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This seems like it should be a pretty well-studied question but I can't seem to find an easy answer:

Is the theory $(\mathrm{Ord}^{<\varepsilon_0}, +, \omega^{\ \cdot}, 0, 1)$ decidable?

From Is the theory of $(\operatorname{Ord}, <)$ decidable? I gather that the theory with just $<$ is decidable, and, of course, $(\mathrm{Ord}^{<\omega}, +, 0, 1)$ is just Presburger arithmetic. $<$ can be written as $x < y \ \Longleftrightarrow\ \exists z, y \neq 0\ \wedge\ x + z = y$, as usual.

The accepted answer to Is ordinal arithmetic more complicated than classical arithmetic? mentions that the theory with addition over all of $\mathrm{Ord}$ can be reduced to that of $\omega^{\omega^\omega}$, though they do not mention whether that is decidable (perhaps the paper does), and this does not include the rather natural $\omega^{\ \cdot}$ operation.

In the other direction it's also not obvious whether $\omega^{\ \cdot}$ allows one to define multiplication (though I suspect not).

What is known in this area?

This seems like it should be a pretty well-studied question but I can't seem to find an easy answer:

Is the theory $(\varepsilon_0, +, \omega^{\ \cdot}, 0, 1)$ decidable?

From Is the theory of $(\operatorname{Ord}, <)$ decidable? I gather that the theory with just $<$ is decidable, and, of course, $(\omega, +, 0, 1)$ is just Presburger arithmetic. $<$ can be written as $x < y \ \Longleftrightarrow\ \exists z, y \neq 0\ \wedge\ x + z = y$, as usual.

The accepted answer to Is ordinal arithmetic more complicated than classical arithmetic? mentions that the theory with addition over all of $\mathrm{Ord}$ can be reduced to that of $\omega^{\omega^\omega}$, though they do not mention whether that is decidable (perhaps the paper does), and this does not include the rather natural $\omega^{\ \cdot}$ operation.

In the other direction it's also not obvious whether $\omega^{\ \cdot}$ allows one to define multiplication (though I suspect not).

What is known in this area?