1 files changed, 16 insertions, 27 deletions
diff --git a/library/algebra/semigroup.tex b/library/algebra/semigroup.tex
index e090a56..be94945 100644
--- a/library/algebra/semigroup.tex
+++ b/library/algebra/semigroup.tex
@@ -5,8 +5,7 @@
 \begin{struct}\label{semigroup}
     A semigroup $A$ is a magma such that
     \begin{enumerate}
-        %\item for all $a, b, c\in \carrier[A]$ we have $\mul[A](a,\mul[A](b,c)) = \mul[A](\mul[A](a,b),c)$.
-        \item\label{semigroup_assoc} for all $a, b, c$ we have $\mul[A](a,\mul[A](b,c)) = \mul[A](\mul[A](a,b),c)$.
+        \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}
 
@@ -18,8 +17,7 @@
 \begin{struct}\label{regularsemigroup}
     A regular semigroup $A$ is a semigroup such that
     \begin{enumerate}
-        %\item for all $a\in \carrier[A]$ there exists $b\in\carrier[A]$ such that $\mul[A](a, \mul[A](b, a)) = a$.
-        \item\label{regularsemigroup_regular} for all $a$ there exists $b\in\carrier[A]$ such that $\mul[A](a, \mul[A](b, a)) = a$.
+        \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}
 
@@ -38,12 +36,18 @@
     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.
@@ -56,27 +60,12 @@
     Then $e = f$.
 \end{proposition}
 \begin{proof}
-    Take $x, y\in\carrier[A]$ such that $e = x\cdot f$ and $f = y\cdot e$ by \cref{idempotents}.
-    %
-    \begin{align*}
-        e
-            &= x \cdot f
-                \explanation{by assumption}\\
-            &= x\cdot (f\cdot f)
-                \explanation{by \cref{idempotents}}\\
-            &= (x\cdot f)\cdot f
-                \explanation{by \cref{semigroup_assoc,inversesemigroup_is_semigroup}}\\
-            &= e\cdot f
-            \explanation{by assumption}\\
-            &= f\cdot e
-                \explanation{by \hyperref[inversesemigroup_comm]{commutativity of idempotent elements}}\\
-            &= (y\cdot e)\cdot e
-                \explanation{by assumption}\\
-            &= y\cdot (e\cdot e)
-                \explanation{by \cref{semigroup_assoc,inversesemigroup_is_semigroup}}\\
-            &= y \cdot e
-                \explanation{by \cref{idempotents}}\\
-            &= f
-                \explanation{by assumption}
-    \end{align*}
+    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}