homomorphism.
\end{proof}
The following lemma, which is claimed in \cite{fritz2021generalization}, is needed in the proof of Observation \ref{prooffhat}. We re-prove it since our method is not exactly the same.
\begin{lemma}\label{t*cont}
For any nonzero $t\in F$, the map given in (\ref{leftmul}) is well-defined and continuous \emph{w.r.t.} the topology in $V$.
\end{lemma}
\begin{proof}
Suppose that $b-d=b'-d'$ in $V$ for some $b,d,b',d'$ in $F$. Then there is $e\in F$ such that $b+d'+e=b'+d+e$ in $F$. Hence
\[
t*b+t*d'+t*e=t*b'+t*d+t*e,
\]
which means $t*b-t*d=t*b'-t*d'$ in $V$. Therefore, the map $t_*$ is well-defined.

The map $t_*$ is clearly linear. To prove the continuity, it is sufficient to show that it is continuous at the point $0_V=\overline{0}_S-\overline{0}_S$: Let $k\in\mathbb{N}$ such that $\overline{u}^k\succcurlyeq t$. For any $\epsilon>0$, we set $\delta=(\min\{\epsilon,1\})^{k+1}$. Then, for any $b\in\mathcal{N}_\delta$, $b\preccurlyeq\sum\limits_{j=0}^m\delta^{j+1}\cdot\overline{u}^j$ for some $m\in\mathbb{N}$. On the other hand,
\[
\left.
\begin{array}{rcl}
t*b&\preccurlyeq&\overline{u}^k*(\sum\limits_{j=0}^m\delta^{j+1}\cdot\overline{u}^j)\\
&=&\sum\limits_{j=0}^m\delta^{j+1}\cdot\overline{u}^{j+k}\\
&\preccurlyeq&\sum\limits_{j=0}^m(\min\{\epsilon,1\})^{k+j+1}\cdot\overline{u}^{j+k}\\
&\preccurlyeq&\sum\limits_{j=0}^{m+k}(\min\{\epsilon,1\})^{j+1}\cdot\overline{u}^j,
\end{array}
\right.
\]
which means $t*b\in\mathcal{N}_{\min\{\epsilon,1\}}\subset\mathcal{N}_\epsilon$. Therefore, for any $b,d\in\mathcal{N}_\delta$, we have $t*b-t*d\in\mathcal{N}_\epsilon-\mathcal{N}_\epsilon$. That is, $t_*$ is continuous at $0_V$.
\end{proof}
\begin{remark}
Theorem \ref{main} is a noncommutative counterpart of Theorem 2.12 in \cite{fritz2021generalization}, although in the noncommutative case, an equivalent characterization for the preorder relation $\preccurlyeq$ via the original relation $\leq$ similar to the one in (\ref{compare}) is presently not available.
\end{remark}
\section{Conclusion}
We provide a noncommutative version of the Vergleichsstellensatz proved by Fritz in \cite{fritz2021generalization}. This is the first Vergleichsstellensatz for noncommutative semialgebras in the literature. It characterizes the relaxed preorder induced by monotone homomorphisms to $\mathbb{R}_+$ by ``asymptotic" inequalities in the semialgebra of the fractions. Finding out how it can be applied to other areas such as probability theory and quantum information theory or how it is related to classical noncommutative real algebraic geometry would be interesting topics in the future.

As a byproduct, we provided a particular method to define the semialgebra of the fractions of a noncommutative semialgebra which generalizes the corresponding concept in the commutative case. One can also apply this method to noncommutative semirings by just getting rid of those expressions containing (formal or non-formal) scalar multiplications in the definition.
\section*{Acknowledgement}
We thank Tobias Fritz for useful discussion and suggestions.
\bibliographystyle{elsarticle-num}

redpajama_set_name: RedPajamaArXiv


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Billiard Blitz Snooker Star
In this very well made online flash snooker you can play against dozens of computer opponents, join few tournaments and many more. Become the true champion. Use your mouse to aim and hit the billiard balls.
Flash Ludo
Maybe you have played this board game with your family or friends. Your task is to move all your dices closer to the middle at the corresponding coloured spots. Spin the dice and move your bones. When you stop at your opponents bone you will remove it and turn his bone into yours.
