20 votes 20 votes Let $R_{1}$ and $R_{2}$ be regular sets defined over the alphabet $\Sigma$ Then: $R_{1} \cap R_{2}$ is not regular. $R_{1} \cup R_{2}$ is regular. $\Sigma^{*}-R_{1}$ is regular. $R_{1}^{*}$ is not regular. Theory of Computation gate1990 normal theory-of-computation regular-language multiple-selects + – makhdoom ghaya asked Nov 22, 2016 edited Apr 17, 2021 by Lakshman Bhaiya makhdoom ghaya 9.0k views answer comment Share Follow See 1 comment See all 1 1 comment reply smsubham commented Oct 21, 2017 reply Follow Share Regular language is closed under Union Intersection Kleene Closure Concatenation Complementation Difference Reversal 3 votes 3 votes Please log in or register to add a comment.
Best answer 33 votes 33 votes Regular Languages are closed under Intersection Union Complement Kleen-Closure $\Sigma^∗−R_1$ is the complement of $R_1$ Correct Options: B;C pC answered Nov 22, 2016 edited May 6, 2021 by soujanyareddy13 pC comment Share Follow See all 0 reply Please log in or register to add a comment.
6 votes 6 votes R1 ∩ R2 is not regular.-FALSE,Regular sets are closed under Intersection R1 ∪ R2 is regular. TRUE,Regular sets are closed under Union ∑∗−R1 is regular,TRUE,Regular sets are closed under Complement R1* is not regular.FALSE,Regular sets are closed under Intersection Prajwal Bhat answered Nov 22, 2016 Prajwal Bhat comment Share Follow See 1 comment See all 1 1 comment reply Ejaz Ali commented Oct 8, 2017 reply Follow Share 4)R1* is not Regular (FALSE) because regular language is closed under kleen closure. 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes Everything is true if they have not used the word not in the question. nikunj answered Aug 31, 2017 nikunj comment Share Follow See 1 comment See all 1 1 comment reply air1ankit commented Dec 20, 2018 reply Follow Share then who told, "Choose the correct alternatives (More than one may be correct)" ? 0 votes 0 votes Please log in or register to add a comment.