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To be more specific, does there exist a decision problem $P$ such that

  • given an oracle machine solving $P$, the Halting problem remains undecidable, and
  • given an oracle machine solving the Halting problem, $P$ remains undecidable?
David Zhang
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2 Answers2

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Yes.

It will be convenient to introduce some jargon and notation. The notion of computing from an oracle gives rise to the preorder of Turing reducibility (or "relative computability," or similar), which we denote "$\le_T$." This in turn gives rise to an equivalence relation on decision problems: $A\equiv_T B$ if $A$ computes $B$ and $B$ computes $A$. The resulting partial order of $\equiv_T$-classes is called the (poset of) Turing degrees, and is generally denoted by "$\mathscr{D}$" or similar.

The properties of this partial order have been extensively studied. Some results include:

  • Given any nonzero (= not the degree of the computable sets) Turing degree $\bf d$, there is a $\bf\hat{d}$ which is incomparable with $\bf d$ (this answers your question).

  • In fact, any (nonzero) Turing degree is contained in an antichain of degrees of size continuum.

  • Every Turing degree is above only countably many degrees, so the "height" of the Turing degrees is $\omega_1$; in case $\mathsf{CH}$ fails, this means the poset of Turing degrees are "wider" than it is "tall" (and even if $\mathsf{CH}$ holds this is still a useful heuristic in certain ways).

  • The Turing degrees form an upper semilattice - given decision problems $A, B\subseteq\omega$, the join of their degrees is the degree of $\{2n: n\in A\}\cup\{2n+1: n\in B\}$. Moreover, this semilattice has no top element (because of the "relativized" halting problem, or Turing jump).

  • However, their exist Turing degrees $\bf d$, $\bf \hat{d}$ with no greatest lower bound. So $\mathscr{D}$ is not a lattice.

  • Moreover, nontrivial infinite joins never exist (Exact Pair Theorem): given an infinite set of degrees $D$, if $D$ is nontrivial (= does not have a finite subset $D_0$ such that $\forall {\bf d}\in D\exists {\bf d_0}\in D_0({\bf d}\le_T {\bf d_0})$) and ${\bf a}$ is an upper bound of $D$, then there is a degree ${\bf b}$ which also is an upper bound of $D$, and which is incomparable with ${\bf a}$; moreover, $\{{\bf d}: {\bf d}\le_T {\bf a}$ and ${\bf d}\le_T {\bf b}\}=D$.

  • If $A'$ and $B'$ are the halting problems of $A$ and $B$, and $A\le_T B$, then $A'\le_T B'$. This means the jump can be thought of as an operation on degrees, not just sets. It turns out this operation is definable just in terms of the partial ordering! This was an extremely surprising result; see these notes of Slaman.

  • The converse of the above bullet point fails extremely badly: the jump is extremely non-injective. For example, for every ${\bf d}\ge_T{\bf 0'}$ there is a ${\bf c}$ such that ${\bf c'}={\bf d}$, ${\bf c}>_T{\bf 0}$, and there is no degree strictly between ${\bf c}$ and ${\bf 0}$ (such degrees are called "minimal," with the more exact phrase "minimal nonzero" being a bit long to say).

The above is all about the global theory of the Turing degrees; people have also studied extensively the local theory of special subclasses of Turing degrees. The best known such class is the class of degrees of domains of partial computable functions, the c.e. degrees. Of course, the degree of the halting problem is the maximal c.e. degree, so in this context the answer to your question is “no;” however, given any c.e. degree which is not the degree of the halting problem or the computable sets, there is an incomparable c.e. degree. This is the Friedberg-Muchnik Theorem, and its proof was a precursor to the method of forcing in set theory.

Let me end by stating my favorite two open questions about the Turing degrees:

  • Does the poset of all Turing degrees have any nontrivial automorphisms?

  • Does the poset of c.e. degrees have any nontrivial automorphisms?

Back in the day both these degree structures were believed to be very homogeneous, with lots of automorphisms; the theme of pure computability theory post-1955ish, however, was quite the opposite: the Turing/c.e. degrees are structurally rich, with lots of definable subclasses, and in fact it is now believed that both partial orders are rigid. All that is currently known however is that the automorphism groups are at most countable.

Finally, having said that the answer to your question is “yes,” let me explain a sense in which the answer to your question is “no.” The only “natural” increasing functions on the Turing degrees which have been discovered so far are essentially iterates of the Turing jump. Martin conjectured that this holds for "reasonable" functions. Ignoring some subtleties around the notion of iteration here there are a few ways to make this conjecture precise, the two most common being the following:

(1) $\mathsf{ZFC}$ should prove that every Borel function which is degree-invariant and increasing is an iterate of the jump, at least when we restrict to some set of the form $\{{\bf d}:{\bf d}\ge_T{\bf c}\}$ (that is, "on a cone").

(2) $\mathsf{ZF+DC+AD}$ should prove that every increasing function whatsoever is an iterate of the jump on a cone - the point being that under $\mathsf{ZF+DC+AD}$ all functions are reasonably nice.

  • This can be "$\mathsf{ZFC}$-ified" to the following (3) which is more ambitious in conclusion but more demanding in hypothesis than (1): "assuming appropriate large cardinals, every increasing function on $\mathscr{D}$ which is in $L(\mathbb{R})$ is an iterate of the jump on a cone," since large cardinals yield determinacy in $L(\mathbb{R})$. See this survey article of Larson for background on determinacy, large cardinals, and tameness properties.

Currently weak versions of Martin’s conjecture have been proved, although the full conjecture is still very much open.

