Are 'numerals' closed under exponentiation?

Question

I have read Edward Nelson's Warning signs of a possible collapse of contemporary mathematics a couple of times, it is a very interesting read, but I do not understand the conclusory paragraph. In particular I am interested in the final 'warning sign', beginning from the fable on page 8 and continuing through to the end (all relevant information for my question is copied below).

The belief that exponentiation, superexponentiation, and so forth, applied to numerals yield numerals is just that—a belief.

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Note: we are working in the standard 7 axioms of PA for this question.

Numerals are defined as follows. $0$ is a numeral; if $x$ is a numeral, so is $\text{S}x$. In a standard way, we introduce addition: $$\begin{align} x+0 &=x,\\ x+\text{S}y &=\text{S}(x+y), \end{align}$$ multiplication: $$\begin{align} x\cdot 0 &=0,\\ x\cdot\text{S}y &=x+(x\cdot y), \end{align}$$ and exponentiation: $$\begin{align} x\uparrow 0 &=\text{S}0,\\ x\uparrow\text{S}y &=x\cdot(x\uparrow y). \end{align}$$ Then we define counting numbers, introducing two new axioms to our system:

8. $0$ is a counting number.
9. If $x$ is a counting number, then $\text Sx$ is a counting number.

additionable numbers:

$x$ is an additionable number in case for all counting numbers $y$, the sum $y + x$ is a counting number.

and finally, multiplicable numbers:

$x$ is a multiplicable number in case for all additionable numbers $y$, the product $y \cdot x$ is an additionable number.

We prove that additionable and multiplicable numbers are closed—i.e. if $x_1$ and $x_2$ are additionable (respectively, multiplicable) numbers, then $x_1+x_2$ is additionable (respectively, $x_1\cdot x_2$ is multiplicable).

But now we come to a halt. If we attempt to define 'exponentiable number' in the same spirit, we are unable to prove that if $x_1$ and $x_2$ are exponentiable numbers then so is $x_1 \uparrow x_2$. [...] The belief that exponentiation, superexponentiation, and so forth, applied to numerals yield numerals is just that—a belief.

It is at this point that I lose the author. If I understand correctly, he is saying that given two numerals $x_1$ and $x_2$ (i.e. $\text S\text S\text S\ldots 0=x_1$ for some number of $\text S$s, and similarly for $x_2$), then $$x_1\uparrow x_2=\underbrace{x_1\cdots x_1}_{x_2\text{-times}}$$ is not a numeral (i.e. we cannot definitely say that there is some number of $\text S$s for which $x_1\uparrow x_2=\text S\text S\text S\ldots 0$). If this is not what was intended, what is the author's point at the end here?

However, if I understand right, then I am not convinced; below I demonstrate my attempt to show this given some $x_1\uparrow x_2$. Can we not 'break this down' as follows? (From here I use the notation $\text S^k 0$ for the numeral $k$, that is the numeral $\underbrace{\text{SS}\ldots\text S}_{k\text{-times}}0$.)

$$\begin{align} x_1\uparrow x_2&=\underbrace{x_1\cdots x_1}_{x_2\text{-times}}\\ &=\underbrace{(\text S^{x_1} 0)\cdots (\text S^{x_1} 0)}_{(\text S^{x_2} 0)\text{-times}}\\ &=(\text S^{x_1} 0\cdot \text S^{x_1} 0)\cdot\underbrace{(\text S^{x_1} 0)\cdots (\text S^{x_1} 0)}_{(\text S^{x_2}0-2 )\text{-times}}, \end{align}$$ now apply the definition of multiplication to $\text S^{x_1} 0\cdot \text S^{x_1} 0$ to find $$\begin{align} \text S^{x_1} 0\cdot \text S^{x_1} 0&=\text S^{x_1} 0+(\text S^{x_1} 0\cdot \text S^{x_1-1} 0)\\ &=\text S^{x_1} 0+\text S^{x_1} 0+(\text S^{x_1} 0\cdot \text S^{x_1-2} 0)\\ &=\cdots=\underbrace{\text S^{x_1} 0+\text S^{x_1} 0+\cdots+\text S^{x_1} 0}_{x_1\text{-times}}+(\text S^{x_1} 0\cdot 0)\\ &=\underbrace{\text S^{x_1} 0+\text S^{x_1} 0+\cdots+\text S^{x_1} 0}_{x_1\text{-times}}. \end{align}$$ Do this a finite number of times to put $x_1\uparrow x_2$ in the form of an addition; at which point applying the definition of addition a finite number of times should yield $$x_1\uparrow x_2=\underbrace{\text{SS}\ldots\text S}_{x_1^{x_2}\text{-times}}0.$$ Now $x_1\uparrow x_2=\underbrace{\text{SS}\ldots\text S}_{x_1^{x_2}\text{-times}}0$ is a numeral by induction.

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The claim 'that exponentiation, superexponentiation, and so forth, applied to numerals yield numerals is just that—a belief' is a strong one, what does Nelson mean and how does he conclude this?

Thank you!

Answer 1

First, the paper's claim

we are unable to prove that if $x_1$ and $x_2$ are exponentiable numbers then so is $x_1 \uparrow x_2$

is clearly very different from the claim

given two counting numbers $x_1$ and $x_2$, $x_1 \uparrow x_2$ is not a counting number

that you paraphrased.

