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Questions tagged [computability-theory]

computable sets and functions, Turing degrees, c.e. degrees, models of computability, primitive recursion, oracle computation, models of computability, decision problems, undecidability, Turing jump, halting problem, notions of computable randomness, computable model theory, computable equivalence relation theory, arithmetic and hyperarithmetic hierarchy, infinitary computability, $\alpha$-recursion, complexity theory.

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If $X'$ computes $\mathcal{O}^{Y}$ must $X$ compute $Y$? If not is there a function $\Gamma$ which guarantees that if $X'$ computes $\Gamma(Y)$ then $X$ computes $Y$? It is easy enough to see that ...
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Take any desired consistent c.e. theory $T$ asserting the basic facts of arithmetic such that $T$ can formalize the operation of Turing machines. There is some Turing machine program $p$ and input $n$ ...
Jack Edward Tisdell's user avatar
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TL;DR: I define a three-player game (Arthur, Nimue, Merlin) where Nimue is shown a hidden bit $b$ chosen by Merlin and tries to communicate it to her ally Arthur, but Arthur must act computably while ...
Gro-Tsen's user avatar
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Do all the Turing degrees agree, in a definable (hyperarithmetic? arithmetic?) way, on an ordering of their representatives? I'll make this precise below but roughly the question is whether there is ...
Peter Gerdes's user avatar
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1 answer
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We call a binary sequence $s:\mathbb{N}\to\{0,1\}$ supernormal if for every injective, increasing and computable function $\iota:\mathbb{N}\to\mathbb{N}$, the binary sequence $s\circ\iota:\mathbb{N}\...
Dominic van der Zypen's user avatar
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This is tangentially related to this old question of mine. Say that a clean well-ordering is a computable well-ordering $\triangleleft$ of $\mathbb{N}$ such that the following additional data is ...
Noah Schweber's user avatar
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1 answer
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Some people already have asked questions concerning Sy Friedman's results: (1) For $x\in\mathbb{R}$ if every $x$-admissible ordinals are stable, then $0^\#\in L[x]$. (2) There can be, by a class-...
Hiroshi Fujita's user avatar
21 votes
1 answer
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In the book, Chi Tat Chong and Liang Yu, "Recursion Theory, Computational Aspects of Definability", De Gruyter 2015, Exercise 5.3.5 (p.98) quotes S.G.Simpson's result as follows: If there ...
Hiroshi Fujita's user avatar
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This question is inspired by a recent Quanta article, which explained that in order to compute BB(6), it is necessary to solve an "antihydra problem" which is somewhat similar to the ...
Gabe K's user avatar
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It is a result of Levin and (independently?) Kautz that if $X \in 2^\omega$ is 1-ML random relative to a computable measure $\mu$ the either $X$ is computable (it is an atom of $\mu$) or $X$ is Turing ...
Peter Gerdes's user avatar
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Is choice needed to construct a free ultrafilter on the boolean algebra of computable sets? Inspiration I ask this purely out of curiosity and because it's a natural follow-up question to GVT's ...
Keith J. Bauer's user avatar
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3 answers
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Is it known to be consistent with ZFC for there to exist a Turing degree invariant projective set which neither contains nor is disjoint from a cone? What about in $L$, i.e., is it known that (the ...
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Lets consider two entities with different hyper-computational strength. Entity A is able to comprehend the whole continuum and hence it decides every arithmetic sentence and, assuming Projective ...
Pan Mrož's user avatar
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The least recursively inaccessible ordinal $I$, is, intuitively, the supremum of the ordinals that come from "recursively" iterating the function $\alpha\mapsto\omega^{CK}_\alpha$. For an ...
Reflecting_Ordinal's user avatar
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0 answers
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What do "Realizability of the axiom of choice in HOL" and "Realizability and the Axiom of Choice" mean when they claim they realize a non extensional version of $\sf AC$? Can they ...
Ember Edison's user avatar
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18 votes
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Basic background: A hydra is a finite rooted tree (with the root usually drawn at the bottom). The leaves of the hydra are called heads. Hercules is engaged in a battle with the hydra. At each step of ...
Monte_carlo's user avatar
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2 answers
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Suppose I want to uniformly represent computation in an infinite series, what is the smallest/most natural set of operations I need to express the $n$-th term that allows me to capture arbitrary ...
Peter Gerdes's user avatar
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1 answer
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Basic setups: A Turing machine M operates on a doubly infinite tape. The tape is divided into cells. Each cell can hold either a 0 or a 1 (meaning, we will consider only TM’s with two symbols). The ...
Monte_carlo's user avatar
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Goodstein's "Recursive Number Theory" (1957) presents a "logic-free" version of primitive recursive arithmetic: all statements in the logic are equalities of expressions involving ...
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Definition: “The” first Kleene algebra $\mathcal{K}_1$ is the set $\mathbb{N}$ of natural numbers endowed with the partial operation $(p,n) \mapsto p\bullet n := \varphi_p(n)$ where $\varphi_p$ is the ...
Gro-Tsen's user avatar
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10 votes
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I begin by recalling Noether's problem over $\mathbb{Q}$: Let $G$ be a finite group that act faithfully by field automorphisms on $\mathbb{Q}(x_1,\ldots,x_n)$, with the action on $\mathbb{Q}$ trivial. ...
jg1896's user avatar
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7 votes
1 answer
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I am looking for a topos that describes realizability by Turing machines with access to a “variable oracle”. I think the construction I want is this. Start with Baire space $\mathcal{N} := \mathbb{N}^...
Gro-Tsen's user avatar
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It is classical that the homeomorphism problem for finite simplicial complexes is unsolvable. All the sources I know for this actually prove something slightly different: Theorem: For $n \geq 5$, ...
Anonymous Math Person's user avatar
8 votes
1 answer
579 views

It is well-known that ZFC cannot prove the non-halting behavior of any more Turing machines than ZF alone can. However, does the addition of AC permit shorter minimum proofs of non-halting behavior in ...
Nick's user avatar
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13 votes
4 answers
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In logic, and I expect in mathematics more broadly, it seems like there is a special role played by notions like measure and (baire) category (as in meeting/avoiding dense sets). Obviously, these ...
Peter Gerdes's user avatar
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