Mushbits
Your task is to help pink and blue bunnies get the same coloured mushrooms. Remember that these bunnies can only jump on the plates of the same colour. After one of them crosses the platform it will change colour. Use your mouse to drag and guide those mushbits.
Cruiser Battleship 2
Remember how you played Battleships using paper and pen with your friends? Now you can play the same game in cool graphics with additional features and options. Destroy enemy before he destroys you. Use mouse to aim and fire.
In this cards game you have to use your Blackjack skills to get as close to 31 point as possible using only one suit. For example, Aces = 11, picture cards = 10. If you collect 3 of one kind you will get 30 points. Click on the deck to pick a card and then select on from yours to switch them. When you're ready click on the Stop the bus.
This is a classic Mahjong game with nice Winter graphics. Your task is to remove all tiles from the screen. To do that you have to find pair of equal tiles and then click on them. You can use only free tiles, it means that there is no other tile on top of it and it has at least 2 free side.
Your mission is to control your army and fight against enemy. Conquer all hexes where's something important located, to improve your army and equipment. Use Mouse to select units and move them around the map. To attack enemy just move your unit to the hexagon which contains an enemy.
Desktop Mini Golf
Your objective is to lead the ball into the hole using as little number of shoots as possible. Use Mouse to set power and direction of your shoot, and click to putt. Use Reset button if you have used to much moves to improve your score.
Big Bucks
If you like monopoly board game you will enjoy this one too. You aim is to travel around the board, buy and sell properties and bankrupt other players. Spend your money wisely or you'll be forced to sell properties to pay your debt. Use mouse to play this game.
Governor of Poker pt. 2
I guess you remember one of the most popular flash games ever Governor of Poker!? Your task is to play poker to earn enough money to purchase each building of the state. This part of the game has more extras, cities and tournaments. Use your poker skills to beat your opponents at the poker table.
Online Multiplayer Pool
Multiplayer Online Poll game. As always in pool games your task is to shoot balls in the holes. After first successful shoot you will be assigned to the Solid colored balls, or the Striped balls. Do not sink down the Black one, you can remove it only when all other your assigned balls are removed. Use Mouse to control the game. Pick your partner and start the game.
Ultimate Chess
I hope you can play chess, because this one has so good animations (~50 animations) on basic Chess moves. In Executions section you can see what you've achieved. Of course if you can't play there are Chess rules available. Use mouse to control the game.
Blackjack Fever
Place your bets. You start the game with two cards. Your task is to pick cards or stand with those you have to make a perfect blackjack or sum less or equal to twenty one. This version of Casino Blackjack game is good because you can see dealer's score all the time.
Big Bomb Blackjack
Blackjack is a straightforward game, once you know the rules. This game is played with one or more standard decks of cards containing 52 cards. The value of each card in the game is straight forward: the royal cards of the Queen, King and Jacks are all valued at 10, and the other cards are at face value, so a 2 is worth 2 points. The Aces, however, are slightly different; they can either be 11 or 1 depending upon how they will best suit the player. The object of the game is for the player to reach the magic number of 21 and achieve blackjack or as close to it as possible without going over the 21 number. The object of the game is to get closer to 21 than the dealer. Online casino blackjack can be played either in a one to one setting with a player and the dealer or between up to 6 players and the dealer. Each player gets dealt two cards face up and the dealer gets one card face up and one face down. The dealer then goes around the table and each player gets the choice of sticking with the cards they have or by taking more cards to try and get close to the number 21. Once each player has made their choice on the cards the dealer then takes their own chances with the cards against those players who have not gone bust - that is taken cards which add up to over 21. If the dealer takes on extra cards which add up to over 21 the players left in the game win. If the dealer sticks with their cards then any player closer to 21 than the dealer wins, but if the dealer is closer to 21 the player looses and the house wins.
Another very nice Blackjack game version with nice graphics and good colors. The table rules are simple as usual - beat dealer by collecting more points than he has but less or equal to 21. Use mouse to control the game. Try to get the best rank and submit your score!
