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Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
test_latex [2019/02/07 18:31] wikiadmin |
test_latex [2019/02/08 11:07] wikiadmin |
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<latex> | <latex> | ||
\begin{array}{l} | \begin{array}{l} | ||
- | a > 2 \wedge res \geq 0 \wedge (n = n0 \lor res > 14) \rightarrow \\ | + | b > 2 \wedge res \geq 0 \wedge (n = n0 \lor res > 14) \rightarrow \\ |
2a \land res + 2a >= 0 \wedge (n - 1 = n_0 \lor res + 3a \geq 0) | 2a \land res + 2a >= 0 \wedge (n - 1 = n_0 \lor res + 3a \geq 0) | ||
\end{array} | \end{array} | ||
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We define concatenation and iteration of languages by | We define concatenation and iteration of languages by | ||
\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 \} \\ | + | K_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^0 &=& \{ \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 \} | ||
\end{eqnarray*} | \end{eqnarray*} | ||
+ | |||
+ | |||
+ | again | ||
+ | |||
+ | $ a * w + b = c $ | ||
End | End |