Intersection of integers and rationals defined by logic

Consider the class of sets of the form $X \cap Y$ where $X \subseteq \mathbb{N}^d$ is defined in FO($\mathbb{N}, +$) and $Y \subseteq \mathbb{Q}^d$ is defined in FO($\mathbb{Q}, +, \leq$). Clearly, this class includes the semilinear sets, since they can be described as $X \cap \mathbb{Q}^d$. Is the converse true? Do all sets of this class are semilinear (i.e. definable in FO($\mathbb{N}, +$))?

flag	asked Oct 26 at 14:43 Steven 11●1

Sure. If $Y\subseteq\mathbb Q^d$ is definable in $(\mathbb Q,+,\le)$, then $Y\cap\mathbb N^d$ is definable in $(\mathbb N,+)$ by quantifier elimination of $(\mathbb Q,+,\le)$. – Emil Jeřábek Oct 26 at 14:57

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