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+\begin{proposition}\label{in_irrefl}
+    For all sets $A$ we have $A\not\in A$.
+\end{proposition}
+\begin{proof}[Proof by \in-induction]
+    %Let $B$ be a set.
+    %Suppose $b\notin b$ for all $b\in B$.
+    %Suppose $B\in B$.
+    %But then
+    %    $B\notin B$.
+    Straightforward.
+\end{proof}