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Prove the existence of regular $C$ so that: $A \prec C \prec B $
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Problem
Given $A,B$ regular languages with $A \prec B$. Prove the existence of $C\in L_{\text{regular}}$ so that: $A \prec C \prec B$.
Here, $A\prec B$ stands for: $A\subset B $ and $B\setminus A $ is infinite.
I tried to go for: $C=\overline{B} \cup A$ and some other options but it didn't work out.
Here, $A\prec B$ stands for: $A\subset B $ and $B\setminus A $ is infinite.
I tried to go for: $C=\overline{B} \cup A$ and some other options but it didn't work out.
Solution
Hint: Note that $B \setminus A$ is regular by closure properties of $\mathsf{REG}$.
Since $B \setminus A$ is also infinite, you can find an infinite language $D \in \mathsf{REG}$ so that $D \prec B \setminus A$.
Since $B \setminus A$ is also infinite, you can find an infinite language $D \in \mathsf{REG}$ so that $D \prec B \setminus A$.
Context
StackExchange Computer Science Q#18646, answer score: 5
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