Is there a $\Delta^0_2$ real with "easy total computability problem"?

Question

This was asked at MSE without success. Granted, a bounty is still ongoing there, but it doesn't look like it will be answered.

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For (noncomputable) $A\subseteq\omega$ let $\tilde{A}=\{e: \varphi_e^A\mbox{ is total and }\exists c(\varphi_e^A\simeq\varphi_c)\}$. A priori $\tilde{A}$ is $\Sigma^0_3(A)$ ("$\varphi_e^A$ is total and there is some $c$ such that on all inputs we eventually see agreement between $\varphi_c$ and $\varphi_e^A$"). However, this bound isn't sharp in general: if $A$ is sufficiently Cohen generic then $\tilde{A}$ is $\Pi^0_2(A)$ (of course we can't do better than this: $\tilde{A}$ is always $\Pi^0_2(A)$-hard).

However, a fair amount of genericity (at a glance, $2$-genericity) is needed for that argument. This raises the question of how hard it must be to compute a real (nontrivially) satisfying "$\tilde{A}$ is not $\Sigma^0_3(A)$-complete." Specifically:

Does every noncomputable $\Delta^0_2$ set $A$ satisfy "$\tilde{A}$ is $\Sigma^0_3$-complete"?

Genericity-based arguments almost certainly won't be useful here, since there aren't even any $\Delta^0_2$ weak $2$-generics.

Answer 1

It seems to me that if $G$ is 1-generic and recursive in $0'$ then $\tilde{G}$ is Boolean $\Sigma^0_2$, which is sufficient to conclude that $\tilde{G}$ is not $\Sigma^0_3(G)$-complete.

First, show that if (1) there is a condition $p$ in $G$ such that there is no $\varphi_e$-splitting pair of conditions extending $p$ and (2) for every condition $q$ in $G$ and every $m$, there is an extension $r$ of $q$ which makes $\varphi_e(m)$ converge, then $e$ is in $\tilde{G}$. We can exhibit a $\varphi_c$ for $\varphi_e(G)$ by observing that the value of $\varphi_e(G)$ at $m$ can be recursively determined by searching for any extension of $p$ that gives a value at $m$. Second, show that if $e$ is in $\tilde{G}$ then (1) and (2) must hold. Obtaining (1) uses the 1-genericity of $G$ and obtaining (2) uses the totality of $\varphi_e(G)$.

Since $G$ is $\Delta^0_2$, Condition (1) is $\Sigma^0_2$ and (2) is $\Pi^0_2$, and so $\tilde{G}$ is Boolean $\Sigma^0_2$.