Presburger Arithmetic with algebraic scalar multiplications

We consider Presburger arithmetic (PA) extended by scalar multiplication by an algebraic irrational number $\alpha$, and call this extension $\alpha$-Presburger arithmetic ($\alpha$-PA). We show that the complexity of deciding sentences in $\alpha$-PA is substantially harder than in PA. Indeed, when $\alpha$ is quadratic and $r\geq 4$, deciding $\alpha$-PA sentences with $r$ alternating quantifier blocks and at most $c\ r$ variables and inequalities requires space at least $K 2^{\cdot^{\cdot^{\cdot^{2^{C\ell(S)}}}}}$ (tower of height $r-3$), where the constants $c, K, C>0$ only depend on $\alpha$, and $\ell(S)$ is the length of the given $\alpha$-PA sentence $S$. Furthermore deciding $\exists^{6}\forall^{4}\exists^{11}$ $\alpha$-PA sentences with at most $k$ inequalities is PSPACE-hard, where $k$ is another constant depending only on~$\alpha$. When $\alpha$ is non-quadratic, already four alternating quantifier blocks suffice for undecidability of $\alpha$-PA sentences

Presburger Arithmetic with algebraic scalar multiplications

We consider Presburger arithmetic (PA) extended by scalar multiplication by an algebraic irrational number $\alpha$, and call this extension $\alpha$-Presburger arithmetic ($\alpha$-PA). We show that the complexity of deciding sentences in $\alpha$-PA is substantially harder than in PA. Indeed, when $\alpha$ is quadratic and $r\geq 4$, deciding $\alpha$-PA sentences with $r$ alternating quantifier blocks and at most $c\ r$ variables and inequalities requires space at least $K 2^{\cdot^{\cdot^{\cdot^{2^{C\ell(S)}}}}}$ (tower of height $r-3$), where the constants $c, K, C>0$ only depend on $\alpha$, and $\ell(S)$ is the length of the given $\alpha$-PA sentence $S$. Furthermore deciding $\exists^{6}\forall^{4}\exists^{11}$ $\alpha$-PA sentences with at most $k$ inequalities is PSPACE-hard, where $k$ is another constant depending only on~$\alpha$. When $\alpha$ is non-quadratic, already four alternating quantifier blocks suffice for undecidability of $\alpha$-PA sentences

A strong version of Cobham's theorem

Let $k,\ell\geq 2$ be two multiplicatively independent integers. Cobham's famous theorem states that a set $X\subseteq \mathbb{N}$ is both $k$-recognizable and $\ell$-recognizable if and only if it is definable in Presburger arithmetic. Here we show the following strengthening: let $X\subseteq \mathbb{N}^m$ be $k$-recognizable, let $Y\subseteq \mathbb{N}^m$ be $\ell$-recognizable such that both $X$ and $Y$ are not definable in Presburger arithmetic. Then the first-order logical theory of $(\mathbb{N},+,X,Y)$ is undecidable. This is in contrast to a well-known theorem of B\"uchi that the first-order logical theory of $(\mathbb{N},+,X)$ is decidable