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Prove/disprove: $\exists L \in RE/R$ such that $L^R \cup L \in R$

Where in my context, $R$ is the turing decidable, and $RE$ is the recursively enurmable. I tried to find such an $L$ but couldn't. What I know for sure is that I need a language in $RE/R$ such that $L \cup L^R = \Sigma^*$, or am I also wrong?

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  • $\begingroup$ Is $L^R$ the set of all reversed words of $L$, or is this something else? $\endgroup$ – G. Bach Jan 24 '14 at 23:35
  • $\begingroup$ Yes. Reversed words of $L$ $\endgroup$ – TheNotMe Jan 24 '14 at 23:46
  • $\begingroup$ Do you use $/$ to denote set difference or right quotient? $\endgroup$ – Raphael Jan 25 '14 at 16:25
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Since this sounds like a homework question to me, I will give the answer as a series of hints. (You should make a serious effort to solve the question after each hint, instead of just reading them all in short succession.)

First: Such an $L$ exists and $L \cup L^R$ is not necessarily $\Sigma^*$.

Try $L \cup L^R = \{a^nb^n\} \cup \{b^na^n\}$.


For each $n$, $L$ should contain exactly one of the words $a^nb^n$, $b^na^n$. How can you do this so that $L$ is in $RE/R$?

And finally the solution:

Let $K$ be any set in $RE/R$ and let $L = \{a^nb^n \mid n \in K\} \cup \{b^na^n \mid n \notin K\}$.

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