--- a/hws/proof.tex	Thu Nov 26 12:55:59 2015 +0000
+++ b/hws/proof.tex	Fri Nov 27 12:08:29 2015 +0000
@@ -188,15 +188,12 @@
 
 \end{itemize}
 
-
-
-
 Let us now analyse $Der\,c\,L(r^*)$, which is equal to $Der\,c\,((L(r))^*)$. Now $(L(r))^*$ is defined
-as $\bigcup_{n \ge 0} L(r)$. We can write this as $L(r)^0 \cup \bigcup_{n \ge 1} L(r)$, where we just 
+as $\bigcup_{n \ge 0} L(r)^n$. We can write this as $L(r)^0 \cup \bigcup_{n \ge 1} L(r)^n$, where we just 
 separated the first union and then let the ``big-union'' start from $1$. Form this we can already infer
 
 \begin{center}
-$Der\,c\,(L(r^*)) = Der\,c\,(L(r)^0 \cup \bigcup_{n \ge 1} L(r)) = (Der\,c\,L(r)^0) \cup Der\,c\,(\bigcup_{n \ge 1} L(r))$
+$Der\,c\,(L(r^*)) = Der\,c\,(L(r)^0 \cup \bigcup_{n \ge 1} L(r)^n) = (Der\,c\,L(r)^0) \cup Der\,c\,(\bigcup_{n \ge 1} L(r)^n)$
 \end{center}
 
 The first union ``disappears'' since $Der\,c\,(L(r)^0) = \varnothing$.