Prove that $\texttt{prefix}(L)$ is regular

Question

Given that $L = \lbrace 0^n1^n : n \geq 0\rbrace$ is a non-regular context-free language, prove that $\texttt{prefix}(L)$ is regular.

So far I have provided that the grammar to produce this language is: $S \rightarrow 0S1 \thinspace | \thinspace \epsilon$

Would you go about proving $\texttt{prefix}(L)$ is regular just like you would any language, proving that $\Sigma^\star$ = $\texttt{prefix}(L)$, or by induction on the length of the words in $\texttt{prefix}(L)$.

Answer 1

It appears that $prefix(L)$ is non-regular.

Let $L = \{0^n 1^n: n \ge 0\}$. Then, $prefix(L)$, defined to be the set of all prefixes of strings in $L$, is the set of all strings that consist of zero or more $0$'s followed by at most the same number of $1$'s, i.e. $prefix(L) = \{0^n 1^m: m \le n\}$. Intuitively, the reason $prefix(L)$ is nonregular is that a machine that can check whether the number of $1$'s does not exceed the number of $0$'s must store a count of the number of $0$'s in the input seen so far, and so this language cannot be recognized by a machine with a finite number of states.

A formal proof that $L' := prefix(L)$ is nonregular is as follows. Consider the subset $S = \{0^1, 0^2, \ldots, 0^p\}$. Any DFA $M$ that recognizes $L'$ must take the inputs $0^i$ and $0^j$ $(i < j)$ to different states because $0^i 1^j \notin L'$ but $0^j 1^j \in L'$. Hence, the machine $M$ contains at least $p$ states. But $p$ was arbitrary. This proves there does not exist a DFA (with a finite number of states) recognizing $L'$.

This proof method requires us to find an infinite subset $S$ of pairwise distinguishable prefixes; see online about the Myhill-Nerode theorem.