186 further $L(der\,c\,r) \,@\, L(r^*)$. By induction hypothesis (I) we know that is equal to |
186 further $L(der\,c\,r) \,@\, L(r^*)$. By induction hypothesis (I) we know that is equal to |
187 $(Der\,c\,L(r)) \,@\, L(r^*)$. (*) |
187 $(Der\,c\,L(r)) \,@\, L(r^*)$. (*) |
188 |
188 |
189 \end{itemize} |
189 \end{itemize} |
190 |
190 |
191 |
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192 |
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193 |
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194 Let us now analyse $Der\,c\,L(r^*)$, which is equal to $Der\,c\,((L(r))^*)$. Now $(L(r))^*$ is defined |
191 Let us now analyse $Der\,c\,L(r^*)$, which is equal to $Der\,c\,((L(r))^*)$. Now $(L(r))^*$ is defined |
195 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 |
192 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 |
196 separated the first union and then let the ``big-union'' start from $1$. Form this we can already infer |
193 separated the first union and then let the ``big-union'' start from $1$. Form this we can already infer |
197 |
194 |
198 \begin{center} |
195 \begin{center} |
199 $Der\,c\,(L(r^*)) = Der\,c\,(L(r)^0 \cup \bigcup_{n \ge 1} L(r)) = (Der\,c\,L(r)^0) \cup Der\,c\,(\bigcup_{n \ge 1} L(r))$ |
196 $Der\,c\,(L(r^*)) = Der\,c\,(L(r)^0 \cup \bigcup_{n \ge 1} L(r)^n) = (Der\,c\,L(r)^0) \cup Der\,c\,(\bigcup_{n \ge 1} L(r)^n)$ |
200 \end{center} |
197 \end{center} |
201 |
198 |
202 The first union ``disappears'' since $Der\,c\,(L(r)^0) = \varnothing$. |
199 The first union ``disappears'' since $Der\,c\,(L(r)^0) = \varnothing$. |
203 |
200 |
204 |
201 |