Consider the following languages over the alphabet $\Sigma = \{a, b, c\}$. Let $L_1 = \{ a^n b^n c^m \mid m, n \ge 0 \}$ and $L_2 = \{ a^m b^n c^n \mid m, n \ge 0 \}$. Which of the following are context-free languages? I. $L_1 \cup L_2$ II. $L_1 \cap L_2$

1. Context-freeness of $L_1$ and $L_2$:

   • $L_1 = \{ a^n b^n c^m \mid m, n \ge 0 \}$ is context-free because a single pushdown automaton (PDA) can push $a$'s and pop them on matching $b$'s, then accept any number of $c$'s.

   • $L_2 = \{ a^m b^n c^n \mid m, n \ge 0 \}$ is context-free because a PDA can skip any number of $a$'s, push $b$'s, and pop them on matching $c$'s.

2. Statement I ($L_1 \cup L_2$):

   • Context-free languages are closed under union.

   • Since both $L_1$ and $L_2$ are CFLs, $L_1 \cup L_2$ is context-free.

3. Statement II ($L_1 \cap L_2$):

   • $L_1 \cap L_2 = \{ a^n b^n c^n \mid n \ge 0 \}$, which is a standard non-context-free language (it requires two independent count comparisons, which a single stack cannot perform).

Therefore, only statement I is context-free, making (A) the correct choice.