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I have few questions about considerations from Edward Nelson's survey Warning Signs of a Possible Collapse of Contemporary Mathematics adressing dealing with countability & finiteness properties of numbers from ultrafinitistic point of view.

More precisely my concern adresses few details from following Sokratian style teacher student dialogue on page 8. It starts with:

S: Will you teach me Arithmetic?
T: Gladly, my boy. $0$ is a numeral; if $x$ is a numeral, so is $Sx$ (where $S$ is the successor function definied on class of all numerals (...not sure if Nelson regards numerals as a set at all).
Numerals are used to count things.
S: I understand.
T: Addition is introduced as follows:
$x + 0 = x$,
$x + Sy = S(x + y)$.
Try some addition.
S: I have encountered no difficulty in reducing sums of numerals to numerals.
T: Multiplication is introduced as follows:
$x \cdot 0 = 0$,
$x \cdot Sy = x + (x \cdot y).$
S: I've tried a few examples and reduced the products to numerals. But what about
$SS0 \cdot ... \cdot SS0 $
with $y$ occurrences of $SS0$, where $y$ is a long numeral?
T: It reduces to the numeral denoted by $SS0 \uparrow y$.
[...]

I'm struggling with precise meaning of both marked in bold statements.

Key Question: What does in this context precisely mean to "reduce sums of numerals to numerals"? What are the performed reduction steps in detail?

I know that Nelson's ultrafinitistic approach (compare with this discussion; especially in comments below Chow's & Nelson's answers) refuses any kind of inductive arguments/constructions and as known any naive attepts do define numerals iteratively like $n+1:= S(n)= S^{\circ(n+1)}(0)$ appears in Nelson's framework problematic as this obviously would face two potential problems:

(1): Either circularity: Namely, to say that $x$ is numeral iff $x=S^{\circ x}(0)$ is so far circular as $x$ is defined in terms including $x$ itself. This appears strange.

(2): Or undesired implicit reference to "classical" set of natural numbers $\Bbb N$ which as I guess Nelson would like to strictly avoid

So say $x,y$ are two numerals by definition above. How their sum $x+y$ in above sense reduces actually to numerals? Or, more specifically, which terms such expression is allowed to include except $0$ and $S$, and which not?

For instance, expressing $x+y$ as "numeral" $S^{\circ(x+y)}(0)$ would appear to be circular as remarked in point (1) and I guess this was not what Nelson would allow/wanted to see there. More generally, wouldn't any attept to reduce sum of numerals to a numeral run in similar circularity issues when trying to define recirsively the set of natural numbers refering finite quantities?

So which expression reducing $x+y$ to numeral was actually asked for? $S^{\circ y}(x)$? If yes, isn't there again appearing implicitly similar kinds of circularity in definition as indicated before in (1)? Moreover, why $x$ as numeral in Nelson's sense should be actually always given of shape $x=S^{\circ x}(0)$? Can it be concluded from definition without any induction principle? I don't see how.

But then - if from above reasons $S^{\circ y}(x)$ was not valid expression reducing sum of numerals $x,y$ to a numeral - after which form was else asked for?

Similarily, what is actually meant by reduction of products to numerals without running into circularity issues as before?

NB: At all my question appears to be quite pedantic, but exactly here - what actually the reduction of sums of numerals to numerals in Nelson's sense should be - I'm failing to see to consistency behind the approach Nelson proposes

user267839
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  • Numerals are symbols used to count things. In usual arithmetic they are symbols used to name numbers. Sum is an operation that, given two numbers, produce as result a new number. Following Nelson, we read the recursive def of sum as an operation that given two numerals produces as output a new numeral. – Mauro ALLEGRANZA Aug 19 '25 at 10:29
  • And yes, if there are no numbers but only symbols, the number used in the iteration of the S operation does not exists, and thus no numerals at all. – Mauro ALLEGRANZA Aug 19 '25 at 10:31
  • @Mauro ALLEGRANZA: yes exactly, it seems (compare with Nelson's answer in linked MO discussion) that Nelson insisted that the reasoning made in his formal system make strictly no usage of "usual numbers" (...what Nelson called "Abstract arguments about a formal system, which are really arguments about its arithmetization"). Maybe what was actually meant in the T-S dialogue by "have encountered no difficulty in reducing sums of numerals to numerals" even the banal – user267839 Aug 19 '25 at 10:50
  • (cont) statement that $x+y$ is by definition a numeral, as Nelson's "$+$" was defined as map $(Nelson's numerals)^2 \to (Nelson's numerals)$. So additional "reducing steps" seem which I up to now fail to understand were actually a red herring there, ie $x+y$ is a numeral, period, no additional reduction steps. Otherwise I not know to which reducing procedure from sum of numerals to numerals Nelson is refering there. – user267839 Aug 19 '25 at 10:51
  • @Mauro ALLEGRANZA: Ad 2nd comment: Well, numeral $0$ exists by definition in Nelson's sense. What one actually can do is to declare $1'$ as shorthand symbol for numeral $S(0)$, $2'$ for $S(1')$, etc, which are numerals by definition. But, this not tells me in Nelson's framework that if $x$ is a numeral, it can be expressed as iteration of $S$ applied to $0$ in prev sense. (And that's the problem I have with formulation what is actually meant by reduce sum of numerals to numeral as I not see what are actually the legal "reduction steps" in Nelson's framewrk – user267839 Aug 19 '25 at 11:00
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    I am not Nelson & I am neither agreeing nor disagreeing with the given arguments. I just wanted to chip-in with my thinking that initially "numerals" might be just $0,S0,SS0,SSS0,\cdots$ , not including $+$ , Eg $S0+SS0$ which should reduce to the numeral $SSS0$. I think later "numerals" will include "Short-hands" with Symbol $\uparrow$ Eg $S0 \uparrow SS0$ , while even later , there will be more compact "Short-hands" with more Symbols ! – Prem Aug 19 '25 at 11:11
  • @Prem - good point. The issue seems to me that Nelson, like every "foundationalist", aim to define everything and this is simply impossible. Does reduction of numbers to string manipulation work without counting symbols and so on? – Mauro ALLEGRANZA Aug 19 '25 at 11:27
  • Indeed , @MauroALLEGRANZA , more-over , there is the "long numeral" which is undefined here !! – Prem Aug 19 '25 at 11:44
  • @MauroALLEGRANZA I'm not certain I know what you're asking, but the existence of Metamath Proof Explorer's approach makes me think the answer might be "yes". – Mark S. Aug 19 '25 at 15:35
  • Taking in addition also further comments, it is quite obvious that string manipulation is adequate for performing numerical calculations: electronic computers do exactly that. The issue IMO is the so-called "foundational" aspects: in what sense counting strings is more "fundamental" that counting tout court? – Mauro ALLEGRANZA Aug 19 '25 at 18:34

