Formalisation of groups and monoids

The test.tex file was deleted and all formalisations of groups and monoids was moved to the fitting document of the library. Some proof steps of the new formalisation were optimized for proof time

1 files changed, 0 insertions, 54 deletions

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-\import{algebra/semigroup.tex}
-\section{monoid}
-
-\begin{struct}\label{monoid}
-    A monoid $A$ is a semigroup equipped with
-    \begin{enumerate}
-        \item $\neutral$
-    \end{enumerate}
-    such that
-    \begin{enumerate} %muss hier ein enumerate hin
-        \item\label{monoid_type} $\neutral[A]\in \carrier[A]$.
-        \item\label{monoid_right} for all $a\in \carrier[A]$ we have $\mul[A](a,\neutral[A]) = a$.
-        \item\label{monoid_left} for all $a\in \carrier[A]$ we have $\mul[A](\neutral[A], a) = a$.
-    \end{enumerate}
-\end{struct}
-
-
-\section{Group}
-
-\begin{struct}\label{group}
-    A group $A$ is a monoid such that
-    \begin{enumerate}
-        \item\label{group_inverse} for all $a \in \carrier[A]$ there exist $b \in \carrier[A]$ such that $\mul[A](a, b) =\neutral[A]$.
-    \end{enumerate} 
-\end{struct}    
-
-\begin{abbreviation}\label{cfourdot}
-    $a\cdot b = \mul(a,b)$.
-\end{abbreviation}
-
-\begin{lemma}\label{neutral_is_idempotent}
-    Let $G$ be a group. $\neutral[G]$ is a idempotent element of $G$.
-\end{lemma}
-
-\begin{proposition}\label{leftinverse_eq_rightinverse}
-    Let $G$ be a group and assume $a \in G$.
-    Then there exist $b\in G$ 
-    such that $a \cdot b = \neutral[G]$ and $b \cdot a = \neutral[G]$.
-\end{proposition}
-\begin{proof}
-    There exist $b \in G$ such that $a \cdot b = \neutral[G]$.
-    There exist $c \in G$ such that $b \cdot c = \neutral[G]$.
-    $a \cdot b = \neutral[G]$.
-    $(a \cdot b) \cdot c = (\neutral[G]) \cdot c$.
-    $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.
-    $a \cdot \neutral[G] = \neutral[G] \cdot c$.
-    $c = c \cdot \neutral[G]$.
-    $c = \neutral[G] \cdot c$.
-    $a \cdot \neutral[G] = c \cdot \neutral[G]$.
-    $a \cdot \neutral[G] = c$ by \cref{monoid_right}.
-    $a = c$ by \cref{monoid_right}.
-    $b \cdot a = b \cdot c$.
-    $b \cdot a = \neutral[G]$.
-\end{proof}
\ No newline at end of file