Second, your attempted proof(?) is not valid. You seem to be mixing the theory with some kind of meta-theory, both in terms of notation (like your underscript braces) and manipulations that aren't part of the theory. You need to work within the theory.

To give you an idea of where you're going wrong, assume PA is consistent. Let $\text{Con}_n \, \text{PA}$ be the sentence "PA does not prove a contradiction in fewer than $n$ symbols." Then, for every $n$, $\text{PA} \vdash \text{Con}_n \, \text{PA}$. But that doesn't mean we can conclude $\text{PA} \vdash \forall n (\text{Con}_n \, \text{PA})$, since that would mean $\text{PA}$ is inconsistent! Your "proof" seems to be making a similar kind of mistake.

Answer 2

I believe it will be helpful to first break down the mathematical claims found in the paper. first work in a system which has a constant $0$ for the number zero and has a function $Sx$ which takes a number and returns a number. It also contains functions $x+y$, $x\times y$ and $x\uparrow y$. Although he is not explicit about the assumed arithmetic axioms it seems clear that he is at least using some of the standard properties of addition and multiplication, so the recursive definitions as you have given them as well as the commutative, assosiative and distributive properties of addition and multiplication.

now to this system he adds a predicate x is a counting number which I will call $C(x)$. he then adds two axioms $C(0)$ and $\forall x.C(x)\rightarrow C(Sx)$. these are the only axioms about $C(x)$ assumed he explicitly does not assume every number is a counting number. He then defines additionable and multiplicable numbers by the definitions $A(x)\leftrightarrow \forall y .C(y)\rightarrow C(y+x)$ and $M(x) \leftrightarrow \forall y .A(y)\rightarrow A(y\times x)$. he then presents proofs using the standard properties of addition that $A(x)\rightarrow C(x)$ and $A(0)\land A(x)\rightarrow A(Sx)$ and that $\forall xy.(A(x)\land A(y))\rightarrow A(x+y)$. He then states the provability without proof of similar results for $M(x)$ including that $\forall xy(M(x)\land M(y))\rightarrow M(x\times y)$. He finally states the claim that it is not provable that $\forall xy.(E(x)\land E(y))\rightarrow E(x\uparrow y)$ for some suitable definition for exponentiable numbers.

These explicitly mathematical claims are all valid. See Fernando and Gilda Ferreria's paper "Interpretabililty in Robinsons Q" for more detail. the claims about additionable and multiplicitable numbers follows from Solovay method of shortening of cuts which in a special case shows that for any cut there are definable subcuts closed under addition and multiplication. $A(x)$ and $M(x)$ are the formulas used to show this. as for the exponentiable numbers the unprovability of there closure under exponentiation follows from this answer.

I think it is also important to note that in the paper Nelson acknowledges that every explicit counting number is exponentiable here is the quote:

For any specific numeral SSS. . . $0$ we can indeed prove that it is an exponentiable number, but we cannot prove that the world of exponentiable numbers is closed under exponentiation. And superexponentiation leads us entirely away from the world of counting numbers.

the issue is he doesn't accept induction for counting numbers so he can not conclude that they are actually closed under exponentiation. as for what he meant by his concluding statement i can think of two reasonable ways to interpret it: that from his implicit axioms you can not prove the totality of exponentiation for counting numbers, even by moving to a definable cut, or the stronger claim that the counting numbers in reality are not closed under exponentiation. Although the weaker claim follows from known results, the stronger claim does not clearly follow from the premises. The only way to show from the basic axioms that some number is not a counting number is to show that there is no way to reduce it's definition to one only involving $S(x)$ and $0$, since $A(x)$ and $M(x)$ contain zero and are closed under successor they satisfy all the basic axioms when you substitute them in for $C(x)$ so the theory can not prove the existance of any counting number that fails to be multiplicitable. Now there is another cut that needs to be mentioned the logarithmic cut. It is defined as $L_2(x)\leftrightarrow A(2\uparrow x)$ note that $2\uparrow0=S0$ so $A(2\uparrow 0)$ is true and since $2\uparrow Sx=2\times (2\uparrow x)=(2\uparrow x)+(2\uparrow x)$ so if $A(2\uparrow x)$ is true so is $A(2\uparrow Sx)$. Since this proof only relies on the fact that $C(x)$ is closed under successor and contains zero we could replace $C(x)$ with any other formula that is also satisfies this and define a logarithmic subcut of that formula including the original logarithmic cut. we can now show that the numbers that satisfy all definable cuts is an undefinable cut that is closed under $x\uparrow y$. first every definable cut is satisfied by zero and if all definable cuts are satisfied by $x$ then for each definable cut it must be the case that $Sx$ also satisfies it. second suppose some definable cut failed to be satisfied by $2\uparrow x$ then the definable logarithmic subcut of that cut would fail to be satisfied by $x$ and so wouldn't satisfy all definable cuts so if all definable cuts are satisfied by $x$ than they are all satisfied by $2\uparrow x$. we get $x\uparrow y$ from the fact that it is always less than $2\uparrow x\times y$. although a definable cut can never be provably closed under exponentiation there is no barrier to one not definable from the base axioms.

in conclusion the mathematical content is accurate, the philosophical conclusions are of course up for debate.