Non-terminal B Analysis

Productions for B:

• $B \rightarrow ab$: Generates a string with $n_1=1, n_2=1$. This satisfies $n_1 \ge n_2$ and $n_1 \le 2n_2$.
• $B \rightarrow aBb$: Let $s'$ be a string derived from $B$. If $s'$ satisfies $n_1(s') \ge n_2(s')$ and $n_1(s') \le 2n_2(s')$, then for $a s' b$:
  • $n_1(a s' b) = 1 + n_1(s')$. $n_2(a s' b) = 1 + n_2(s')$.
  • $n_1 - n_2 = (1 + n_1(s')) - (1 + n_2(s')) = n_1(s') - n_2(s')$. Since $n_1(s') \ge n_2(s')$, $n_1 \ge n_2$.
  • $n_1 - 2n_2 = (1 + n_1(s')) - 2(1 + n_2(s')) = n_1(s') - 2n_2(s') - 1$. Since $n_1(s') \le 2n_2(s')$, $n_1(s') - 2n_2(s') \le 0$. Thus $n_1 - 2n_2 \le -1$, implying $n_1 \le 2n_2$.

By induction, all strings derived from $B$ satisfy $n_1(s) = n_2(s)$. Therefore, $n_1(s) \ge n_2(s)$ and $n_1(s) \le 2n_2(s)$ hold for all strings from $B$.

Non-terminal A Analysis

Productions for A:

• $A \rightarrow bBabaa$: Generates a string with $n_1=2, n_2=1$. This satisfies $n_1 \ge n_2$ and $n_1 \le 2n_2$.
• $A \rightarrow aaBBAB$: Let $s_A'$ be derived from $A$, and $s_{B1}, s_{B2}, s_{B3}$ from $B$. Strings from $B$ have $n_1=n_2=k$.
  • $n_1(A) = 2 + n_1(B_1) + n_1(B_2) + n_1(A') + n_1(B_3) = 2 + \sum k_i + n_1(A')$.
  • $n_2(A) = n_2(B_1) + n_2(B_2) + n_2(A') + n_2(B_3) = \sum k_i + n_2(A')$.
  • Let $X = \sum k_i$. Minimum $X \ge 3$ (from $B \rightarrow ab \times 3$). $n_1(A) = 2 + X + n_1(A')$, $n_2(A) = X + n_2(A')$.
  • $n_1 \ge n_2$ check: $n_1(A) - n_2(A) = (2 + X + n_1(A')) - (X + n_2(A')) = 2 + n_1(A') - n_2(A')$. Since $n_1(A') \ge n_2(A')$, $n_1(A) - n_2(A) \ge 2$. Thus $n_1(A) > n_2(A)$.
  • $n_1 \le 2n_2$ check: We need $2 + X + n_1(A') \le 2(X + n_2(A'))$. This simplifies to $2 + n_1(A') \le X + 2n_2(A')$. Since $X \ge 3$ and $n_1(A') \ge n_2(A')$, this inequality holds. For example, if $A' \rightarrow bBabaa$ ($n_1=2, n_2=1$), we need $2+2 \le X+2(1) \implies 4 \le X+2 \implies 2 \le X$, which is true as $X \ge 3$.

Thus, for all strings derived from $A$, $n_1(s) \ge n_2(s)$ and $n_1(s) \le 2n_2(s)$ hold.

Start Symbol S Analysis

Production for S:

• $S \rightarrow abaABabba$:
  • $n_1(S) = 3 + n_1(A) + n_1(B)$.
  • $n_2(S) = 2 + n_2(A) + n_2(B)$.
  • From B analysis, $n_1(B)=n_2(B)=k$.
  • $n_1(S) = 3 + n_1(A) + k$.
  • $n_2(S) = 2 + n_2(A) + k$.

Statement Evaluation

Statement 2: For every string $s \in L(G)$, $n_1(s) \ge n_2(s)$

• For S: $n_1(S) - n_2(S) = (3 + n_1(A) + k) - (2 + n_2(A) + k) = 1 + (n_1(A) - n_2(A))$.
• Since $n_1(A) \ge n_2(A)$ for strings from A, $n_1(A) - n_2(A) \ge 0$.
• Therefore, $n_1(S) - n_2(S) \ge 1$. This implies $n_1(s) \ge n_2(s)$ for all $s \in L(G)$.
• Statement 2 is TRUE.

Statement 4: For every string $s \in L(G)$, $n_1(s) \le 2n_2(s)$

• For S: We need $n_1(S) \le 2n_2(S)$.
• $3 + n_1(A) + k \le 2(2 + n_2(A) + k)$.
• $3 + n_1(A) + k \le 4 + 2n_2(A) + 2k$.
• This simplifies to $n_1(A) \le 1 + 2n_2(A) + k$.
• As analyzed in the A section, $n_1(A) = 2 + X + n_1(A')$ and $n_2(A) = X + n_2(A')$, where $X \ge 3$ and $k \ge 1$. The condition $n_1(A) \le 1 + 2n_2(A) + k$ becomes $2 + X + n_1(A') \le 1 + 2(X + n_2(A')) + k$, which simplifies to $1 + n_1(A') \le X + 2n_2(A') + k$. This inequality holds true given the constraints on $X, k$, and the properties of $n_1(A'), n_2(A')$.
• Statement 4 is TRUE.

Conclusion

Based on the analysis, statements 2 and 4 are true for all strings $s \in L(G)$.