Casino Blackjack
Blackjack is the most popular casino card game in the world, and for a very good reason - blackjack offers many advantages to the player that the majority of casino games don't. Gain 21 point by collecting cards to beat the Dealer and win some money.
Very funny Texas Hold'em Poker
The object of Texas Hold 'Em is to win everyone else's money. The game ends when you run out of money, or when all the other players run out of money. Place bets, make 5 card combination as usual and become best player of Texas Poker.
Power Pinball
In this Pinball there are such things as ports, breakable blocks, ball storms and other interesting things. Use left and right arrow keys to control the flippers. Press and hold Down arrow key to launch ball and Up arrow key to tilt board.
Good Ol' Poker
Get the best score in a poker round. A round consists of 15 classic sets of dealing, betting and drawing. Choose your character by looking at his unique attributes and special ability. Use it in the game when the Confidence bar is loaded to maximum. Game is based on all Texas Hold 'Em poker rules, anyway go through starting tutorial.
Realtime Chess
This Chess game is an online version. No more waiting while your opponent makes his turn. Just move your pieces in real time as quick as possible to beat his king and win the game. You can move any piece at any time. Play the game locally against your friend on one computer or join lobby and play with somebody around the world. Use Mouse to make the moves.
Texas Hold 'em Poker: Heads Up
In this Great Texas Poker game You play One on One against Artificial Intelligence, also known as Computer! Choose difficulty level and become the greatest Poker Player. Rules stays the same, make card combination having Your two cards and 5 on common desk. Place Your bets, check, call, raise or fold.
Defeat your opponents in battle to conquer every spot on the board. Build and upgrade your deck, form a strategy, protect your castle, and crush all who stand in your way! You can place your unit using Space button if clicking is hard. Sometimes sending back your units is good.
Our gladiators step out from the arena and set out in a wider and more dangerous world. Up to four gladiators compete in a combination of mini games, adventures and battles across mysteries of danger and adventure. Use mouse to control the game. Check out in game Help Guide before You start the game.
Power Pool Frenzy
Here's 3rd game of Power Pool. Power Pool Frenzy is ten levels of total chaos, and potentially big, huge and massive scores. Use mouse to click the white ball, hold left mouse button down and drag around to set the power and direction of your shoot, release to hit the ball.
Puzzle Freak 2
Your challenge in this game is to go around the playing board as quickly as possible by completing various puzzles. The faster you complete a puzzle, the more IQ points you get. IQ points are also affected by the relative difficulty of a puzzle. There are 4 types of puzzle squares you can land on, ranging from easy to very hard. There are also chance squares on the board, which may help you on your journey to the finish line. Use mouse to control the game.
Block Drop
Your objective in this game is to control the gem to remove all of the blocks by jumping off of them leaving only the checkered block behind. The increasingly complex levels are all randomly generated, so no two games will ever be the same. Use the arrow keys to jump one space to the next block. Press the Shift + arrow key will give You two space jump. If you don't know what to do you can use the menu to show the solution.
Power Pool 2
2 new table shapes to play on. There's also a new power-up ball that makes the cueball so big it can't get sunk so you can smack it round with impunity. Each table has 20 different unique levels to play, and on each table it's possible to rack up gigantic combos and points tallies. Use mouse to click the white ball, hold click down and drag around to set the power and direction of your shoot.
Obama Chess
Like chess? Love Obama? With Barack's investiture as the new president of USA, here's a special edition of Flash Chess game. Meet the prominent political figures from US Administration and Congress and have fun!
Flash Chess 3
I'm a fan of Chess games. Only PC chess games. So – here's one incredibly high quality Chess flash game. Great graphics, smart bots, beautiful gameplay, hints and more. For game instructions use your mouse to control Your positions. But to play this not so easy desk game and win computer You must study Chess rules, sorry. Enjoy the game!
Desk games represent one of the most widely played genre of computer games. They do not take much time, as are quite easy to understand. Desk games are founded on board games which were played during long centuries of human history and, therefore, are familiar to nearly everyone.
Generally desk games are divided into two large categories. The first one includes logical games: chess, domino, monopoly, card games, etc. These games require logical thinking, attention and concentration on the game proceeding. System of scoring is just the same as in the real ones. Although often some specific in-game currency is present.
