\import{algebra/magma.tex}

\section{Semigroups}

\begin{struct}\label{semigroup}
    A semigroup $A$ is a magma such that
    \begin{enumerate}
        \item\label{semigroup_assoc} for all $a, b, c\in \carrier[A]$ we have $\mul[A](a,\mul[A](b,c)) = \mul[A](\mul[A](a,b),c)$.
    \end{enumerate}
\end{struct}



\section{Regular semigroups}


\begin{struct}\label{regularsemigroup}
    A regular semigroup $A$ is a semigroup such that
    \begin{enumerate}
        \item\label{regularsemigroup_regular} for all $a\in \carrier[A]$ there exists $b\in\carrier[A]$ such that $\mul[A](a, \mul[A](b, a)) = a$.
    \end{enumerate}
\end{struct}



\section{Inverse semigroups}

\begin{struct}\label{inversesemigroup}
    An inverse semigroup $A$ is a regular semigroup such that
    \begin{enumerate}
        \item\label{inversesemigroup_comm} for all $a,b\in\idempotents{A}$ we have $\mul[A](a, b) = \mul[A](b, a)$.
    \end{enumerate}
\end{struct}

\begin{proposition}\label{inversesemigroup_is_semigroup}
    Suppose $A$ is an inverse semigroup.
    Then $A$ is a semigroup.
\end{proposition}
\begin{proof}
    Follows by \cref{inversesemigroup,regularsemigroup}.
\end{proof}


\begin{proposition}\label{inversesemigroup_is_regularsemigroup}
    Suppose $A$ is an inverse semigroup.
    Then $A$ is a regular semigroup.
\end{proposition}
\begin{proof}
    Follows by \cref{inversesemigroup}.
\end{proof}

\begin{proposition}\label{idempotentelems_eq_iff_orbits_eq}
    Let $A$ be an inverse semigroup.
    Let $e,f\in\idempotents{A}$.
    % TODO use Orbits explicitly?
    Suppose for all $x\in\carrier[A]$ there exists $y\in\carrier[A]$
    such that $x\cdot e = y\cdot f$.
    Suppose for all $x\in\carrier[A]$ there exists $y\in\carrier[A]$
    such that $x\cdot f = y\cdot e$.
    Then $e = f$.
\end{proposition}
\begin{proof}
    Take $x,y$ such that $x,y\in\carrier[A]$ and
        $e = x\cdot f$ and $f = y\cdot e$ by \cref{idempotents}.
    We have $e=e\cdot f$ by
        \cref{idempotents,semigroup_assoc,inversesemigroup_is_semigroup}.
    We have $f=f\cdot e$ by
        \cref{idempotents,semigroup_assoc,inversesemigroup_is_semigroup}.
    We have $e\cdot f=f\cdot e$ by \cref{inversesemigroup_comm}.
    Follows by assumption.
\end{proof}