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Could someone explain me how to form a context-free grammar with all rules R by this example language, please? \begin{equation} L:=\left\{w c v c \overleftarrow{w} | w, v \in\{a, b\}^{+}\right\} \end{equation} The arrow over w means, that the word w is written backwards.

I already know that \begin{equation} \Sigma=\{a, b, c\} \end{equation} and V (non terminal symbols) maybe have to be \begin{equation} V=\{S, A, B, C\} \end{equation}

Thank you for helping.

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I think you can solve this using this construct:

$$S \rightarrow ASA|BSB|ATA|BTB $$ $$T \rightarrow CAZC|CBZC$$ $$Z \rightarrow AZ|BZ|\epsilon$$ $$A \rightarrow a , B \rightarrow b, C \rightarrow c$$

First rule provides the $w$ and $w^R$ in the language, second rule makes sure $v \ge 1$ and has 1 c in both sides and third is just constructing the $v$ further. My assumption was your arrow on top of last $w$ was just meant it is reversed.

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  • $\begingroup$ Could you explain what T and Z mean in your rules? Do you mean the word backwards with \begin{equation} w^{R} \text { ?} \end{equation} $\endgroup$
    – mathflower
    Dec 1 '19 at 12:38
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    $\begingroup$ I just use those arbitrary non-terminals to impose extra rules. Using "S" to impose the $w$ and $w^R$ rule then using T to make sure we got $|v| \ge 1$ and Z to create any $ |v| > 1$. As I said they are arbitrary non-terminals , you can use this kind of rules as long as they abide by the rules of your language. $\endgroup$
    – Yiğit Aras Tunalı
    Dec 1 '19 at 12:41
  • $\begingroup$ So you mean, v can not be greater than 1, because it can not be ϵ because of the +. Right? $\endgroup$
    – mathflower
    Dec 1 '19 at 12:53
  • $\begingroup$ v has to be greater than 1, because of $v \in \{a,b\}^+$. I think I made a mistake, lemme fix it, because I forgot $w \in \{a,b\}^+$ as well $\endgroup$
    – Yiğit Aras Tunalı
    Dec 1 '19 at 13:00
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    $\begingroup$ It should be correct now, by $w^R$ I meant if w=ab then $w^R$= ba. $\endgroup$
    – Yiğit Aras Tunalı
    Dec 1 '19 at 13:44

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