We know the Halting Problem is in the 1st level of arithmetical hierarchy. We know that the Totality Problem is in the 2nd level of arithmetical hierarchy. What are some examples of problems in the 3rd (or higher) level of arithmetical hierarchy. +1 for problems which instances are solved in the practical world (programmers, mathematicians, etc.).
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$\begingroup$ Any more help to make this an understandable question is appreciated. $\endgroup$– Otakar Molnár LópezJul 6, 2017 at 20:25
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1$\begingroup$ I'm not sure, but maybe deciding if two Turing (unequal) machines will generate same outputs for all inputs is harder problem than totality. $\endgroup$– rus9384Jul 6, 2017 at 20:37
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$\begingroup$ en.wikipedia.org/wiki/… $\endgroup$– D.W. ♦Jul 7, 2017 at 1:47
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$\begingroup$ @rus9384 I think it should be equivalent: you can write it in the form $\forall x.\exists y.p(x,y)$ with $p$ recursive, as Totality. Here $x$ is the input, and $y$ is a number of steps large enough to make both TMs halt (once $y$ is known, checking that the output is the same is trivial). $\endgroup$– chiJul 7, 2017 at 9:01
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$\begingroup$ What about $n^{th}$-order logic? Maybe evaluating if statement in such logic is true is a complete problem for $\Sigma^0_n$? $\endgroup$– rus9384Jul 10, 2017 at 0:03
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1 Answer
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Let $\{W_e\}_{e\in\mathbb N}$ be a canonical list of the recursively enumerable, i.e., computably enumerable, i.e., Turing recognizable, languages (or sets).
The property that $W_e$ is actually recursive, i.e., computable, is $\Sigma^0_3$: $$\exists d\forall n(n\in W_e\leftrightarrow n\not\in W_d).$$
It is at the 3rd level of the hierarchy: it is not of lower complexity than $\Sigma^0_3$. See e.g. Soare: Recursively enumerable sets and degrees, Theorem 3.4, for the proof.
answered Sep 30, 2018 at 6:59