in Theory of Computation edited by
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20 votes
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Let $R_{1}$ and $R_{2}$ be regular sets defined over the alphabet $\Sigma$ Then:

  1. $R_{1} \cap R_{2}$ is not regular.
  2. $R_{1} \cup R_{2}$ is regular.
  3. $\Sigma^{*}-R_{1}$ is regular.
  4. $R_{1}^{*}$ is not regular.
in Theory of Computation edited by
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Regular language is closed under

  • Union
  • Intersection
  • Kleene Closure
  • Concatenation
  • Complementation
  • Difference
  • Reversal
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3 Answers

33 votes
33 votes
Best answer

Regular Languages are closed under

  1. Intersection
  2. Union
  3. Complement
  4. Kleen-Closure

$\Sigma^∗−R_1$ is the complement of $R_1$
 

Correct Options: B;C

edited by
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6 votes
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  1. R1 ∩ R2 is not regular.-FALSE,Regular sets are closed under Intersection
  2. R1 ∪ R2 is regular. TRUE,Regular sets are closed under Union
  3. −R1 is regular,TRUE,Regular sets are closed under Complement
  4. R1* is not regular.FALSE,Regular sets are closed under Intersection

1 comment

4)R1* is not Regular (FALSE) because regular language is closed under kleen closure.
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1 vote
1 vote
Everything is true if they have not used the word not in the question.

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then who told, "Choose the correct alternatives (More than one may be correct)" ?
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Answer:

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