Let R_1 and R_2 be regular sets defined over the alphabet Sigma. Then:
A. R_1 intersection R_2 is not regular.
B. R_1 union R_2 is regular.
C. Sigma* - R_1 is regular.
D. R_1* is not regular.
GATE 1990 · Theory of Computation · Regular Language · medium
Answer: B and C are TRUE. R_1 union R_2 is regular (option B), and Sigma* - R_1 is regular (option C).
Check each option for regular language closure: A. R_1 intersection R_2 is NOT regular — FALSE. Intersection of two regular languages IS always regular (closed under intersection).
B. R_1 union R_2 is regular — TRUE. Regular languages are closed under union.
C. Sigma* - R_1 is regular — TRUE. Regular languages are closed under complement (swap accept/reject states in DFA).
D. R_1* is NOT regular — FALSE. Kleene star of a regular language is always regular.
Confirm TRUE statements: B is TRUE: If DFA M_1 accepts R_1 and DFA M_2 accepts R_2, the product construction gives a DFA for R_1 union R_2.
C is TRUE: If DFA M accepts R_1, swapping accepting and non-accepting states gives a DFA for Sigma* - R_1.