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Skolem Arithmetic is the multiplication-flavored cousin of Presburger Arithmetic.

Presburger Arithmetic is a complete theory and listed as an example of a complete theory on the Wikipedia article. I was wondering whether Skolem Arithmetic was complete as well.

This article includes some interesting facts about Skolem Arithmetic and mentions in passing that it's a complete theory, but doesn't contain a proof that it's complete.

The article does mention that Skolem Arithmetic is the theory of the monoid $(M, \cdot)$, but since Peano Arithmetic isn't complete, I don't see a reason to believe a priori that the multiplicative fragment of PA would be either.

Is there a reference for the completeness of Skolem Arithmetic (or if the proof of completeness of SA is simple, that would also be fine)?

Greg Nisbet
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As is the case with Presburger arithmetic, Skolem arithmetic satisfies a weak version of quantifier elimination (see e.g. Emil Jerabek's old MO answer), and this can be used to prove completeness. However, I think it's also instructive to at least outline an approach via EF-games, if only because in this case that approach runs into a difficulty which really highlights the efficacy of syntactic (= quantifier-elimination-style) methods.


Suppose I have models $\mathcal{S}_1,\mathcal{S}_2$ of Skolem arithmetic. Let $P_1,P_2$ be their respective sets of primes. In each model $\mathcal{S}_i$, the powers of a given prime $p$ form a model of Presburger arithmetic $\mathcal{A}_p^i$, and the appropriate version of prime factorization ("two numbers divisible by the same powers-of-primes are equal") is provable in Skolem arithmetic, so we can conflate $\mathcal{S}_1$ with a particular substructure of $\prod_{p\in P_1}\mathcal{A}_p^1$ and similarly conflate $\mathcal{S}_2$ with a particular substructure of $\prod_{p\in P_2}\mathcal{A}^2_p$.

Now WLOG (Lowenheim-Skolem!) both $\mathcal{S}_1$ and $\mathcal{S}_2$ are countable, so we can assume that $P_1=\{p_i:i\in\omega\}$ and $P_2=\{q_i: i\in\omega\}$. Moreover, completeness of Presburger arithmetic tells us that for each $n,i$ $\mathsf{Duplicator}$ has a winning strategy $\Sigma_{n,i}$ in the length-$n$ EF-game $\mathsf{EF}_n(\mathcal{A}^1_{p_i}, \mathcal{A}^2_{q_i})$. Fixing $n\in\omega$ there is a natural way to try to paste the $\Sigma_{n,i}$s together to form a winning strategy $\Sigma$ for $\mathsf{Duplicator}$ in $\mathsf{EF}_n(\mathcal{S}_1,\mathcal{S}_2)$: basically, we just "play coordinatewise." E.g. if $n=1$ and player $\mathsf{Spoiler}$ opens with the element $f\in\mathcal{S}_1$, player $\mathsf{Duplicator}$ should respond with the element $g\in\mathcal{S}_2$ satisfying $$g(i)=\Sigma_{1,i}(f(i)).$$

Unfortunately, this doesn't quite work: how do we know that the $g$ this suggests is actually an element of $\mathcal{S}_2$ after all? The issue is supports: one structure may have elements divisible by (externally) infinitely many primes while the other may not.

This can be fixed by showing that each model $\mathcal{S}$ of Skolem arithmetic is elementarily equivalent to its "finite-support substructure," that is, the substructure of $\mathcal{S}$ generated by the powers-of-primes of $\mathcal{S}$ itself. But offhand I don't see a way to do this that doesn't essentially use the same ideas of the quantifier elimination argument. So, while I love EF-games and do think that the picture of decomposing a model of Skolem arithmetic into a bunch of models of Presburger arithmetic is important, in this particular case quantifier-elimination-type results are probably the way to go.

Noah Schweber
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