Answer :
L₂ is not regular
To answer this question, we need to understand a few concepts from the theory of formal languages and automata. One of these concepts is the notion of a regular language. Regular languages are a class of languages that can be recognized by finite automata. There are several closure properties related to regular languages, which tell us about the regularity of languages obtained by performing certain operations on other regular languages.
Let's go over these properties step by step:
Closure under union, concatenation, and Kleene star:
- Regular languages are closed under the operation of union. If L₁ and L₂ are regular languages, then L₁ ∪ L₂ is also regular.
- They are also closed under the operation of concatenation. If L₁ and L₂ are regular, then the concatenation L₁L₂ is regular.
- Lastly, regular languages are closed under the Kleene star operation. If L is a regular language, then L* (the set of all strings that can be formed by concatenating zero or more strings from L) is also regular.
However, we are particularly interested in whether regular languages are closed under the operation of taking quotients. The quotient of L₁L₂ by L₁ is basically the language L₂ (though this is an intuitive rather than formal explanation). If L₁L₂ is regular and L₁ is regular, one might ask whether L₂ must be regular.
The class of regular languages is *not* closed under the operation of taking quotients with arbitrary languages. Therefore, knowing that L₁ is regular and that L₁L₂ is regular does not give us enough information to conclude that L₂ is regular. To see why, consider a counterexample where L₂ is an inherently non-regular language (for instance, a language that requires a non-constant amount of memory to recognize, such as the classic {a^n b^n | n ≥ 0}, which is the set of strings of 'a's followed by an equal number of 'b's). It is possible to construct a regular language L₁ such that the concatenation L₁L₂ is regular, but L₂ by itself is not regular.
