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
---
 Error.txt                  | 16 ---------
 latex/stdlib.tex           |  3 +-
 library/algebra/group.tex  | 83 +++++++++++++++++++++++++++++++++++++++++++++-
 library/algebra/monoid.tex | 19 +++++++++++
 library/everything.tex     |  3 +-
 library/test.tex           | 54 ------------------------------
 6 files changed, 105 insertions(+), 73 deletions(-)
 delete mode 100644 Error.txt
 create mode 100644 library/algebra/monoid.tex
 delete mode 100644 library/test.tex

diff --git a/Error.txt b/Error.txt
deleted file mode 100644
index 7aa836a..0000000
--- a/Error.txt
+++ /dev/null
@@ -1,16 +0,0 @@
-stack exec zf -- library/test.tex 
-Verification failed: prover found countermodel
-fof(leftinverse_eq_rightinverse,conjecture,fa=fc).
-fof(monoid_right,axiom,![XA]:(monoid(XA)=>![Xa]:(elem(Xa,s__carrier(XA))=>apply(s__mul(XA),pair(Xa,s__neutral(XA)))=Xa))).
-fof(leftinverse_eq_rightinverse1,axiom,apply(s__mul(fG),pair(fa,s__neutral(fG)))=fc).
-fof(leftinverse_eq_rightinverse2,axiom,apply(s__mul(fG),pair(fa,s__neutral(fG)))=apply(s__mul(fG),pair(fc,s__neutral(fG)))).
-fof(leftinverse_eq_rightinverse3,axiom,fc=apply(s__mul(fG),pair(s__neutral(fG),fc))).
-fof(leftinverse_eq_rightinverse4,axiom,fc=apply(s__mul(fG),pair(fc,s__neutral(fG)))).
-fof(leftinverse_eq_rightinverse5,axiom,apply(s__mul(fG),pair(fa,s__neutral(fG)))=apply(s__mul(fG),pair(s__neutral(fG),fc))).
-fof(leftinverse_eq_rightinverse6,axiom,apply(s__mul(fG),pair(apply(s__mul(fG),pair(fa,fb)),fc))=apply(s__mul(fG),pair(fa,apply(s__mul(fG),pair(fb,fc))))).
-fof(leftinverse_eq_rightinverse7,axiom,apply(s__mul(fG),pair(apply(s__mul(fG),pair(fa,fb)),fc))=apply(s__mul(fG),pair(s__neutral(fG),fc))).
-fof(leftinverse_eq_rightinverse8,axiom,apply(s__mul(fG),pair(fa,fb))=s__neutral(fG)).
-fof(leftinverse_eq_rightinverse9,axiom,apply(s__mul(fG),pair(fb,fc))=s__neutral(fG)&elem(fc,s__carrier(fG))).
-fof(leftinverse_eq_rightinverse10,axiom,apply(s__mul(fG),pair(fa,fb))=s__neutral(fG)&elem(fb,s__carrier(fG))).
-fof(leftinverse_eq_rightinverse11,axiom,elem(fa,s__carrier(fG))).
-fof(leftinverse_eq_rightinverse12,axiom,group(fG)).
\ No newline at end of file
diff --git a/latex/stdlib.tex b/latex/stdlib.tex
index 4921363..e545395 100644
--- a/latex/stdlib.tex
+++ b/latex/stdlib.tex
@@ -36,6 +36,8 @@
     \input{../library/cardinal.tex}
     \input{../library/algebra/magma.tex}
     \input{../library/algebra/semigroup.tex}
+    \input{../library/algebra/monoid.tex}
+    \input{../library/algebra/group.tex}
     %\input{../library/algebra/quasigroup.tex}
     %\input{../library/algebra/loop.tex}
     \input{../library/order/order.tex}
@@ -43,5 +45,4 @@
     \input{../library/topology/topological-space.tex}
     \input{../library/topology/basis.tex}
     \input{../library/topology/disconnection.tex}
-    \input{../library/test.tex}
 \end{document}
diff --git a/library/algebra/group.tex b/library/algebra/group.tex
index 48934bd..a79bd2f 100644
--- a/library/algebra/group.tex
+++ b/library/algebra/group.tex
@@ -1 +1,82 @@
-\section{Groups}
+\import{algebra/monoid.tex}
+\section{Group}
+
+\begin{struct}\label{group}
+    A group $G$ is a monoid such that
+    \begin{enumerate}
+        \item\label{group_inverse} for all $g \in \carrier[G]$ there exist $h \in \carrier[G]$ such that $\mul[G](g, h) =\neutral[G]$.
