Consider the following context-free grammars: $G_1: S \to aS | B, B \to b | bB$ $G_2: S \to aA | bB, A \to aA | B | \varepsilon, B \to bB | \varepsilon$ Which one of the following pairs of languages is generated by $G_1$ and $G_2$, respectively?

Let us analyze the languages generated by each grammar:

1. Grammar $G_1$:

   • $S \to aS \mid B$

   • $B \to b \mid bB$

   The variable $B$ generates one or more $b$'s: $L(B) = b^+ = \{b^n \mid n > 0\}$. The start variable $S$ generates zero or more $a$'s followed by a string from $B$: $L(G_1) = a^* b^+ = \{a^m b^n \mid m \ge 0 \text{ and } n > 0\}$.

2. Grammar $G_2$:

   • $S \to aA \mid bB$

   • $A \to aA \mid B \mid \varepsilon$

   • $B \to bB \mid \varepsilon$

   $B$ generates zero or more $b$'s: $L(B) = b^*$. $A$ generates zero or more $a$'s followed by $B$: $L(A) = a^* b^*$. Then for the start variable $S$:

   • Through branch $S \to aA$: strings of the form $a a^* b^* = a^+ b^* = \{a^m b^n \mid m > 0 \text{ and } n \ge 0\}$.

   • Through branch $S \to bB$: strings of the form $b b^* = b^+ = \{a^0 b^n \mid n > 0\}$.

   Taking the union:

   $$L(G_2) = (a^+ b^*) \cup (b^+) = \{a^m b^n \mid m > 0 \text{ or } n > 0\}$$

   (i.e., all strings of the form $a^m b^n$ where at least one of $m, n$ is strictly greater than 0).

Matching these gives:

• $L(G_1) = \{a^m b^n \mid m \ge 0 \text{ and } n > 0\}$

• $L(G_2) = \{a^m b^n \mid m > 0 \text{ or } n > 0\}$

This matches option (D).