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Prove/ Disprove:

  • For every nontrivial $A,B\in R$, $A\le_m B$
  • For every nontrivial $A,B\in RE$, $A\le_m B$

trivial set is the empty-set or $\Sigma^*$.

So basically the question is if for every two nontrivial sets there is a computable function from $A$ to $B$.

I'd be glad to get hint/guidance since I'm kinda stuck and not sure how to attack this problem.

Thanks.

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closed as unclear what you're asking by D.W., David Richerby, Juho, Thomas Klimpel, André Souza Lemos Jun 15 '15 at 0:38

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ What have you tried? Where did you get stuck? We want to help you understand concepts, but just doing your exercise for you won't be helping you (or anyone else). We expect you to make a serious effort on your own before asking and show us in the question what you tried. Also, please ask only one question per question. $\endgroup$ – D.W. Jun 14 '15 at 6:06
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    $\begingroup$ Incidentally, you've been given this feedback before: 1, 2, 3, 4. $\endgroup$ – D.W. Jun 14 '15 at 6:11
  • $\begingroup$ FWIW, here's a duplicate with answers. $\endgroup$ – Raphael Jun 15 '15 at 21:38
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HINT:

Go again over the definition of reductions. The many-one reduction from $A$ to $B$ is a function $f: \Sigma^* \to \Sigma^*$ that satisfies:

  1. $f$ is complete (=defined on any input)
  2. $f$ is computable (hint, hint).
  3. $f$ is valid: $x\in A$ if and only if $f(x) \in B$.

The "non trivial" part means there exists $x \in A$ and $x' \notin A$, and more importantly - the same holds for $B$. Now it should be fairly easy for you to complete the answer of your question.

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  • $\begingroup$ Got it. we choose $a\in B$ and $b\notin B$ and the rest is simple... $\endgroup$ – Elimination Jun 14 '15 at 12:12

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