LARA

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strings_and_languages [2012/09/19 16:56]
vkuncak 		strings_and_languages [2019/07/02 16:16]
fabien [Strings and languages]
Line 28: Line 28:
 \begin{eqnarray*} \begin{eqnarray*}
   L_1 \cdot L_2  &=& \{ s_1 \cdot s_2 \mid s_1 \in L_1 \land s_2 \in L_2 \} \\   L_1 \cdot L_2  &=& \{ s_1 \cdot s_2 \mid s_1 \in L_1 \land s_2 \in L_2 \} \\
-  L^0 &=& \{ \epsilon \} \\+  L^ZZ0 &=& \{ \epsilon \} \\
   L^{n+1} &=& L \cdot L^n \\   L^{n+1} &=& L \cdot L^n \\
   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 \}
Line 36: Line 36:
  
 Observe that Observe that
-\[+\begin{equation*}
    ​\emptyset \cdot L = \{ s_1 \cdot s_2 \mid s_1 \in \emptyset \land s_2 \in L \} = \{s_1 \cdot s_2 \mid \mbox{false} \} = \emptyset    ​\emptyset \cdot L = \{ s_1 \cdot s_2 \mid s_1 \in \emptyset \land s_2 \in L \} = \{s_1 \cdot s_2 \mid \mbox{false} \} = \emptyset
-\]+\end{equation*}
 Similarly, $L \cdot \emptyset = \emptyset$. Similarly, $L \cdot \emptyset = \emptyset$.
  
 Also directly from definition follows: Also directly from definition follows:
-\[+\begin{equation*}
   \{ w_1 \} \cdot \{ w_2 \} = \{ w_1\cdot w_2 \}   \{ w_1 \} \cdot \{ w_2 \} = \{ w_1\cdot w_2 \}
-\]+\end{equation*}