Double Posner-Robinson Join (or a cupping analog of minimal pair)

Question

Are there incompatible degrees $D_0, D_1 <_T 0'$ such that for all $X$ if $D_0 \oplus X \equiv_T D_1 \oplus X \equiv_T 0'$ then $X \equiv_T 0'$? So kinda like a cupping analog of a minimal pair.

To put this another way, the Posner-Robinson join theorem gives us that for any $0 <_T A \leq_T 0'$ there is a degree $B <_T 0'$ such that $A \oplus B \equiv_T 0'$. Is there a two degree version of Posner-Robinson, i.e., for any incompatible degrees $D_0, D_1 <_T 0'$ there is a degree $B <_T 0'$ such that $B \oplus D_i \equiv_T 0'$

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I suspect that given $n$ incompatible degrees below $0'$ there is a degree below $0'$ joining all of them to $0'$ and that there is some nice simple proof. Let me observe that the two degree version would follow from combining Posner-Robinson with upper cone avoidance.

For, suppose that given $D_0, D_1 <_T 0'$ we can produce a degree $B_0$ via Posner-Robinson joining $D_0$ to $0'$ with $B_0 \ngeq_T D_1$. We could then apply Posner-Robinson again relativized to $B_0$ (as $D_1$ isn't computable in $B_0$) to produce $B >_T B_0$ which joins $D_1$ to $B'_0 \equiv_T 0'$ (as the degree produced by Posner-Robinson is low).

Answer 1

To be not so cheeky, I am giving a detailed answer for the following multiple version of PR-theorem.

Theorem: For any countable sequence $\{\mathbf{a}_n\}_{n\in \omega}$ non-zero Turing degrees, there is some degree $\mathbf{g}$ so that for any $n$, $\mathbf{a} _n\vee \mathbf{g}=\mathbf{g'}$.

I know four proofs of the theorem.

The first is by a slightly modification of Posner-Robinson's original argument. It was Joe who told me that $\Delta^0_2$ version of the theorem can be found in the P-R's original paper. Then Frank Stephan observed that, by a slight modification of their method, one may obtain the (almost-)full version.

The second proof is about Jockusch-Shore style proof, which is due to Kirill Gura, who was an undergraduate student in Madison. I learned the proof from Joe too. In Kirill's proof, a notion called Kalimullin pair, which is from the $e$-degree theory, was used.

The third proof is due to Stpehan, Tanuwijaya, Yang and me. It is also a Jockusch-Shore style proof. The key fact we used is that for any countably many non-zero degrees, each of them contains a set so that the union of them is immune but the intersection is infinite.

The fourth-proof is due to Slaman. And it is a pretty straightforward application of Kumabe-Slaman forcing. The proof can also be found in Kirill's draft.

Notice that $\mathbf{g}$ can be 1-generic in the both the second and third proofs. And no way in the other proofs. Ted told me that he and Kucera tried very hard to find a method to replace Kumabe-Slaman forcing, which, so far, has very limited applications such as to prove higher P-R theorem.

I got interested in the theorem when I was in a Dagstuhl meeting in 2017. Joe told me that he was thinking about a question, which was raised by Andrew Marks, about simultaneously randomness relative to multiple-degrees. Then I just recognized that maybe a multiple version of PR-theorem useful. Then he told me that it is already in P-R's paper.

That was all I remember.

Answer 2

As pointed out in the comments, this is theorem 2 (relativized in theorem 3) in "Degrees joining to 0'" by Posner-Robinson. I summarize the method used below.

In the 1 degree version they use the fact that if $0 <_T C <_T 0'$ there is some function $g(m)$ computable in $C$ that gives the first stage $s$ at which $C_s|_m$ is equal to $C|_m$. They then build a one generic $G = \cup G_k$ by letting $G_{k+1}= G_k\hat{}0^m\hat{}1\hat{}\sigma\hat{}0'(k+1)$ with either $\sigma = \langle\rangle$ and $G_{k+1}$ forcing divergence or with $\lvert G_{k+1}\rvert = g(m)$ and forcing $\phi_{k+1}$ to converge. Thus, given $C$, it's possible to recover the length of $\sigma$ at each level and thus recover $0'$. The existence of such a $\sigma$ is given by the fact that the function $f(m)$ giving the length of the least $G_k\hat{}0^m\hat{}1\hat{}\sigma$ forcing convergence would dominate $g(m)$.

Note that any degree computing a function which dominates $g$ is able to recover $0'$ since it can check if there is an appropriately short $\sigma$ forcing convergence.

The trick they use in that paper to get the $n$-degree version is to observe that if $0_T < C_i, i \leq n$ and $g_i$ is as above then and function $f$ dominating $g(x) = \min_{i \leq n} g_i(x)$ computes one of the sets $C_i$ (by induction, basically either $f$ dominates $g_{n}$ or you can enumerate evidence of this fact letting you compute a bound on $\max_{i < n } g_i$). Now the same argument works as above and any of the $C_i$ can recover $0'$. This extends to infinite sequences that are uniformly computable in $0'$ as well.

As I guessed, this can be combined with e-splitting to avoid computing any degree $0 <_T A_i, i \in \omega$ where $A_i$ is uniformly computable in $0'$.