Sell Closure Property For Regular Language In Fulton
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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.
The set of regular languages is closed under complementation. The complement of language L, written L, is all strings not in L but with the same alphabet. The statement says that if L is a regular lan- guage, then so is L. To see this fact, take deterministic FA for L and interchange the accept and reject states.
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).
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.
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.
Regular languages are closed under reversal, meaning if L is a regular language, then its reversed language LR is also regular. This is proven by creating a new automaton that reverses the transitions of the original DFA. Thus, the reversed language is also accepted by a finite automaton, confirming its regularity.
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.
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.
