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Let $\Sigma$ be an alphabet, and suppose that $A$, $B \subseteq \Sigma^*$ are Turing recognizable languages where both $A \cup B$ and $A \cap B$ are decidable. Prove that $A$ is decidable.

Is this true? I have tried to prove it by considering the intersection and complement of the sets, but I think I am missing something.

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Given an $x$ (for which you want to check membership in $A$), I would start by checking if $x \in A \cap B$, which is possible, since $A \cap B$ is decidable. If $x \in A \cap B$ you also know that $x \in A$.

If $x \not\in A \cap B$, you can continue and ask whether $x \in A \cup B$.

If so, either $x \in A$ or $x \in B$. Since $A$ and $B$ are Turing recognizable, you can enumerate both sets in an interleaved fashion. Let $a_i \in A$ and $b_i \in B$ be the respective elements in the enumeration. Since $x \in A \cup B$, there is an $i$, such that $a_i = x$ or $b_i = x$, but not both. If $a_i = x$, then $x \in A$, else $x \not\in A$.

If $x \not\in A\cup B$, then $x \not\in A$.

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  • $\begingroup$ You might consider not encouraging undesirable posting behaviour in the future. Up to you. Your decision -- I respect whatever you decide, either way. $\endgroup$ – D.W. Jun 27 '16 at 16:50

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