Prove the existence of $C\in L_{regular}$ so that: $A \prec C \prec B $

Question

Given $A,B$ regular languages. Prove the existence of $C\in L_{regular}$ so that: $A \prec C \prec B $
Whereas $A\prec B$ stands for: $A\subset B $ and $B\setminus A $ is infinite regular language.
I tried to go for: $C=\overline{B} \cup A$ and some other options but it didn't work out.

EDIT: It is also given that: $A\prec B$.

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

The problem statement is not clear, but I assume $A≺B$, So $D=B-A$ is an infinite regular language. Now as shown here, $D$ can be written as $D=L_1\cup L_2$, where $L_1$ and $L_2$ are infinite regular languages and $L_1 \cap L_2=\varnothing$. Consider $C=A \cup L_1$.

edited Apr 13 '17 at 12:20

Community

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answered Dec 05 '13 at 16:27

hhsaffar

2,937

Thanks for the answer, I checked again and your assumption is correct it is given that: $A\prec B$. I edited the question accordingly. – Dec 05 '13 at 22:16

Prove the existence of $C\in L_{regular}$ so that: $A \prec C \prec B $

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