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Oct 21, 2011 at 12:49 comment added Joel David Hamkins Yes, I agree. It seems that the parameter-free definable sets in $\langle\mathbb{Z},+\rangle$ are precisely the finite unions of arithmetic progressions that are symmetric about $0$. Once you allow parameters, then you can also define all finite translations of these sets and thus get all finite unions of arithmetic progressions.
Oct 21, 2011 at 5:14 comment added user6976 Yes, I did not notice "-". It does not change the situation very much though.
Oct 21, 2011 at 4:00 comment added Joel David Hamkins This answer is not correct, since the OP's structure has an automorphism taking $x$ to $-x$, and so every parameter-free definable subset must be symmetric with respect to $0$. But not all arithmetic progressions are like this. In the link provided, the version of Presburger arithmetic includes $1$ in the language, which in effect allows parameters into the definitions.
Oct 21, 2011 at 2:14 history answered user6976 CC BY-SA 3.0