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A naive question about Edward Nelson's ultrafinitistic way of reasoning arising from reading this discussion:

Say we start with setup $F$ being a formal system by adjoining an unary predicate symbol $\psi$, imposing the axiom $\psi(0)$, and following rule of inference: from $\rm\psi(a)$ infer $\rm\psi(Sa)$ (for any term $a$ in $F$). (... as notation suggests the underlying language includes symbol $0$ and succession function $S$)

Nelson stresses that using this inference rule its not possible to proof that the proposition $\psi(80^{5000})$ would be actually a theorem of $F$?

More precisely, in a comment below his answer he stated that the key issue behind this phenomenon that altough in "ordinary" language/formal system (including implementation of arithmetiation ) the theorem $\exists p[p$ is an arithmetized proof in F of $\psi(80^{5000})]$ is doubtless true.
But that reasoning cannot be expressed in $F$.

Question: Could somebody elaborate the point stated in last sentence, ie why one can't formulate a correct proof inside $F$?

(If the question is too vague, let go a step back. Can it be made precise in which sense reasoningexpressible in $F$ substantially differs from reasoning in arithmetized $F$?)

My naive guess up now (but not sure if that's the key problem): Does it have exclusively to do with lack of internal proof (so inside $F$!) that $80^{5000}$ is actually finite?

If we would be able prove the later, then we could simply iterate application of successor function $S$ to $0$ exactly $80^{5000}$ times and exploiting the given rule of inference. If I'm not missing something, this would give a correct proof inside $F$, right? Or is the problem a bit more subtle ...as ultrafinitists usually abandon not only infinities, but also finite, but "big" numbers)
(Of course, if we a priori not know within the formal system that $80^{5000}$ is finite, we cannot apply this argument. )

Is this the key problem behind non existence of proof of claim $\psi(80^{5000})]$ Nelson refered to? In other words, the inability to produce a formal proof in $F$ seems to depend only on potential inability to proof internally that $80^{5000}$ is finite without running is circularity making unavoidably use of invoking implicitly finiteness phrased in terms of "classical" natural numbers?

user267839
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    From Ultrafinitism: "some ultrafinitists are concerned with acceptance of objects in mathematics that no one can construct in practice because of physical restrictions in constructing large finite mathematical objects." The article then gives an example of denial of the existence of the floor of Skewes number due to the lack of actual computation of its floor. So the objection may well be one of physical effectiveness rather than theoretical effectiveness. – Eric Towers Aug 18 '25 at 18:41
  • @Eric Towers: Ok, but so far I understand it is still possible to write formally the term $50^{8000}$. What a ultrafinitist may deny is that its possible to express it as a term in $S$ and '$0$' because it may "to large". But why does this indicate that it's not possible to prove $\psi(80^{5000})$ as Nelson stated? – user267839 Aug 18 '25 at 18:54
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    How many states of matter in the universe are there? Are there $80^{5000}$? If not, there there is no way to perform this computation because this universe can't. (There's also the straightforward problem that even taking one Planck time per "$S$", there hasn't been nearly enough time to write that many "$S$"s.) These guys tend to argue that exponentiation (and other operations) are only partial functions due to honest computability (not just theoretical computability in a larger universe with way more time). – Eric Towers Aug 18 '25 at 20:33

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