Question 52

Analysis of grammar $G_1$:

• $B \to b \mid bB$: The non-terminal $B$ generates one or more $b$'s, i.e., $L(B) = b^+ = \{b^n \mid n > 0\}$.
• $S \to aS \mid B$: The start symbol generates zero or more $a$'s followed by a derivation from $B$. Thus, $L(G_1) = a^* b^+ = \{a^m b^n \mid m \ge 0 \text{ and } n > 0\}$.

Analysis of grammar $G_2$:

• $B \to bB \mid \varepsilon$: Non-terminal $B$ generates zero or more $b$'s, i.e., $L(B) = b^* = \{b^n \mid n \ge 0\}$.
• $A \to aA \mid B \mid \varepsilon$: $A$ generates zero or more $a$'s followed by $B$ or $\varepsilon$, which simplifies to $L(A) = a^* b^*$.
• $S \to aA \mid bB$:
  • The branch $S \to aA$ generates $a(a^* b^*) = a^+ b^*$, which is all strings of $a$'s and $b$'s with at least one $a$ ($m > 0, n \ge 0$).
  • The branch $S \to bB$ generates $b(b^*) = b^+$, which is all strings with only $b$'s and at least one $b$ ($m = 0, n > 0$).
  • The union $a^+ b^* \cup b^+$ contains all strings of the form $a^m b^n$ where at least one of $m$ or $n$ is strictly positive ($m > 0 \text{ or } n > 0$). Note that the empty string $\varepsilon$ cannot be generated by $S$.

Comparing the languages:

• $L(G_1) = \{a^m b^n \mid m \ge 0 \text{ and } n > 0\}$
• $L(G_2) = \{a^m b^n \mid m > 0 \text{ or } n > 0\}$
  This matches option D.