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.
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.
Concatenation. If L1 and L2 are languages, then the concatenation of the two languages, L = L1 · L2, is the set of all strings of the form x1x2 where x1 ∈ L1 and x2 ∈ L2. Theorem If L1 and L2 are regular languages, then the new language L = L1 · L2 is regular.
Closure properties on regular languages are defined as certain operations on regular language that are guaranteed to produce regular language. Closure refers to some operation on a language, resulting in a new language that is of the same “type” as originally operated on i.e., regular.
This can be achieved by combining the finite automata or regular expressions for L1 and L2 appropriately. The closure under concatenation is a property of regular languages that states if we concatenate two regular languages together, the resulting language will also be regular.
Recursively enumerable languages are also closed under intersection, concatenation, and Kleene star. Suppose that M1 and M2 accept the recursively enumerable languages L1 and L2. We need to show that if w is in our new language, it will be accepted.
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.
Closure properties on regular languages are defined as certain operations on regular language that are guaranteed to produce regular language. Closure refers to some operation on a language, resulting in a new language that is of the same “type” as originally operated on i.e., 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).
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.