Which natural number predicates can be defined in Robinson arithmetic?

Question

I'm especially wondering about the order relation, subtraction, division and exponentiation here:

• $x \leq y \quad \Leftrightarrow \quad \exists u\ y=x+u$
• $z= x-y \quad \Leftrightarrow \quad x=y+z\lor (x\leq y \land z=0)$
• $z=x/y \quad \Leftrightarrow \quad \exists u\ x+u=yz\land Su\leq y$
• $z=x^y \quad \Leftrightarrow \quad E(x,y,z)$

I know that $E(x,y,z)$ can be defined for Peano arithmetic, but that any explicit formula for $E(x,y,z)$ is ridiculously complicated. Because Peano arithmetic and Robinson arithmetic have the same language, any formula for $E(x,y,z)$ which works for Peano arithmetic is also a valid formula for Robinson arithmetic.

These look like valid possible definitions for the requested predicates, so where is the problem? I'm pretty certain that exponentiation cannot be defined in Robinson arithmetic! This means that no matter which formula for $E(x,y,z)$ that works in Peano arithmetic I use, I won't be able to prove $E(x,y,z)\land E(x,y,z')\ \rightarrow\ z=z'$ in Robinson arithmetic.

I'm not sure whether there are definitions for subtraction and division for which it can be proved (in Robinson arithmetic) that the corresponding formulas define functions. I'm looking for a "clear definite" answer regarding this "uncertainty". I don't really believe that the usual order relation can be defined in Robinson arithmetic. The usual order relation would satisfy the following formulas:

• $x\leq y \land y\leq z\ \rightarrow\ x\leq z$ (transitivity)
• $x\leq y \land y\leq x\ \rightarrow\ x=y$ (antisymmetry)
• $x\leq x$ (reflexivity)
• $x\leq y \lor y\leq x$ (totality)

Clarification Here one might ask which of these properties have to be provable before we can claim that we defined "the usual order relation". Clearly transitivity and reflexivity cannot be waived. I wouldn't want to waive antisymmetry either if it can be avoided, because this condition is quite similar to the condition which shows that a predicate defines a function. There is no need to prove totality, because totality is not a "Horn property", and because the line has to be drawn somewhere.

In order to show that a predicate $P(x,y,z)$ defines a function, at least the condition $P(x,y,z)\land P(x,y,z')\ \rightarrow\ z=z'$ must be provable. It would be nice if $\exists z\ P(x,y,z)$ would be provable as well, but because this is not a "universal Horn property" and the line has to be drawn somewhere, it can be waived. Then we just have a partial function instead of a total function, but at least we still have a function.

Insight Asaf and Peter raise a very valid point, which I didn't even notice before. By which criteria do I decide that a predicate (given by an arbitrary first order formula in $Q$) corresponds to a certain natural number predicate? Peter suggests an answer, which I would slightly weaken such that it also applies to predicates which don't define functions. For $E(x,y,z)$ my condition reads

If $m^n = k$ then $Q \vdash E(\overline{m}, \overline{n}, \overline{k})$ and
if $m^n \neq k$ then $Q \vdash \lnot E(\overline{m}, \overline{n}, \overline{k})$

where $\overline{m}$ is $Q$'s formal numeral for $m$.

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If it should turn out that some of the "harmless" natural number predicates can't be defined in Robinson arithmetic, then it would be interesting to know whether there is a conservative extension (="equiconsistent") of Robinson arithmetic where all canonical "harmless" natural number predicates can be defined. Exponentiation is not "harmless", because if $x$, $y$ and $z$ are binary encoded, the length of $z=x^y$ will be of the order $|x|y$, which is exponentially larger than $|x|+|y|$.

Answer 1

Any explicit formula for $E(x,y,z)$ [in the language of PA] is ridiculously complicated.

Well, actually it's not that complicated! It is a tedious but easy exercise, once you've grasped how to use Gödel's beta-function (which itself can be written in primitive notation in half a line or so) to write down a candidate $E$ in primitive notation in a few lines.

Exponentiation cannot be defined in Robinson arithmetic!

Well, it depends what you mean by defined! Different authors mean different things by "define" (one of the mildly annoying things in this area is that is little consistency accross textbooks here).

Certainly, the following holds for $Q$ (Robinson Arithmetic): there is a formula $E(x, y, z)$ such that

if $m^n = k$ then $Q \vdash E(\overline{m}, \overline{n}, \overline{k})$ and

for every $m, n$, $Q \vdash \exists!zE(\overline{m}, \overline{n}, z)$

where $\overline{m}$ is $Q$'s formal numeral for $m$. And plenty of authors will call that defining (even "strongly defining") exponentiation. Indeed, in this sense, $Q$ can (initially surprisingly) define all the primitive recursive functions.

But yes, $Q$ is absolutely lousy at proving generalizations, and in particular (as I think you are pointing out)

$Q \nvdash \forall x\forall y\exists!zE(x, y, z)$

and so, $Q$ can't show that exponentiation is (as they say) a provably total function. Is this what you mean by defining?

Probably so: still, before proceeding further with your questions, we perhaps need an explicit statement of what exactly you mean by "definition" here.

