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  1. If $G$ is not a regular grammar, then $L(G)$ is infinte.

  2. If $L^*$ is context free then $L$ is definitely context free.

  3. If $G$ is a context free grammar that is language is $L$ (meaning $L(G) = L$),

    then there exists a context free grammar $G^r$ such that $L(G^r) = L^r$

    • $L^r =\{w^r \mid w \in L\}$
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Please make the title more descriptive. Thanks. – Babak S. Dec 21 '12 at 14:13
@BabakSorouh I went ahead and changed the title a bit as part of the task of cleaning up all of the spelling mistakes. user14988, please check to see that I didn't alter the meaning of your question with my grammar tweaks. – rschwieb Dec 21 '12 at 14:20
@rschwieb: Yes, and I’ve made the change. – Brian M. Scott Dec 21 '12 at 14:24
What does "definitely context free" mean? Note that for any language $L$, $(\{0,1\} \cup L)^* = \{0,1\}^*$ is context free. – Carl Mummert Dec 21 '12 at 14:35

This isn’t a full answer, but it’s too long for a comment.

(1) is actually false as stated. The grammar $G$ with initial symbol $S$, terminal symbols $0$ and $1$, non-terminal symbols $S,A,B,C$, and the productions below is clearly not regular, but $L(G)=\{01\}$.

$$\begin{align*} &S\to ABC\\ &A\to 0\\ &B\to 1\\ &C\to C\epsilon C\mid\epsilon \end{align*}$$

What is true is that if $L$ is a finite language, then there is a regular grammar $G$ such that $L(G)=L$; this is probably what you were really meant to prove.

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