Are $A$ and $B$ necessarily be decidable if $(A∩\overline{B})∪(\overline{A}∩B)$ is decidable and $A$ & $B$ being exhaustive? I found the following question 


  Suppose A and B are recursively enumerable languages such that $A∪B=Σ^∗$. Further, suppose $(A∩\overline{B})∪(\overline{A}∩B)$ is decidable. Which of the following is true? 
  A- It is possible that either A or B is decidable but not both 
  B- None of them can be decidable 
  C- Both of them are regular 
  D- Both of them are decidable 


My attempt: It is possible for $A$ &amp; $B$ to be mutually exclusive and yet be exhaustive. Now this can happen even when one of them is not decidable. So $A$.

But answer given id $D$. Are my arguments flawed? If yes, how and where?



The source of the question is reputable test series, but still there are chances of it being wrong.
It is possible for $A$ &amp; $B$ to be mutually exclusive and yet be exhaustive. Now this can happen even when one of them is not decidable. So option A.


The analysis above is correct except the last conclusion "So option A". We cannot exclude the possibility that $A$ and $B$ are both decidable. In fact, it happens in the current situation because of the extra condition, as shown in the latter part of this answer.



Hints, try using the following facts. If you do not know them, prove them first.


For any two languages $X$ and $Y$, if $X$ is exhaustive and $Y$ is decidable, then $X\cap Y$ is exhaustive. (Moreover, $X\setminus Y$ is exhaustive, too. We do not need this fact, though.)
For any language $X$, $X$ is decidable if (and only if) both $X$ and $\overline X$ are exhaustive.




Here is how we can find both $A$ and $B$ are decidable.

Since $A$ is exhaustive and $(A\cap \overline B)\cup(\overline A\cap B)$ is decidable, $A\cap ((A\cap \overline B)\cup(\overline A\cap B))=A\cap\overline B$ is exhaustive. Since $A\cup B=\Sigma^*$, $A\cap\overline B=\overline B$. Since $B$ and $\overline B$ are exhaustive, $B$ is decidable. 

Symmetrically, $A$ is decidable.