10 votes 10 votes Let $R_1$ and $R_2$ be regular sets defined over the alphabet, then $ R_1 \cap R_2$ is not regular $R_1 \cup R_2$ is not regular $\Sigma^* - R_1$ is regular $R_1^*$ is not regular Theory of Computation isro2015 theory-of-computation regular-language + – go_editor asked Jun 19, 2016 • retagged Jul 4, 2017 by Arjun go_editor 3.1k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 10 votes 10 votes option C is correct Regular sets are closed under union, intersection, complement, and kleen closure But regular sets are not closed under infinite union rameshbabu answered Jun 19, 2016 • selected Jun 27, 2016 by Anu rameshbabu comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes "Regular languages are closed under Intersection, Union ,Kleens Closure ,Compliment" According to this point option C is perfect akash.dinkar12 answered Apr 14, 2017 akash.dinkar12 comment Share Follow See all 4 Comments See all 4 4 Comments reply Srinivas Rao commented Apr 27, 2017 reply Follow Share Sir can you give example where $\sum* - R1$ is not regular 0 votes 0 votes akash.dinkar12 commented Apr 27, 2017 reply Follow Share ∑∗−R1 is regular, this statement is true, not false. 0 votes 0 votes Srinivas Rao commented Apr 27, 2017 reply Follow Share Oh soory Sir, i didn't read the options carefully 0 votes 0 votes akash.dinkar12 commented Apr 27, 2017 reply Follow Share ok sir 0 votes 0 votes Please log in or register to add a comment.