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.