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strings_and_languages [2009/04/28 20:39]
vkuncak
strings_and_languages [2019/07/02 16:16]
fabien [Strings and languages]
Line 18: Line 18:
 \end{eqnarray*} \end{eqnarray*}
 Therefore, $(\Sigma^*, {\ \cdot\ }, \epsilon)$ is a [[monoid]]. Therefore, $(\Sigma^*, {\ \cdot\ }, \epsilon)$ is a [[monoid]].
- 
-We sometimes omit ${} \cdot{} $ and write $w_1 \cdot w_2$ simply as $w_1 w_2$. 
  
 If $w$ is a word, we define $w^0 = \epsilon$ and $w^{n+1} = w \cdot w^n$. If $w$ is a word, we define $w^0 = \epsilon$ and $w^{n+1} = w \cdot w^n$.
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   L^* &=& \bigcup_{n \geq 0} L^n = \{ w_1 \ldots w_n \mid w_1,​\ldots,​w_n \in L \}   L^* &=& \bigcup_{n \geq 0} L^n = \{ w_1 \ldots w_n \mid w_1,​\ldots,​w_n \in L \}
 \end{eqnarray*} \end{eqnarray*}
 +
 +==== Simple Consequences ====
 +
 +Observe that
 +\begin{equation*}
 +   ​\emptyset \cdot L = \{ s_1 \cdot s_2 \mid s_1 \in \emptyset \land s_2 \in L \} = \{s_1 \cdot s_2 \mid \mbox{false} \} = \emptyset
 +\end{equation*}
 +Similarly, $L \cdot \emptyset = \emptyset$.
 +
 +Also directly from definition follows:
 +\begin{equation*}
 +  \{ w_1 \} \cdot \{ w_2 \} = \{ w_1\cdot w_2 \}
 +\end{equation*}
 +