Question:

Suppose \( D_1 = (S_1, \Sigma, q_1, F_1, \delta_1) \) and \( D_2 = (S_2, \Sigma, q_2, F_2, \delta_2) \) are finite automata accepting languages \( L_1 \) and \( L_2 \), respectively. Then, which of the following languages will also be accepted by the finite automata: 

(A) \( L_1 \cup L_2 \) 
(B) \( L_1 \cap L_2 \) 
(C) \( L_1 - L_2 \) 
(D) \( L_2 - L_1 \) 
Choose the correct answer from the options given below:

Show Hint

The union, intersection, and difference of two regular languages are always regular. Finite automata can be constructed for these operations.
Updated On: Sep 25, 2025
  • A, B and D only
  • A, B and C only
  • A, B, C and D
  • B, C and D only
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The Correct Option is C

Solution and Explanation

Step 1: Language Operations. 
The union, intersection, and difference of languages are all regular operations. If \( L_1 \) and \( L_2 \) are accepted by finite automata \( D_1 \) and \( D_2 \), respectively, then the languages resulting from the union, intersection, and difference of these languages will also be accepted by finite automata. 

- \( L_1 \cup L_2 \): The union of two regular languages is regular. 

- \( L_1 \cap L_2 \): The intersection of two regular languages is regular. 

- \( L_1 - L_2 \): The difference of two regular languages is regular. 

- \( L_2 - L_1 \): Similarly, the difference of two regular languages is regular.

Step 2: Conclusion. 
Since all the given operations (union, intersection, and difference) result in regular languages, the correct answer is (3), as all four languages can be accepted by finite automata. 
 

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