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For A,B that are not decidable, does AB U BA not necessarily decidable?

I think that the answer is NO. Not necessarily. I thought about the following example, but it does not refute exactly:

If we take A and A complement, and without losing generality the empty word is in A, than every word in A' is also in AA' U A'A, but I can't use it to show that every word in A is also in AA' U A'A.

If someone can help me how can I show that also A is in AA' U A'A the problem will be solved.

thank you very much.

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Take some undecidable language $A$ and choose $B=A'\bigcup \{\epsilon\}$. $B$ is also undecidable, and the rest of the proof is just as what you have already done.

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