Match LIST-I with LIST-II \[\begin{array}{|c|c|c|}\hline \text{ } & \text{LIST-I} & \text{LIST-II} \\ \hline \text{A.} & \text{A Language L can be accepted by a Finite Automata, if and only if, the set of equivalence classes of $L$ is finite.} & \text{III. Myhill-Nerode Theorem} \\ \hline \text{B.} & \text{For every finite automaton M = $(Q, \Sigma, q_0, A, \delta)$, the language L(M) is regular.} & \text{II. Regular Expression Equivalence} \\ \hline \text{C.} & \text{Let, X and Y be two regular expressions over $\Sigma$. If X does not contain null, then the equation $R = Y + RX$ in R, has a unique solution (i.e. one and only one solution) given by $R = YX^*$.} & \text{I. Arden's Theorem} \\ \hline \text{D.} & \text{The regular expressions X and Y are equivalent if the corresponding finite automata are equivalent.} & \text{IV. Kleen's Theorem} \\ \hline \end{array}\]
\[\text{Matching List-I with List-II}\]
Choose the correct answer from the options given below:
Step 1: Matching the Concepts.
- \( A \): The statement is related to **Arden's Theorem**, which deals with regular expressions and finite automata, so the correct match is \( A \) - I.
- \( B \): This is related to the fact that the language of a finite automaton is regular, which is part of the **Myhill-Nerode Theorem**, so the correct match is \( B \) - III.
- \( C \): This matches with **Regular Expression Equivalence**, where the equation \( R = Y + RX \) has a unique solution, which is described by the regular expression equivalence theory, so the correct match is \( C \) - II.
- \( D \): The equivalence of regular expressions \( X \) and \( Y \) when their corresponding finite automata are equivalent is given by **Kleen's Theorem**, so the correct match is \( D \) - IV.
Step 2: Conclusion.
Thus, the correct matching is \( A - I, B - III, C - II, D - IV \), which corresponds to answer (3).
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