Add theorem that no anomalous pairs means commutative to blueprint

blueprint/src/content.tex

@@ -185,4 +185,53 @@ \section{Content}
 
 \begin{theorem}
 An ordered semigroup without anomalous pairs is an Archimedean commutative semigroup.
-\end{theorem}
+\end{theorem}
+\begin{proof}
+Let $a$ and $b$ be elements of an ordered semigroup $S$.
+If either $a$ or $b$ is zero, then they commute.
+
+We begin with the case that $a$ and $b$ are positive.
+If $a + b < b + a$, then for all $n\in \mathbb{N}^+$, we have that
+\begin{align}
+(n+1)(a+b) &= a + n(b+a) + b \\
+&> n(b+a) + b \\
+&> n(b+a) \\
+&> n(a+b)
+\end{align}
+And so $a+b$ and $b+a$ form an anomalous pair.
+
+The same for the case that $b + a < a + b$.
+Therefore, we must have that $a+b = b+a$.
+
+The case where $a$ and $b$ are negative follows similarly.
+
+We now look at the case where $a$ is negative and $b$ is positive.
+If the element $0$ exists and $a+b = 0$, then $a + b + a = a$ and so $b+a = 0$.
+Therefore, $a$ and $b$ commute.
+
+If $a + b$ is positive, then
+\begin{align}
+(a+b) + (a+b) &> a + b \\
+(b + a) + b &> b \\
+b + a &> 0
+\end{align}
+
+We already showed that positive elements commute and so
+\[ (b+a) + b = b + (b + a)\]
+
+Now we look at the case where $a+b < b+a$.
+Then we have that
+\begin{align}
+2(a + b) &= a + ((b+a) + b) \\
+&= a + (b + (b + a)) \\
+&= (a + b) + (b + a) \\
+&> (a + b) + (a + b) \\
+&= 2(a + b)
+\end{align}
+Which is a contradiction.
+We get a similar contradiction in the case that $b+a < a+b$.
+Therefore, $a+b = b+a$.
+
+The case where $a+b$ is negative follows similarly.
+The case where $b$ is negative and $a$ is positive is symmetric.
+\end{proof}

blueprint/web/dep_graph_document.html

@@ -547,7 +547,7 @@ <h1 id="doc_title">Dependencies</h1>
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blueprint/web/index.html

@@ -491,6 +491,9 @@ <h1 id="a0000000003">0.2 Content</h1>
 
     <a class="icon proof" href="index.html#a0000000027">#</a>
 
+    <a class="icon proof" href="index.html#a0000000028"><svg  class="icon icon-cogs "><use xlink:href="symbol-defs.svg#icon-cogs"></use></svg>
+</a>
+
 
 
         </div>
@@ -500,6 +503,40 @@ <h1 id="a0000000003">0.2 Content</h1>
 
   </div>
 </div>
+<div class="proof_wrapper" id="a0000000028">
+  <div class="proof_heading">
+    <span class="proof_caption">
+    Proof
+    </span>
+    <span class="expand-proof">▶</span>
+  </div>
+  <div class="proof_content">
+  <p>Let \(a\) and \(b\) be elements of an ordered semigroup \(S\). If either \(a\) or \(b\) is zero, then they commute. </p>
+<p>We begin with the case that \(a\) and \(b\) are positive. If \(a + b {\lt} b + a\), then for all \(n\in \mathbb {N}^+\), we have that </p>
+<div class="displaymath" id="a0000000029">
+  \begin{align}  (n+1)(a+b) & = a + n(b+a) + b \\ & {\gt} n(b+a) + b \\ & {\gt} n(b+a) \\ & {\gt} n(a+b) \end{align}
+</div>
+<p> And so \(a+b\) and \(b+a\) form an anomalous pair. </p>
+<p>The same for the case that \(b + a {\lt} a + b\). Therefore, we must have that \(a+b = b+a\). </p>
+<p>The case where \(a\) and \(b\) are negative follows similarly. </p>
+<p>We now look at the case where \(a\) is negative and \(b\) is positive. If the element \(0\) exists and \(a+b = 0\), then \(a + b + a = a\) and so \(b+a = 0\). Therefore, \(a\) and \(b\) commute. </p>
+<p>If \(a + b\) is positive, then </p>
+<div class="displaymath" id="a0000000030">
+  \begin{align}  (a+b) + (a+b) & {\gt} a + b \\ (b + a) + b & {\gt} b \\ b + a & {\gt} 0 \end{align}
+</div>
+<p>We already showed that positive elements commute and so </p>
+<div class="displaymath" id="a0000000031">
+  \[  (b+a) + b = b + (b + a) \]
+</div>
+<p>Now we look at the case where \(a+b {\lt} b+a\). Then we have that </p>
+<div class="displaymath" id="a0000000032">
+  \begin{align}  2(a + b) & = a + ((b+a) + b) \\ & = a + (b + (b + a)) \\ & = (a + b) + (b + a) \\ & {\gt} (a + b) + (a + b) \\ & = 2(a + b) \end{align}
+</div>
+<p> Which is a contradiction. We get a similar contradiction in the case that \(b+a {\lt} a+b\). Therefore, \(a+b = b+a\). </p>
+<p>The case where \(a+b\) is negative follows similarly. The case where \(b\) is negative and \(a\) is positive is symmetric. </p>
+
+  </div>
+</div>
 
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