Language $L_1$ is defined by the grammar: $S_1 → aS_1b|ε$ Language $L_2$ is defined by the grammar:…

Question

Language $L_1$ is defined by the grammar: $S_1 → aS_1b|ε$ Language $L_2$ is defined by the grammar: $S_2 → abS_2|ε$ Consider the following statements: $P$: $L_1$ is regular $Q$: $L_2$ is regular Which one of the following is TRUE?

Accepted Answer

Correct answer: C — Answer: P is false and Q is true. Reasoning: L1: The grammar S1 → a S1 b | ε generates L1 = { a^n b^n | n ≥ 0 }. This language is not regular. Sketch using the pumping lemma: assume a pumping length p and take s = a^p b^p. Any decomposition s = xyz with |xy| ≤ p forces y to consist only of a's and y ≠ ε. Pumping y down (taking y^0) gives fewer a's than b's, a string not in L1, contradicting regularity. L2: The grammar S2 → ab S2 | ε generates L2 = { (ab)^n | n ≥ 0 }, which equals the regular expression (ab)*. This is regular — for example, a 2-state DFA can alternate expecting 'a' then 'b' and accept at the start/end. Therefore P is false (L1 non-regular) and Q is true (L2 regular).

Answer

A. Both $P$ and $Q$ are true

Answer

B. $P$ is true and $Q$ is false

Answer

C. $P$ is false and $Q$ is true

Answer

D. Both $P$ and $Q$ are false

Language \(L_1\) is defined by the grammar: \(S_1 → aS_1b|ε\) Language \(L_2\)…