Closure under Union For any regular languages L and M, then L ∪ M is regular. Proof: Since L and M are regular, they have regular expressions, say: Let L = L(E) and M = L(F). Then L ∪ M = L(E + F) by the definition of the + operator.
Intersection. Theorem If L1 and L2 are regular languages, then the new language L = L1 ∩ L2 is regular. Proof By De Morgan's law, L = L1 ∩ L2 = L1 ∪ L2. By the previous two theorems this language is regular.
Let L be a regular language, and M be an NFA that accepts it. Here, δR is δ with the direction of all the arcs reversed. Thus, it is proved that L is closed under reversal.
Proof: Observe that L \ M = L ∩ M . We already know that regular languages are closed under complement and intersection.
Regular Languages are closed under intersection, i.e., if L1 and L2 are regular then L1 ∩ L2 is also regular. L1 and L2 are regular • L1 ∪ L2 is regular • Hence, L1 ∩ L2 = L1 ∪ L2 is regular.
Regular languages are closed under concatenation - this is demonstrable by having the accepting state(s) of one language with an epsilon transition to the start state of the next language. If we consider the language L = {a^n | n >=0}, this language is regular (it is simply a).
What's more, we've seen that regular languages are closed under union, concatenation and Kleene star. This means every regular expression defines a regular language.
Closure Properties of Regular Languages Given a set, a closure property of the set is an operation that when applied to members of the set always returns as its answer a member of that set. For example, the set of integers is closed under addition.
3 The Regular Languages are Closed under Reverse Homomorphism. A reverse homomorphism replaces entire strings in a language by individual symbols. This is fairly easy to envision in a “set of strings” view, e.g., if I had a language of all strings ending in “aa”: {aa,aaa,baa,aaaa,abaa,baaa,bbaa,…}
Reversal. Statement: Under reversal, the set of regular languages is closed. Proof: Let M be a deterministic finite automaton that accepts L; we will create M' from M so that M and M' states are the same. Make the final state of M the accepting state of M' and the final state of M the beginning state of M'.