Consider a context-free grammar $G$ with the following 3 rules. $$S \rightarrow aS, \; S \rightarrow aSbS, \; S \rightarrow c$$ Let $w \in L(G)$. Let $n_a(w)$, $n_b(w)$, $n_c(w)$ denote the number of times $a, b, c$ occur in $w$, respectively. Which of the following statements is/are TRUE?

Let $k_1, k_2, k_3$ denote the number of times the production rules $S \rightarrow aS$, $S \rightarrow aSbS$, and $S \rightarrow c$ are applied in the derivation of a terminal word $w \in L(G)$, respectively.

1. Tracking the Non-terminal $S$:

   • Initial count of $S = 1$.

   • Rule $S \rightarrow aS$ replaces $1$ instance of $S$ with $1$ instance of $S$ ($Δ S = 0$).

   • Rule $S \rightarrow aSbS$ replaces $1$ instance of $S$ with $2$ instances of $S$ ($Δ S = +1$).

   • Rule $S \rightarrow c$ replaces $1$ instance of $S$ with $0$ instances of $S$ ($Δ S = -1$).

   Since the final string $w$ has no non-terminals ($0$ instances of $S$):

   $$1 + 0 \cdot k_1 + 1 \cdot k_2 - 1 \cdot k_3 = 0 \implies k_3 = k_2 + 1$$

2. Counting Terminals:

   • $c$ is only introduced by rule $S \rightarrow c$, so $n_c(w) = k_3$.

   • $b$ is only introduced by rule $S \rightarrow aSbS$, so $n_b(w) = k_2$.

   • $a$ is introduced by both $S \rightarrow aS$ and $S \rightarrow aSbS$, so $n_a(w) = k_1 + k_2$.

3. Evaluating the Options:

   • Option C: $n_c(w) = k_3 = k_2 + 1 = n_b(w) + 1$. Thus, (C) is TRUE.

   • Option D: $n_c(w) = n_b(w) + 1 \neq 2 \cdot n_b(w)$ in general (e.g., for $w = c$, $n_c=1, n_b=0$). Thus, (D) is FALSE.

   • Option A: $n_a(w) = k_1 + k_2 \ge k_2 = n_b(w)$. For $w = c$, $k_1 = k_2 = 0$, giving $n_a(c) = 0$ and $n_b(c) = 0$. Since $0 > 0$ is false, (A) is FALSE.

   • Option B: From $n_a(w) \ge n_b(w)$ and $n_b(w) = n_c(w) - 1$, we have:

     $$n_a(w) \ge n_c(w) - 1 > n_c(w) - 2$$

     Thus, $n_a(w) > n_c(w) - 2$ holds for all $w \in L(G)$. So (B) is TRUE.

Therefore, the correct options are B and C.