Consider the following two languages over the alphabet $\{a, b, c\}$, where $m$ and $n$ are natural numbers. $$L_1 = \{a^m b^m c^{m+n} \mid m, n \ge 1\}$$ $$L_2 = \{a^m b^n c^{m+n} \mid m, n \ge 1\}$$ Which ONE of the following statements is CORRECT?

### Analysis of Language $L_1$:

In $L_1$, there are two simultaneous conditions tied to the same parameter $m$:

1. The number of $a$'s must equal the number of $b$'s ($m$).

2. The number of $c$'s must be at least $m + 1$ (i.e., $m+n$ with $n \ge 1$).

A Pushdown Automaton (PDA) has only one stack. If it pushes $a$'s and matches them against $b$'s, the stack becomes empty and cannot remember $m$ to ensure that the number of $c$'s is $m+n$.

Using the Pumping Lemma for CFLs:

• Choose $w = a^p b^p c^{p+1} \in L_1$, where $p$ is the pumping length.

• Any decomposition $w = uvxyz$ with $|vxy| \le p$ and $|vy| \ge 1$ cannot maintain both the equality between $a$'s and $b$'s and the condition that the count of $c$'s is greater than the count of $a$'s and $b$'s upon pumping.

• Hence, $L_1$ is not a context-free language.

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### Analysis of Language $L_2$:

In $L_2$, the total number of $c$'s equals the sum of the number of $a$'s and $b$'s. A PDA can easily recognize this language:

1. Push all $a$'s onto the stack.

2. Push all $b$'s onto the stack.

3. For each $c$ read from the input, pop one symbol from the stack (first popping all $b$'s, then all $a$'s).

4. Accept if the stack is empty after consuming the input (with $m, n \ge 1$).

A Context-Free Grammar (CFG) for $L_2$ can be given as:

• $S \to aSc \mid aAc$

• $A \to bAc \mid bc$

Hence, $L_2$ is a context-free language.

Therefore, $L_1$ is not context-free but $L_2$ is context-free, which corresponds to option (C).