# Proving OP(L) is regular [closed]

I did some searching before I decided to ask this question and there was nothing similar to my question that helped me. So I came to CS stack-exchange for hints. So, I am currently working on a proof to show that if $L$ is regular then $OP(L)$ is also regular without the use of pumping lemma. For the sake of this question, I am using $\{0,1\}$ as alphabet.

$$OP(L) = \{w \mid wz ∉ L \text{ for every } z ∈ \{0,1\}^+ \}.$$

At the moment, I have a hard time starting the proof. What I understand is that, if $L$ is regular then we can define a DFA that can decide $L$. And then we can define another machine for $OP(L)$ with every transition from $M$ along with $\epsilon$-transitions.

Am I on the right track? I could really use some hints.

• – J.-E. Pin Oct 23 '17 at 10:09
• Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted. – D.W. Oct 23 '17 at 13:57
• I'm voting to close this question because it was cross-posted. – D.W. Oct 23 '17 at 13:57

Hint: Consider a DFA for $L$, and some word $w$. Suppose that upon reading $w$, the DFA reaches some state $q$. Is knowing $q$ enough for deciding whether $w \in OP(L)$?
A possibility is the use closure properties of the regular languages. The new language $\mathrm{OP}(L)$ can be written using concatenation and Boolean operations using the original $L$ and simple regular languages.