+    \end{enumerate} 
+\end{struct}   
+
+\begin{corollary}\label{group_implies_monoid}
+    Let $G$ be a group. Then $G$ is a monoid.
+\end{corollary}
+
+\begin{abbreviation}\label{cfourdot}
+    $g \cdot h = \mul(g,h)$.
+\end{abbreviation}
+
+\begin{lemma}\label{neutral_is_idempotent}
+    Let $G$ be a group. $\neutral[G]$ is a idempotent element of $G$.
+\end{lemma}
+
+\begin{lemma}\label{group_divison_right}
+    Let $G$ be a group. Let $a,b,c \in G$.
+    Then $a \cdot c = b \cdot c$ iff $a = b$.
+\end{lemma}
+\begin{proof}
+    Take $a,b,c \in G$ such that $a \cdot c = b \cdot c$.
+    There exist $c' \in G$ such that $c \cdot c' = \neutral[G]$.
+    Therefore $a \cdot c = b \cdot c$ iff $(a \cdot c) \cdot c' = (b \cdot c) \cdot c'$.
+    \begin{align*}
+        (a \cdot c) \cdot c'
+            &= a \cdot (c \cdot c')
+                \explanation{by \cref{semigroup_assoc,group_implies_monoid,monoid_implies_semigroup}}\\
+            &= a \cdot \neutral[G]
+                \explanation{by \cref{group_inverse}}\\
+            &= a
+                \explanation{by \cref{group_implies_monoid,monoid_right}}
+    \end{align*}
+    \begin{align*}
+        (b \cdot c) \cdot c'
+            &= b \cdot (c \cdot c')
+                \explanation{by \cref{semigroup_assoc,group_implies_monoid,monoid_implies_semigroup}}\\
+            &= b \cdot \neutral[G]
+                \explanation{by \cref{group_inverse}}\\
+            &= b
+                \explanation{by \cref{group_implies_monoid,monoid_right}}
+    \end{align*}
+    $(a \cdot c) \cdot c' = (b \cdot c) \cdot c'$ iff $a \cdot c = b \cdot c$ by assumption. 
+    $a = b$ iff $a \cdot c = b \cdot c$ by assumption.
+\end{proof}
+
+
+\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,group_divison_right}.
+    $a = c$ by \cref{monoid_right,group_divison_right,neutral_is_idempotent}.
+    $b \cdot a = b \cdot c$.
+    $b \cdot a = \neutral[G]$.
+\end{proof}
+
+\begin{definition}\label{group_abelian}
+    $G$ is an abelian group iff $G$ is a group and for all $g,h \in G$ $\mul[G](g,h) = \mul[G](h,g)$. 
+\end{definition}
+
+
+\begin{definition}\label{group_automorphism}
+    Let $f$ be a function. $f$ is a group-automorphism iff $G$ is a group and $\dom{f}=G$ and $\ran{f}=G$. 
+\end{definition}
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
diff --git a/library/everything.tex b/library/everything.tex
index 599bc27..a5166af 100644
--- a/library/everything.tex
+++ b/library/everything.tex
@@ -20,13 +20,14 @@
 \import{cardinal.tex}
 \import{algebra/magma.tex}
 \import{algebra/semigroup.tex}
+\import{algebra/monoid.tex}
+\import{algebra/group.tex}
 \import{order/order.tex}
 %\import{order/semilattice.tex}
 \import{topology/topological-space.tex}
 \import{topology/basis.tex}
 \import{topology/disconnection.tex}
 \import{topology/separation.tex}
-\import{test.tex}
 
 \begin{proposition}\label{trivial}
     $x = x$.
diff --git a/library/test.tex b/library/test.tex
deleted file mode 100644
index d30bbba..0000000
--- a/library/test.tex
+++ /dev/null
@@ -1,54 +0,0 @@
-\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
-- 
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