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I'm preparing an Formal language exam, One question from previous year's final is:

Prove or disprove:If L is a context free language, then there exists a language P that is generated by a pure context-free grammar and a regular language R so that $L = P \cap R$.

My solution to this question is quite simple. Assume context-free language P is equal to context free language L, and let regular language R to be $\Sigma^*$, where $\Sigma$ is the alphabet set. $\Sigma^*$ is definitely a regular language. So the answer to this question is yes.

However, I do came up with a more difficult problem. What if I add a limitation $P \neq L?$ Will this statement still hold? I current have no idea about this.

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The question in the title of your contribution "Can every context free language written as a intersection of another context free language and a regular language?" is the one you answer above, by taking $P=L$ and $R=\Sigma^*$.

However that is not the question you quote from your test: "If $L$ is a context free language, then there exists a language $P$ that is generated by a pure context-free grammar and a regular language $R$ so that $L=P\cap R$." This additionally requires $P$ to be pure context-free, that is, generated by a grammar that has no dictinction between terminals and nonterminals.

Let $G$ be a context-free grammar with alphabet $\Sigma$ (terminals and nonterminals) and terminal alphabet $\Delta$. As an ordinary context-free grammar the language of $G$ equals $L(G) = \{ w\in \Delta^* \mid S\Rightarrow^* w\}$. Interpreted as pure grammar it cannot "see" terminals and the language is $P(G) = \{ w\in \Sigma^* \mid S\Rightarrow^* w\}$. Or, $L(G) = P(G) \cap \Delta^*$, which gives the answer to your final exam. The elements in $P(G)$ are usually called sentential forms.

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  • $\begingroup$ Thank you Hendrik, I didn't realize that this "pure" will make such a difference. We didn't talk about pure context grammar in this semester but this might taught in previous year. Thanks again! $\endgroup$ – jinxuan Wu Dec 6 '14 at 18:27
  • $\begingroup$ Glad I could help you. Good luck with your exam. $\endgroup$ – Hendrik Jan Dec 7 '14 at 0:51
  • $\begingroup$ I disagree (mildly) with your first paragraph: to me, "another context-free language" implies that you're not allowed to take $P=L$. OTOH, if that was the intended reading, then $\Sigma^*$ is a CFL that is not the intersection of a CFL distinct from $\Sigma^*$ and a regular language. So maybe I agree with you after all. $\endgroup$ – David Richerby Apr 23 '15 at 16:55
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If you are allowed to change the alphabet, then the answer is trivially yes. For example, let $\Sigma$ be the original alphabet, and let $\alpha$ be a new symbol. Then $P = L \cup \{\alpha\}$ is context-free, and $L = P \cap \Sigma^*$.

If you are not allowed to change the alphabet, then the answer is trivially no. For if you take $L = \Sigma^*$, you must take $P = \Sigma^*$.

What if you are not allowed to change the alphabet and $L \neq \Sigma^*$? Again the answer is trivially yes. Choose some word $w \notin L$. Then $P = L \cup \{w\}$ is context-free and $R = \overline{\{w\}}$ is regular, and $L = P \cap R$.

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