Question

Show that if the languages L1 and L2 are regular, then the following language is regular:

1) L1 ∩ L2 results in a regular language because we can picture the result as a combination of L1 AND L2 which is concatenation and regular languages are closed under concatenation.
2) L1 ∩ L2 = (LI' ∪ L2").Since regular languages are closed under complement and union, they are also closed under intersection.
3) The languages created by intersection is never regular.
4) Regular languages are not closed intersection, therefore this cannot be guaranteed.

Answer 1

Intersection of two regular languages L1 and L2 yields another regular language because regular languages are closed under the operations of complement, union, and thus, intersection.

Step-by-step explanation:

To show that the intersection of two regular languages L1 and L2 is also a regular language, we can rely on the properties of regular languages. Regular languages are closed under various operations, including intersection.

The first assertion that L1 ∩ L2 is a result of concatenation is incorrect; intersection operation is not the same as concatenation. Instead, the correct process is explained in statement 2, where we use the concept of complement and union to demonstrate the closure of regular languages under intersection.

To show that L1 ∩ L2 is regular, taking complements of both L1 and L2, which gives us L1' and L2', is a valid step because regular languages are closed under complementation. Then we can take the union of L1' and L2', denoted as L1' ∪ L2', which is also regular as regular languages are closed under union. Finally, taking the complement of this union results in the intersection of L1 and L2 (De Morgan's laws), which must be regular because the regular languages are closed under complementation.

Therefore, assertion 3 and 4 are incorrect. Regular languages are closed under intersection, and it is possible to guarantee that the language created by intersecting two regular languages is regular.