From 15deff4df111d86c84d808f1c9cc4e30013287d0 Mon Sep 17 00:00:00 2001
From: Simon-Kor <52245124+Simon-Kor@users.noreply.github.com>
Date: Thu, 11 Apr 2024 13:37:03 +0200
Subject: 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
---
 library/algebra/monoid.tex | 19 +++++++++++++++++++
 1 file changed, 19 insertions(+)
 create mode 100644 library/algebra/monoid.tex

(limited to 'library/algebra/monoid.tex')

diff --git a/library/algebra/monoid.tex b/library/algebra/monoid.tex
new file mode 100644
index 0000000..06fcb50
--- /dev/null
+++ b/library/algebra/monoid.tex
@@ -0,0 +1,19 @@
+\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}
+        \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}
+
+\begin{corollary}\label{monoid_implies_semigroup}
+    Let $A$ be a monoid. Then $A$ is a semigroup.
+\end{corollary}
\ No newline at end of file
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