Answer 2

If it should turn out that some of the "harmless" natural number predicates can't be defined in Robinson arithmetic, then it would be interesting to know whether there is a conservative extension (="equiconsistent") of Robinson arithmetic where all canonical "harmless" natural number predicates can be defined. Exponentiation is not "harmless", because if $x$, $y$ and $z$ are binary encoded, the length of $z=x^y$ will be of the order $|x|y$, which is exponentially larger than $|x|+|y|$.

Asaf's comment "Robinson arithmetic can't even prove that $Sx\neq x$ for all $x$" shows that some of the "harmless" natural number predicates can't be defined in Robinson arithmetic, at least if "define" is read according to the criteria described in the "clarification section". But because these criteria only require provability of "universal Horn properties", one could try to add a restriction of the induction axiom to universal predicates to Robinson arithmetic. However, this extension is "almost certainly" not conservative. A better strategy might be to start by adding the axiom for antisymmetry of the order relation (using the implicit definition of $\leq$), because it is needed in any case. Then one can try to verify whether this gives the requested (conservative?) extension. However, the given description of canonical "harmless" natural number predicate is "too informal" for being able to verify this.

One feature is shared by the suggested definitions of the order relation, subtraction and division. The predicates $P(x,y,z)$ are all of the form $P(x,y,z)=\exists u_1\ldots u_r:\phi(u_1,\ldots,u_r,x,y,z)$ with a quantifier free formula $\phi$. The explicit formula for $E(x,y,z)=\operatorname{pow}(x,y,z)$ given by Hagen von Eitzen on the other hand doesn't have this form:

Using the following abbreviations

$$\begin{align}a\le b&\equiv\exists n\colon a+n=b\\ a< b&\equiv Sa\le b\\ \operatorname{mod}(a,b,c)&\equiv \exists n\colon a=b\cdot n+c\land c<b\\ \operatorname{seq}(a,b,k,x)&\equiv \operatorname{mod}(a,S(b\cdot Sk),x)\\ \operatorname{pow}(a,b,c)&\equiv\exists x\exists y\colon\operatorname{seq}(x,y,0,S0)\land\operatorname{seq}(x,y,b,c)\land \\&\quad\forall k\forall z\colon((k<b\land \operatorname{seq}(x,y,k,z))\to \operatorname{seq}(x,y,Sk,a\cdot z))\end{align}$$ we have $\operatorname{pow}(a,b,c)$ if and only if $c=a^b$. Intriguingly, you need some elementary number theory (such as the Chinese remainder theorem) to meta-appreciate this.

This seems to have the form $P(x,y,z)=\exists u_1\ldots u_r\forall v_1\ldots v_s:\phi(u_1,\ldots,u_r,v_1,\ldots,v_s,x,y,z)$ or even $P(x,y,z)=\exists u_1\ldots u_r\forall v_1\ldots v_s\exists w_1\ldots w_t:\phi(u_1,\ldots,u_r,v_1,\ldots,v_s,w_1,\ldots,w_t,x,y,z)$ if we insist that modulo is not a primitive operation.

Hence it might be a good idea to interpret "harmless" natural number predicate as the predicates which can be defined by existential predicates.

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Edit That conclusion is wrong. The correct conclusion is to interpret "harmless" natural number predicate as the predicates which can be defined by bounded quantification. I feel that the order relation should just be part of the language, to avoid unnecessary complications and confusion.

All quantifications in the definition of $\operatorname{mod}(a,b,c)$ and $\operatorname{seq}(a,b,k,x)$ can be replaced by bounded quantifications, and most quantifications for $\operatorname{pow}(a,b,c)$ can be replaced too: $$\begin{align}a< b&\equiv Sa\le b\\ \operatorname{mod}(a,b,c)&\equiv \exists n\leq a\colon a=b\cdot n+c\land c<b\\ \operatorname{seq}(a,b,k,x)&\equiv \operatorname{mod}(a,S(b\cdot Sk),x)\\ \operatorname{pow}(a,b,c)&\equiv\exists x\,\exists y\colon\operatorname{seq}(x,y,0,S0)\land\operatorname{seq}(x,y,b,c)\land \\&\quad\forall k\leq b\,\forall z\leq c\colon((k<b\land a\cdot z\leq c\land\operatorname{seq}(x,y,k,z))\to \operatorname{seq}(x,y,Sk,a\cdot z))\end{align}$$

The remaining quantifiers $\exists x\,\exists y$ cannot be replaced by bounded quantifications, because they are not polynomially bounded. (Even alternations between bounded existential and bounded universal quantifiers are not great, because they closely match the complexity of the polynomial hierarchy.)

Edit 2 The required bound for $\exists x\,\exists y$ would be approximately $c^b$. This is because the used encoding of the computation is essentially a fixed width encoding. If a variable width binary encoding would be used, then the computation would fit into a number bounded approximately by $c\cdot c$. Such an encoding is really possible (but even more cumbersome than the encoding using $\operatorname{mod}(a,b,c)$ and $\operatorname{seq}(a,b,k,x)$ given above), so even $\operatorname{pow}(a,b,c)$ is a "harmless" natural number predicate. It is not harmless as a function, because the result $c$ must be used to bound the quantifiers.