Noah Schweber
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  • Very thorough answer! If you'll allow me to go one step further, are you aware of any explicit examples of decision problems with independent undecidability from the Halting problem? – David Zhang Jun 16 '15 at 15:16
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    No, and in fact none are known: this is part of the motivation behind Martin's conjecture, that we don't know of any natural degrees "off the beaten path" of iterates of the jump of 0. – Noah Schweber Jun 16 '15 at 15:21
  • However, such degrees do have explicit descriptions: you can write down a $\Sigma^0_2$ formula $\varphi$ such that ${x: \varphi(x)}$ is Turing incomparable with $0'$. Unfortunately, it's not unique even up to $\equiv_T$. So, one might say that there are explicit examples, but no natural examples. – Noah Schweber Jun 16 '15 at 15:22
  • Looking instead at natural examples of c.e. sets, about which more is known: there are some precise negative results known in this direction. For example, I believe (I'm blanking on the reference) that Sacks showed there is no $e$ such that for all $X$, $X<_TW_e^X<_TX'$, that is, we can't find intermediate c.e. degrees uniformly. On the other hand, this leaves "unrelativizing" candidates. For example, it is possible (but extremely unlikely) that the set of (codes for) polynomials with rational solutions is of intermediate c.e. degree; see http://qcpages.qc.cuny.edu/~rmiller/slides/IMS2015.pdf. – Noah Schweber Jun 16 '15 at 16:12
  • @NoahSchweber I'm also curious to see this $\Sigma_2^0$ formula. In what sense is it not "unique"? (One presumes that a formula describes a set with a well-defined Turing degree.) – Mario Carneiro Jun 17 '15 at 15:18
  • This falls with the halting problem. Only with possession of the solution to the halting problem will you be able to see this. – Joshua Jun 17 '15 at 15:28
  • @MarioCarneiro it's not unique in the sense that there are infinitely many $\Sigma^0_2$ formulas, each defining Turing degrees incomparable with the halting problem, which define different Turing degrees. So no one of these $\Sigma^0_2$ degrees is, on the face of things, particularly special. We can pick the one with the shortest $\Sigma^0_2$ description, but since that depends on how we encode formulas, even that doesn't seem natural. – Noah Schweber Jun 17 '15 at 15:32
  • @Joshua, what do you mean exactly? I don't understand your comment. – Noah Schweber Jun 17 '15 at 15:32
  • @NoahSchweber How are they parameterized? You mention that there is an antichain of size continuum, but they can't all have descriptions because there are only countably many formulas. Also for the purposes of this question it would seem appropriate to just pick one rather than give an overview of the field (which is interesting but doesn't really answer the question). (Or at least link to something that contains such an explicit description.) – Mario Carneiro Jun 17 '15 at 15:48
  • The minimal solution to the self-halting problem is extremely powerful, and allows all statements to be classified as true, false, paradoxical, independent, or syntax error. – Joshua Jun 17 '15 at 15:51
  • Um what? First, what is "the minimal solution?" The halting problem (or "self-halting problem" as you write) is a uniquely defined set; what is "minimal" doing here? More importantly, no, it does not let you determine the truth value of all first-order statements; rather, it just gives a uniform procedure for telling if a sentence is provable from a computable set of axioms. (And since we're looking at first-order logic, you don't need to include "paradoxical" in there.) But this isn't enough to answer general $\Sigma^0_2$ questions, so the halting problem doesn't help here at all. – Noah Schweber Jun 17 '15 at 16:30
  • There are plenty of natural examples of $\Sigma^0_2$ sets which are not computable relative to the halting problem: for instance, $\overline{Tot}={e: W_e\not=\omega}$, $Fin={e: \vert W_e\vert<\aleph_0}$, or ${e: \exists x\forall y(\langle x, y\rangle\not\in W_e)}.$ These aren't relevant to the OP, since they're not Turing-incomparable with the halting problem; they each have Turing degree ${\bf 0''}$, the halting problem's halting problem. But they illustrate the weakness of the halting problem beyond the $\Delta^0_2$ realm. – Noah Schweber Jun 17 '15 at 16:34
  • And, looking beyond first-order formulas: if we take, say, second-order logic with the standard semantics, the set of validities (not "provable statements" - there is no proof system for standard-semantics second-order logic) is mind-bogglingly complex; even if we only ask about sentences with one second-order quantifier, we're already looking at a set more complicated the jump iterated any computable number of times - $\omega$-many jumps, $\epsilon_0$-many jumps, etc. won't reach it. (See https://en.wikipedia.org/wiki/Hyperarithmetical_theory.) Halting is not all-powerful! – Noah Schweber Jun 17 '15 at 16:49
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    @MarioCarneiro Sorry, I missed your comment. The construction is a nice one, and goes as follows: we build an infinite perfect subtree $T$ of the full binary tree $2^{<\omega}$ such that any two paths through $T$ have incomparable Turing degree. Now since $T$ is perfect, there is a natural bijection $f$ between $2^\omega$ and $[T]$ (the set of paths through $T$). Moreover, $f$ is computable relative to $T$, in the sense that there is a natural number $e$ such that for all reals $X$, $$f(x)=\Phi_e^{T\oplus X}.$$ – Noah Schweber Jul 27 '15 at 02:59
  • Alternatively, here's a really silly argument: add a perfect set of mutually generic reals, then use Shoenfield absoluteness. :P – Noah Schweber Jul 27 '15 at 03:02
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The Wikipedia article on Turing degree states that for every non-zero degree there is an incomparable degree. In your case, this means that there is some degree which is incomparable with 0', and any member of this degree answers your question.

Yuval Filmus
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