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strings_and_languages [2012/09/19 16:55] vkuncak |
strings_and_languages [2019/07/02 16:16] fabien [Strings and languages] |
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\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, $L \cdot \emptyset = \emptyset$. | ||
- | Of course, $\{ w_1 \} \cdot \{ w_2 \} = w_1\cdot w_2$. | + | Also directly from definition follows: |
+ | \begin{equation*} | ||
+ | \{ w_1 \} \cdot \{ w_2 \} = \{ w_1\cdot w_2 \} | ||
+ | \end{equation*} | ||