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Questions related to computability theory, a.k.a. recursion theory

1
vote
The flaw is that $R_\Gamma$ includes not just the natural number $10$, but any natural number which gives rise to the same partial recursive function. It will be infinite, and not recursively enumerab …
answered Apr 12 '17 by 6005
0
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I'm not sure why I thought Zorn's lemma doesn't work, because it does, easily, and gives a stronger result than just a single example. Let's say that a set of Turing machines $S$ is self-deciding if, …
answered Feb 28 '17 by 6005
1
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Typically, no, this is not possible. It depends on how the infinite set is represented. (I'm assuming it's a set of integers.) The usual way to represent a possibly-infinite set $S$ is as a Turing mac …
answered Jan 13 by 6005
4
votes
It is known that the halting problem is undecidable even when we fix either the Turing machine $M$ or the input $w$. You have to be more careful about this statement. It's not true for any fixed …
answered Apr 2 '20 by 6005
1
vote
The halting program, $halt?$, does not exist, and neither does $D$. In reality, then, your $A'$ and your $A$ are identical. The halting problem cannot be solved for either of them. However, on second …
answered Apr 12 '17 by 6005
0
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Regarding your answer, here are some corrections. For part (1), your argument is partially correct but incorrectly stated. For part (2), your argument is not correct. Your biggest mistake is in the wa …
answered Jan 7 by 6005
5
votes
It will just be too hard, because you can exploit this computability to then solve hard problems by encoding them as part of a TM input to the effective reduction. …
answered Feb 25 '20 by 6005
3
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The question in the title is subjective, but I suspect no. Hilbert's 10th is one of the most important, but also one of the most complex, results of the last century. The proof itself spans 21 years o …
answered Dec 7 '21 by 6005