The second category involves skill-based games. These include any kinds of tile-matching games, sport games, etc. Therefore, speed and quick reaction is needed more than long, strategic thinking, so the gamer has to make a decision very quickly.
Often desk games have much in resemblance with gambling games. Thus, the player has to be cute and cunning, must know the rules well and know when it is possible to fraud the opponent. Intuition can also be very helpful.
Desk games are of different levels of complexity. There enough games both for beginners and advanced players. Most desk games provide some help for beginners and contain different options which hint at a possible solution. But it also frequently happens that the gamer loses points if using those hints.
Desk games are not simulators, but they create a maximally truthful game environment. When the player has no one to play with, he can always turn on the computer and start playing desk games which are always available and present quite good opportunity to rest and relax. Usually they have time restrictions. The more complicated the level is, the less time the gamer has to succeed.
Naturally, there is an innumerable number of desk games available online and hundreds of new ones are appearing all the time. The most convenient thing about desk games is that they are not distracting for a long time, like role-playing games or strategies do, but can be stopped and started again anytime. They also do not require special knowledge, just a certain abilities for logical thinking and quick reaction. Therefore, these games can also be played by children, facilitating their education. Actually, this genre is updating constantly and players of all ages can always find a lot of interesting and absorbing games.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Disable Download Link in Gridview when no filename is available I'm a newbie, but need some assistance. I have researched the request but only found one response, which looks like it can be done, but I'm not sure how to do it physically in vb and VS2010.
Basically, I have a webgrid with a download link along with a filename. I want the download link to be disabled if there is no filename as there is nothing to download (obviosuly).
This resolution I found was this - Here - However, don't really follow it, specifically because im doing it in vb too.
Any assistance on this would be great.
Thanks
My code for the gridview.
<asp:GridView ID="SearchErrorGridView" runat="server" AutoGenerateColumns="False" AllowPaging="True"
        AllowSorting="True" DataKeyNames="FileName" DataSourceID="SearchHelpDeskErrorMessages"
        PageSize="10" CssClass="mGrid" PagerStyle-CssClass="pgr" AlternatingRowStyle-CssClass="alt"
        EmptyDataText="No Applicable Data" SelectedRowStyle-BackColor="#99CCFF">
        <AlternatingRowStyle CssClass="alt"></AlternatingRowStyle>
        <Columns>
            <asp:BoundField DataField="RequestID" HeaderText="RequestID" ReadOnly="True" SortExpression="RequestID" />
            <asp:BoundField DataField="Name" HeaderText="Client Name" SortExpression="Name" />
            <asp:BoundField DataField="EmpName" HeaderText="Employee Name" SortExpression="EmpName" />
            <asp:BoundField DataField="Module" HeaderText="Module" SortExpression="Module" />
            <asp:BoundField DataField="Category" HeaderText="Category" SortExpression="Category" />
            <asp:BoundField DataField="ErrorMessage" HeaderText="Error Message" SortExpression="Error Message" />
            <asp:BoundField DataField="ShortDescription" HeaderText="ShortDescription" SortExpression="ShortDescription" />
            <asp:BoundField DataField="Resolution" HeaderText="Resolution" SortExpression="Resolution" />
            <asp:BoundField DataField="DateOfCall" HeaderText="Date Of Call" SortExpression="DateOfCall" />
              <asp:BoundField DataField="FileName" HeaderText="File Name" />
              <asp:TemplateField HeaderText="File Download">
                                      <ItemTemplate>
                            <asp:LinkButton ID="lnkDownload" runat="server" Text="Download" OnClick="lnkDownload_Click"></asp:LinkButton>
                        </ItemTemplate>
                    </asp:TemplateField>
            <asp:ButtonField CommandName="Select" Text="Select" />
        </Columns>
        <EmptyDataRowStyle CssClass="mGrid" />
        <PagerStyle CssClass="pgr"></PagerStyle>
    </asp:GridView>

redpajama_set_name: RedPajamaStackExchange


text:
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California's landmark bill on gig economy called 'watershed moment' for vulnerable workers
By Sara MojtehedzadehWork and Wealth Reporter
Wed., Sept. 11, 2019timer4 min. read
For those wondering why it's so hard to find a middle-class job or agonizing over rising income inequality, pay attention to a new California bill that tackles those issues — by taking aim at the gig economy in a way that could drastically change the face of precarious work.
