Consider the first-order language $\mathcal{L}\_{\text{OA}}:=(+,\cdot,0,1)$; in this language, we can formulate statements of ordinal arithmetic. Clearly, the theory $T\_{\text{OA}}$ of $(\text{On},+,\cdot,0,1)$ is not recursive, as $\omega$ is definable in $\mathcal{L}\_{\text{OA}}$ (as being the only ordinal $\gamma$ different from $0$, not having a predecessor and having no $\alpha$ and $\beta$ different from $0$ such that $\alpha+\beta=\gamma$ with $\alpha$ not having a predecessor) and hence the theory TA (true arithmetic) of $\mathbb{N}$ is computable from $T\_{\text{OA}}$. Note that $T\_{\text{OA}}$ is absolute between transitive class models of ZFC.


My question is whether the reverse reduction holds: Can we decide $T\_{\text{OA}}$, given TA? More precisely, what is the Turing degree of $T\_{\text{OA}}$?



Is ordinal arithmetic more complicated than classical arithmetic?
Using what are now called Ehrenfeucht–Fraïssé Games and extensions thereof, Ehrenfeucht showed that


$$(\mathrm{Ord},{<}) \sim (\omega^\omega,{<})$$
$$(\mathrm{Ord},{<},{+}) \sim (\omega^{\omega^\omega},{<},{+})$$
$$(\mathrm{Ord},{<},{+},{\cdot}) \sim (\omega^{\omega^{\omega^\omega}},{<},{+},{\cdot})$$


where $\sim$ denotes elementary equivalence. Since $(\omega^{\omega^{\omega^\omega}},{<},{+},{\cdot})$ is a computable structure, its first-order theory is computable from $0^{(\omega)}$ ($\equiv\_T TA$).


The question omits the relation $<$ but that doesn't matter for the last two results since $x < y$ is definable by $\exists z(x + 1 + z = y)$. [Thanks to Andrés Caicedo for pointing out the correct definition.]


*A. Ehrenfeucht*, [**An application of games to the completeness problem for formalized theories**](http://www.ams.org/mathscinet-getitem?mr=126370), *Fund. Math.* **49** (1960-1961), 129–141.


