Essential incompleteness without Robinson arithmetic? One main method for proving essential incompleteness of a theory is by interpreting Q (or perhaps just R). Many theory weaker than those have been found to be essentially incomplete, but they all turned out to be capable of interpreting them.
So is there a theory that is essentially incomplete that is impossible to interpret the above theories? Perhaps something that is incomplete but is not arithmetic in nature?
 A: We can create very artificial essentially incomplete theories, which have really no interpretability strength.
Work in a language with infinitely many unary predicate symbols $U_i$ ($i\in\mathbb{N}$) and one constant symbol $c$ (we don't have to work in this language but it makes things simpler), let $\psi$ be the sentence "There is exactly one element," and let $\varphi_i$ be the sentence $U_i(c)$. We build a theory $S$ consisting of $\psi$, some instances of $\varphi_i$, and some instances of $\neg\varphi_j$, as follows:
Let $T_e$ denote the $e$th recursively enumerable theory, and let $T_{e, s}$ be an enumeration of $T_e$ such that $T_{e, s+1}\setminus T_{e, s}$ has at most one element (so we're putting things into $T_e$ one by one). We let:


*

*$S_{yes}=\{\neg\varphi_e: \exists s(\varphi_e\in T_{e,s}, \neg\varphi_e\not\in T_{e, s})\}$,

*$S_{no}=\{\varphi_e: \exists s(\neg\varphi_e\in T_{e,s}, \varphi_e\not\in T_{e, s})\}$,

*$S=S_{yes}\cup S_{no}\cup\{\psi\}$.
$S$ is recursively enumerable, hence by Craig's trick is recursively axiomatizable, and is consistent; but has no recursively enumerable completions. And it's easy to show that it does not interpret $Q$ - or really, anything interesting at all.