The legislation means companies like Uber will no longer be able to exclude workers in California from basic rights like minimum wage by calling them independent contractors, a category that has no protection under workplace laws. And in Ontario and beyond, labour advocates are watching closely.
"In terms of the changing labour market and workplace today this is probably the biggest issue for precarious, non-standard workers," said Mary Gellatly of the Toronto-based Parkdale Community Legal Clinic, "because employers are increasingly looking at how to shift their liability or responsibilities for work being done."
California's Assembly Bill 5, approved Tuesday, is the most significant blow yet to employee misclassification, which worker advocates have long described as the bedrock of the gig economy. The practice sees companies categorizing workers as self-employed to evade workplace protection obligations, workers' compensation premiums and payroll taxes. While it's popularly associated with businesses like Uber, which steadfastly maintains its drivers are not employees, the practice is also common in sectors like cleaning, construction, nail salons and courier services.
Now, at least a million Californian workers in these fields could become employees with the right to minimum wage, unemployment and disability insurance, paid sick days and family leave.
Lawmakers wrote the impetus for the new protections into the bill's text.
"The misclassification of workers as independent contractors," the legislation reads, "has been a significant factor in the erosion of the middle class and the rise in income inequality."
David Doorey, Associate Professor of Work Law at York University, calls the bill "potentially a watershed moment."
"For 'gig' companies, it will require a fundamental rethinking of how their businesses operate. The bill rejects entirely their operating premise, which is that they are just technology companies that link service providers to customers. The bill places gig companies in the position of an employer and all that entails in terms of legal compliance and human resources practices," he said.
Ride-hailing companies like Uber and Lyft and DoorDash, a food delivery firm, have opposed the measures arguing that they would compromise worker flexibility and inflate operating expenses.
In 2017, an extensive report by two independent experts to the Ontario ministry of labour described misclassification as having a "significant adverse impact" on workers here too, and recommended that the problem become "a priority enforcement issue" for government.
An estimated 12 per cent of Ontario's 5.25 million workers are independent contractors with no paid employees of their own.
The report said a portion of those workers "are misclassified as they are actually employees," based on the ministry of labour's experience and "significant anecdotal evidence."
That's because misclassified workers, especially those in vulnerable or low-wage jobs, often don't know that their job doesn't meet the test of an independent contractor. True independent contractors are defined under the law as people who are in business for themselves, stand to make a profit or loss as a result of their work, control the terms and conditions of their work, and have the power to subcontract out some of the work to others.
Employee misclassification is illegal in Ontario, but right now, it's up to workers to prove their case by filing complaints to the ministry of labour. California's bill is radical because it flips that obligation and assumes all workers are employees, unless businesses can prove otherwise.
In response to the 2017 report, Ontario's previous Liberal government passed legislation that required employers to prove they hadn't misclassified a worker when complaints were filed — a significantly paler version of California's new legislation. Premier Doug Ford's Progressive Conservative government has since reversed the measure.
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Given that, Doorey said he doesn't anticipate California's legislation will create "any immediate spillover effects" closer to home. But the bill is still a significant bellwether, he said.
"We are witnessing early tremors in a coming seismic shift in law that will eventually refocus social protections on economic vulnerability and precarity," he said.
There are other testing grounds for the gig economy too, although none so sweeping as California's legislation. In Toronto, for example, Foodora couriers are challenging their employment status through a union drive. The Ontario Labour Relations Board is currently deciding whether the workers are independent contractors with no right to unionize, as Foodora contends, or not.
With a federal election underway, Gellatly hopes political leaders will commit to replicating California's legislation federally. Although most workers in Canada are governed by provincial labour laws, some sectors like trucking — where misclassification is common — are regulated by Ottawa.
"Work is work," she said. "And people should have basic protections."
Sara Mojtehedzadeh is a Toronto-based reporter covering work and wealth. Follow her on Twitter: @saramojtehedz


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