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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.
The closure property states that if L1 and L2 are regular languages, then their union L1 ∪ L2 is also a regular language. This means that any string belonging to either L1 or L2, or both, can be recognized by a finite automaton or expressed using a regular expression.
Closure under intersection If L and M are regular languages, then so is L ∩ M. We assume that the alphabets of both automata are the same ; that is Σ is the union of the alphabets of L and M, if they are different.
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.
To prove if a language is a regular language, one can simply provide the finite state machine that generates it. If the finite state machine for a given language is not obvious (and this might certainly be the case if a language is, in fact, non-regular), the pumping lemma for regular languages is a useful tool.
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.
A regular language satisfies the following equivalent properties: it is the language of a regular expression (by the above definition) it is the language accepted by a nondeterministic finite automaton (NFA) it is the language accepted by a deterministic finite automaton (DFA)
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.