Which one of the following languages is CFL and what is the grammar for it? Can we use pumping lemma for the not CFL?
\begin{align*} L_1 &= \{a^mb^ma^n \mid m,n \geq 0\}\\ L_2 &= \{a^mb^ma^n \mid m \geq n \geq 0\} \end{align*}
Which one of the following languages is CFL and what is the grammar for it? Can we use pumping lemma for the not CFL?
\begin{align*} L_1 &= \{a^mb^ma^n \mid m,n \geq 0\}\\ L_2 &= \{a^mb^ma^n \mid m \geq n \geq 0\} \end{align*}
It’s very easy to write a context-free grammar that generates $L_1$. You might start with a production $S\to PA$ and add productions ensuring that $P$ generates $\{a^nb^n:n\ge 0\}$ and $A$ generates $a^*$.
Yes, you can use the pumping lemma for context-free languages to show that $L_2$ is not context-free; the proof is very similar to the usual example of such a proof, the proof that $\{a^nb^nc^n:n\ge 0\}$ is not context-free.