\section{Structures of riegs}
\subsection{Canonical preorder of a rieg}
\begin{definition}[Canonical preorder of a monoid]
    The \emph{canonical preorder} of a monoid $(M, \ast, e)$ is defined by \[x\leq y \iff \exists z\in M ( x \ast z = y).\]
\end{definition}


\begin{definition}[Canonical preorder of a rig]
    The \emph{canonical preorder} of a rig or rieg is the canonical preorder with respect to the addition.
\end{definition}

\begin{example}
    The canonical preorder of the natural number rieg $\N$ is the usual total order.
\end{example}

This preorder is used in several contexts.
\begin{remark}[Dioids]\label{RemarkDefinitionOfDioids}
    A rig $R$ is called \emph{dioid} if the canonical preorder is antisymmetric (i.e., partial order). This structure is used for algorithm theory \cite{gondran2008graphs}. As mentioned in Remark \ref{RemarkDioidsandRiegs}, until now, the author does not know the direct connection between dioids and riegs.
\end{remark}



\begin{definition}[Minimal element]
    An element $x\in R$ of a rieg $R$ is said to be minimal if it satisfies the following two conditions:
    \begin{enumerate}
        \item $\lnot(x\leq 0)$
        \item For any $0\leq y \leq x$, $y=0$ or $y=x$.
        % \item If $x=y+y'$, then $y=0$ or $y'=0$.
    \end{enumerate}
\end{definition}

The second condition is equivalent to saying that ``
If $x=y+y'$, then $y=0$ or $y'=0$." This condition is some sense of indecomposability.

\begin{question}
    Can we replace the first condition with a more simply looking condition $x \neq 0$? In other words,
    Is $0$ the only minimum element in a rieg?  Is $0$ the only element that has the additive inverse?

    No! by the dual number rieg, see subsubsection \ref{SubsubsectionDualNumbers}.
\end{question}



\subsection{Connected Element}
\begin{definition}
    An element $x\in R$ of a rieg $R$ is said to be \emph{connected} if it satisfies the following two conditions:
    \begin{enumerate}
        % \item $(-)^{x}\colon R \to R$ is a rig homomorphism.
        \item $0^x= 0$ 
        \item $(a+b)^x=a^x + b^x$ .
    \end{enumerate}
\end{definition}

In other words, $x$ is connected if $(-)^{x}\colon R \to R$ is a rig homomorphism.

\begin{proposition}[Internally connected objects and connected elements]
    
\end{proposition}


\subsection{Connectedly based rieg}

\begin{proposition}
    For a connectedly based rieg, 
    \[0^x = \]
\end{proposition}
\memo{Heyting, locally connected, strict initial}

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