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Context: Presburger arithmetic is the theory $\tau$ of structure $$ A = (\mathbb{N},0,1,+,\{c|\cdot\}_{c\in\mathbb{N}})$$ where for each integer $c > 1$, the unary predicate c|n holds if and only if n is divisible by c. Recall that $\tau$ has quantier-elimination.

The first part is to show that suppose a set $S \in \mathbb{N}$ is ultimately periodic if there exist positive integers $n_0$ and $p$ such that for all $n > n_0$, $n \in S$ iff $ n + p \in S$. Show that any quantifier-free formula that mentions a single variable $x$ defines an ultimately periodic subset of $\mathbb{N}$.

Afterwards we should be able to use this result to show that there is no formula on free variables $x$, $y$ and $z$ that defines the multiplication relation $M = \{(a,b,c)\in \mathbb{N}^3: ab = c\}$ on the structure $A$.

I'm completely lost on this question and I cannot see a relation on the two parts of the question. I have read the post here Presburger arithmetic but don't think it is too helpful. Any help would be appreciated!

Olivier Roche
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RnHdw
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    The connection between the two parts is that the set of squares, ${a^2:a\in\mathbb N}$ is definable using the predicate $M$ but is not ultimately periodic. – Andreas Blass May 30 '20 at 14:06
  • @AndreasBlass I see! In this case do you have a clue of showing part 1? – RnHdw May 30 '20 at 16:17
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    I think part 1 should follow directly from quantifier-elimination, by induction on quantifier-free one-variable formulas. But before doing that, check that "Recall that $\tau$ has quantifier-elimination" is really true; I think the vocabulary should also include $\leq$. – Andreas Blass May 30 '20 at 16:56
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    More precisely, the absence of $\leq$ from the vocabulary doesn't affect the questions much, but it does affect the claim of quantifier-eliminability. I see no way to express $(\exists z),x+z=y$ without quantifiers in the given vocabulary. – Andreas Blass May 30 '20 at 16:59
  • @AndreasBlass Thank you for your comments! The definition of Presburger arithmetic is given specifically as context and the fact that it has quantifier-elimination is proven previously so it is indeed true. – RnHdw May 31 '20 at 06:24
  • So how does that proof eliminate quantifiers from $(\exists z),x+z=y$ using only the specified vocabulary? – Andreas Blass May 31 '20 at 12:52

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Clearly, the unions, intersections and complements of ultimately periodic sets are again ultimately periodic. Since a quantifier-free formula is a boolean combination of atomic formulas, all we have to do is to prove the following claim :

Claim Atomic formulas in one variable define ultimately periodic sets.

Up to equivalence, there are of two kinds of atomic formulas in the variable $x$ :

  1. $x + n = m$ for $n, m \in \mathbb{N}$.
  2. $c \mid (x + n)$ for $n, c \in \mathbb{N}$.

Atomic formulas of kind 1. define singletons (or the empty set if $n>m$, or the whole set if $n=m$), which are clearly ultimately periodic.
Atomic formulas of kind 2. define a ultimately periodic set with period $p = c$ because if $c \mid (x + n)$, then $c \mid (x + n + c)$. $\square$

Hence, every quantifier-free formula in a single variable defines a ultimately periodic set.

Olivier Roche
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