Let R1 and R2 be regular sets defined over the alphabet, then
(A) R1 ∩ R2 is not regular
(B) R1 ∪ R2is not regular
(C) Σ * – R1 is regular
(D) R1 * is not regular
Explanation: Regular languages are closed under Union (∪), Intersection (∩) and Kleene Closure (*), which make options (A), (B) and (D) incorrect.
Option (C): Σ* – R1 = Σ* ∩ R1’= Regular languages are closed under complement operation, it is regular. So, option (C) is correct.
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