1 Answers1

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Is $S(S(0))$ a numeral? Let's check.

$0$ is a numeral. Since $0$ is a numeral, $S(0)$ is a numeral. Since $S(0)$ is a numeral, $S(S(0))$ is a numeral. Great!

Is $S(S(0))+S(0)$ a numeral? Let's check.

Everything that's a numeral has the form $0$ or the form $S(x)$ where $x$ is a numeral. Since the expression $S(S(0))+S(0)$ has neither of these forms, it is not a numeral.

But $S(S(0))+S(0)$ is an expression involving $+$, and we have a definition of $+$. Let's apply it.

$S(S(0))+S(0)$ has the form $x+Sy$, where $x$ is $S(S(0))$ and $y$ is $0$. So $S(S(0))+S(0)$ reduces to $S(S(S(0))+0)$.

Now the inner expression $S(S(0))+0$ has the form $x+0$, where $x$ is $S(S(0))$. So $S(S(S(0))+0)$ reduces to $S(S(S(0)))$, which is a numeral.

More compactly, we could write the reduction steps as $$S(S(0))+S(0)\rightarrow S(S(S(0))+0) \rightarrow S(S(S(0)))$$

Note that we have not introduced any general rule here like $S^{x}(0)+S^y(0) = S^{x+y}(0)$. We're simply taking an expression involving $+$, pattern matching, and applying the definition until we arrive at a numeral.

When S says "I have encountered no difficulty in reducing sums of numerals to numerals", he means just this: in all the examples S has tried, expressions of the form $x+y$, where $x$ and $y$ are numerals, have indeed reduced to numerals by following the pattern matching algorithm described above.

Later, when T talks about "the numeral denoted by $S(S(0))\uparrow y$", he's not saying that $S(S(0))\uparrow y$ is a numeral - it's not! He's saying that there is a particular numeral which is named by the expression $S(S(0))\uparrow y$ in the same way that the expression $S(S(0))+S(0)$ names the numeral $S(S(S(0)))$.

Alex Kruckman
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  • maybe a trifle, but is it clear at all that using Nelson's definition every numeral is either $0$ or of form $Sx$ for $x$ numeral? The rule of inference says if $y$ numeral, then $Sy$ too, but can it be derived formally that every numeral has a predecessor? (...drawing analogy to ordinals, where not every ordinal has a predecessor, I not see any obstruction forbidding existence of that kind of sloppy dubbed "long numerals"...). Note, that we are not allowed to apply any kind of induction precedure there – user267839 Aug 19 '25 at 16:55
  • Furthermore, even assuming the problem indicated in prefious comment can be somehow resolved, it is also not clear if the "algorthm" you proposed finishes: taking an expression involving $+$, pattern matching, and applying the definition until (!) we arrive at a numeral may run for eternity, doesn't it? – user267839 Aug 19 '25 at 17:29
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    (1) Perhaps Nelson should have added "and nothing else is a numeral" to his description of numerals. Then it's clear that every numeral either has the form $0$ or the form $Sx$. If you're presented with an expression which has neither of these forms, you can tell at a glance that it's not a numeral. What is more involved is checking whether an expression of the form $Sx$ is a numeral, since this requires you to recursively answer the question of whether $x$ is a numeral. @user267839 – Alex Kruckman Aug 20 '25 at 19:07
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    (2) Neither Nelson nor I have claimed (at this point) that the reduction algorithm for $+$ always terminates. S has only said that he has tried some examples and encountered no difficulties. Of course, classically it's easy to show that it terminates. But the ultrafinitist may doubt this claim. This is the whole point of Nelson's "additionable numbers" introduced later in the essay. @user267839 – Alex Kruckman Aug 20 '25 at 19:11