Can exponentiation be defined in Robison's Q by use of the exponential Diophantine equation? Robinson's Q is an axiomatization of arithmetic which only defines addition and multiplication, and does not have the axiom schema of induction (unlike Peano axioms).
I was wondering, given Matiyasevich's result on Diophantine equations, and the fact that exponentiation is Diophantine, i.e., there is a polynomial $f$ with integer coefficients such that
$$
\forall x,y,z \: \exists w_1,\ldots, w_n \: : \: f(x,y,z,w_1,\ldots,w_n) = 0 \iff x=y^z
$$
Can this be used to define exponentiation in Robinson's Q?
My question was inspired by Why is it impossible to define multiplication in Presburger arithmetic yet possible to define exponentiation in Peano Arithmethic? and Why is it impossible to define multiplication in Presburger arithmetic? and https://math.stackexchange.com/a/4142/386635 .
 A: It's not really proper to speak of "definability in a theory." Definability really is a notion appropriate to structures: it does make sense to ask whether (for example) multiplication is definable in the structure $\mathcal{N}=(\mathbb{N};+)$, and this question amounts to asking whether there is a formula $\varphi(x,y,z)$ in the language of $\mathcal{N}$ such that for each $a,b,c\in\mathbb{N}$ we have $$[\mathcal{N}\models\varphi(a,b,c)]\quad\iff\quad a\cdot b=c.$$ (And the answer to this question is no.) See this old answer of mine for more on this.
That said, you'll often hear people say things like "multiplication is not definable in Presburger arithmetic" (which is known to be the set of sentences true in the structure $(\mathbb{N};+)$). So, what the heck does that mean?
Well, the intuitive idea is that there is no formula in the language of Presburger arithmetic which "Presburger-arithmetic-provably acts like" multiplication. This is made precise by the notion of representability: given a first-order theory $T$ in the language of arithmetic or some sublanguage thereof (like Presburger), a function $f:\mathbb{N}^k\rightarrow\mathbb{N}$ is $T$-representable iff there is some formula $\theta(x_1,...,x_k,y)$ in the language of $T$ such that for each $a_1,...,a_k,b\in\mathbb{N}$ we have $$[T\vdash \theta(\underline{a_1}, ..., \underline{a_k}, \underline{b})]\quad\iff\quad f(a_1,...,a_k)=b$$ and $$[T\vdash \neg\theta(\underline{a_1}, ..., \underline{a_k}, \underline{b})]\quad\iff\quad f(a_1,...,a_k)\not= b$$ where "$\underline{n}$" is the numeral corresponding to the number $n$ (actually, there is some unfortunate flexibility on this notion in the literature - see e.g. Salehi, and  always carefully check the definition being used by your text!). Note that a fortiori anything representable in such a theory is definable in the corresponding version of $\mathbb{N}$, so non-definability results imply non-representability results. This is why the conflation between structures and theories is so common: it's often justified "under the hood" (see e.g. the italicized parenthetical in this old answer of mine).

OK, so now we have two questions:

*

*Is the structure $(\mathbb{N};+,\cdot)$ - the structure on the natural numbers corresponding to Robinson arithmetic - rich enough to define exponentiation?

*

*The answer is yes. Indeed, every computable function is definable over $(\mathbb{N};+,\times)$, and Post showed that that's just the tip of the iceberg. MRDP is massive overkill here.



*Is exponentiation representable in $\mathsf{Q}$?

*

*The answer is also yes! Essentially due to Godel (as part of his proof of the first incompleteness theorem), we have that "representable-over-$\mathsf{Q}$" is equivalent to "computable."



Exceeept ...

*

*There is a lot that $\mathsf{Q}$ cannot prove about exponentiation, which in practice means that the answer to the question one means to ask when one asks "Is exponentiation definable over $\mathsf{Q}$?" is often negative. For example, the usual formula defining exponentiation in terms of addition and subtraction does not $\mathsf{Q}$-provably define a function in the first place!

Ultimately, here's what I consider to be the right answer to your question:

The intuitive notion of "definability in a theory" is much more flexible than one may expect. In some weak senses, $\mathsf{Q}$ is strong enough to talk about exponentiation; in others, it isn't. Ultimately you'll need to be very careful and precise about what you want to do with exponentiation in $\mathsf{Q}$!


As a coda, let me make a remark contrasting definability and (any version of) representability. One of the things that makes logic difficult to learn is that different fundamental ideas "live at" different levels of generality, and this is an especially subtle case in point:
Definability makes sense for any structure. Representability, however, does not automatically make sense for any theory: representability leans heavily on the idea that theories of arithmetic "talk about" $\mathbb{N}$ and every element of $\mathbb{N}$ has a "canonical name." There is no obvious way to make sense of representability for general theories, such as the theory of groups or $\mathsf{ZFC}$. If you want to develop analogues of representability for such theories, you need to do some careful work. For example, maybe "representability over $\mathsf{ZFC}$" should mean "representability in the expansion of $\mathsf{ZFC+V=L}$ by the canonical definable Skolem functions" (whatever that means).
