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Tennenbaums' theorem does not apply to Robinson Arithmetic ($Q$). There is a computable, nonstandard model of $Q$ "consisting of integer-coefficient polynomials with positive leading coefficient, plus the zero polynomial, with their usual arithmetic."

I recently asked How does induction fail in computable nonstandard models? One example was $\forall x \exists y(x=y+y \lor x=S(y+y))$. In the model above this fails for the infinite nonstandard number represented by the polynomial $x+1$ which is neither even nor odd. This predicate is true for the nonstandard number represented by $2x$ and successors of $2x$.

It is tempting to think this predicate can be used to pick out the standard natural numbers. We choose numbers that are even or odd and less than the number represented by the polynomial $x$. The problem is there is no finite numeral for the number represented by the polynomial $x$. We can not write $\forall x < SSS...SSS0$ in the language of $Q$. We could ask for the smallest number that is neither even nor odd but this would be like asking for the smallest nonstandard number.

Is this predicate in this model an example of overspill? If not, how is it different from overspill?

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Russell Easterly
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You're right, "is either even or odd" doesn't kill overspill - yet overspill fails in that model via a closely related formula: specifically, the formula "Every $y<x$ is either even or odd," or $$\varphi(x)\equiv\forall y[y<x\rightarrow[(\exists z(y=2z))\vee(\exists z(y=2z+1))]],$$ characterizes the natural numbers. (To be $100\%$ precise: if $M$ is the model you describe and $N\subset M$ is the initial segment corresponding to $\mathbb{N}$, then $\varphi^M=N$.)

Proof sketch: Clearly $\varphi(n)$ holds for every $n\in\mathbb{N}$. So the important point is the converse. First, show that for any non-constant polynomial $p$ we have $p>x-n$ for some natural number $n$. Then, just note that no polynomial of the form $x-n$ is either even or odd.

Note that this is just a sharpening of the statement that "every number is either even or odd" witnesses the failure of induction. As far as I know, every natural sentence which witnesses the failure of induction in $M$ can be easily tweaked (by the same trick of saying "For every smaller number . . .") to also witness the failure of overspill - that is, to define $\mathbb{N}$.

EDIT: Just for completeness, here's a counterexample to induction which doesn't convert to a counterexample to overspill in the manner outlined above: $$\psi(x)\equiv\exists y, z<x\forall w<x\exists k\in\mathbb{N}[(w=y+k)\vee (w+k=y)\vee(w=z+k)\vee(w+k=z)].$$ Here, "$\exists k\in\mathbb{N}$" is an abbreviation for "$\exists k[\varphi(k)\wedge . . .]$," where $\varphi$ is the formula defining $\mathbb{N}$ given above.

The formula $\psi$ translates to "There are two elements $y, z<x$ such that everything $<x$ is within a finite distance from either $y$ or $z$." This picks out exactly the naturals, and polynomials of the form $X\pm j$ (e.g. set $y=X$ and $z=0$ - there are other choices). It also clearly satisfies the hypotheses of induction.

BASICALLY: any counterexample to induction yields a definable cut (the longest initial segment of the model in which the counterexample holds without exception); these definable cuts need not, however, be exactly the $\mathbb{N}$-piece, and it is consistent that there is an easily definable proper cut yet $\mathbb{N}$ is not a definable cut. Note that by contrast a counterexample to overspill is (by definition) an instance of the $\mathbb{N}$-piece being a definable cut.

Noah Schweber
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