Consider the following languages: $$L_1 = \{a^n w a^n \mid w \in \{a,b\}^*\}$$ $$L_2 = \{w x w^R \mid w, x \in \{a,b\}^*, |w|, |x| > 0\}$$ Note that $w^R$ is the reversal of the string $w$. Which of the following is/are TRUE?

Let us analyze both languages:

1. Language $L_1$:

   $$L_1 = \{a^n w a^n \mid w \in \{a,b\}^*\}$$

   For $n = 0$, $a^0 w a^0 = w$. Since $w$ can be any string in $\{a,b\}^*$, setting $n = 0$ generates all strings in $\{a,b\}^*$. Thus:

   $$L_1 = \{a,b\}^*$$

   This is clearly a regular language (and consequently context-free).

2. Language $L_2$:

   $$L_2 = \{w x w^R \mid w, x \in \{a,b\}^*, |w|, |x| > 0\}$$

   Since $|w| \ge 1$, let the first symbol of $w$ be $c \in \{a, b\}$. Then $w^R$ ends with the same symbol $c$. Because $x$ has length at least $1$, the total length of any string in $L_2$ is at least $|w| + |x| + |w^R| \ge 1 + 1 + 1 = 3$.

   Conversely, any string $s \in \{a, b\}^*$ of length $\ge 3$ that starts and ends with the same symbol can be written with $|w| = 1$ (where $w$ is simply the first character) and $x$ as the remaining inner substring of length $\ge 1$. Thus:

   $$L_2 = a(a+b)^+ a \cup b(a+b)^+ b$$

   This language is generated by a regular expression and is recognized by a finite automaton, so $L_2$ is regular (and consequently context-free).

Since every regular language is also context-free:

• (A) $L_1$ and $L_2$ are regular: TRUE

• (B) $L_1$ and $L_2$ are context-free: TRUE

• (C) $L_1$ is regular and $L_2$ is context-free: TRUE

• (D) $L_1$ and $L_2$ are context-free but not regular: FALSE (they are regular)