Consider the context-free grammar $G$ below $$\begin{aligned} S &\to aSb \mid X X &\to aX \mid Xb \mid a \mid b\text{ ,} \end{aligned}$$ where $S$ and $X$ are non-terminals, and $a$ and $b$ are terminal symbols. The starting non-terminal is $S$. Which one of the following statements is CORRECT?

To determine the language generated by the context-free grammar $G$, let us analyze the productions step by step:

1. Language generated by $X$:

   • The non-terminal $X$ can add any number of leading $a$'s using $X \to aX$ and any number of trailing $b$'s using $X \to Xb$.

   • The base productions are $X \to a$ and $X \to b$.

   • Thus, every derivation from $X$ yields a string of the form:

     $$a^i (a \mid b) b^j \quad \text{where } i, j \ge 0$$

   • This gives the regular language:

     $$L(X) = a^* (a + b) b^*$$

2. Language generated by $S$:

   • The production $S \to aSb \mid X$ allows wrapping the string produced by $X$ with an equal number of $a$'s and $b$'s:

     $$S \Rightarrow^* a^n X b^n \quad \text{for } n \ge 0$$

   • Substituting the language generated by $X$:

     $$L(G) = \{ a^n (a^i (a \mid b) b^j) b^n \mid n, i, j \ge 0 \} = \{ a^{n+i} (a \mid b) b^{n+j} \mid n, i, j \ge 0 \}$$

   • Since $k = n + i \ge 0$ and $m = n + j \ge 0$ can independently take any non-negative integer values, this simplifies to:

     $$L(G) = \{ a^k (a \mid b) b^m \mid k, m \ge 0 \} = a^*(a + b)b^*$$

Hence, the language generated by $G$ is regular and equal to $a^*(a + b)b^*$, making (B) the correct statement.