Finitism / potential infinity requires the paraconsistent logic NAFL

Matthias matthias.eberl at mail.de
Mon Mar 13 23:50:25 EDT 2023

```For me, too, finite means "standard finite". My approach does not
enlarge the syntax of FOL, as e.g. First Order Dynamic Logic mentioned
by Vaughan Pratt, but it is purely model theoretic.

Regarding R. Srinivasan question "So how would your model theory (that
uses only potentially infinite sets) establish the existence of a
nonstandard c such that ~phi(c) holds?":

Henkin's construction to show completeness for FOL makes sense from a
potentialist point of view. The model consists of (open) formulas, which
are added step by step and are identified if they are provable equal.
This can be seen as a direct system, and all one has to do from a
potentialist's view is to skip the last completion step, i.e., switching
to an infinite limit set. In such a first-order model of PA the
"non-standard numbers" are all formulas that are not provable equal to a
closed term Suc(...Suc(0)...). In particular, if one adds the formulas
(exists m. m = c) and c > n, where n = 0,1,... to the PA axioms, then
one can use a non-standard number in such a model to interpret c.

When I said that the nonstandard numbers are indefinitely large finite
numbers, I mean the following: Both, the career set of the model and the
set of formulas in the theory, if they are infinite, are potential
infinite. Any reference to them then refers to some finite set (of
elements in the model and formulas resp.). So for a PA model, at each
state of the potential infinite set of formulas (exists m. m = c) and c
> n, where n = 0,1,... there are always standard natural numbers that
satisfy this finite set of formulas. These numbers will eventually occur
in the standard model of PA (*). So you do not need non-standard numbers
if you consider the set of formulas as a potential infinite set (i.e.,
as an increasing finite set).

In short, you have non-standard models also from a potentialist point of
view, but each non-standard number occurs at a specific stage and not at
the limit step (e.g. the union of all finite sets). And since you do not
have a completed, infinite set of formulas (exists m. m = c) and c > n,
where n = 0,1,..., you do not need non-standard numbers in this case.

Regards,
Matthias

(*) In this argument the state of the PA model (its size) depends on how
many formulas (exists m. m = c) and c > n you take. You may want to
introduce other dependecies as well, e.g. for each element n in the
model there is a formula (exists m. m = c) and c > n. But both
dependecies cannot be met at the same time. The whole point of limit
constructions is to get rid of these dependencies, so that you have both
conditions at the same time. But then you have the paradoxes of
infinity. As a potentialist, you have to stay with these dependencies
and consider them further.

------ Originalnachricht ------
Von "Radhakrishnan Srinivasan" <rk_srinivasan at yahoo.com>
An fom at cs.nyu.edu
Datum 10.03.2023 02:03:26
Betreff Finitism / potential infinity requires the paraconsistent logic
NAFL

>Matthias said:
>
>***************************************
>But from a potentialists perspective, the
>formulas in (*) exists only up to some greatest n (not all infinitely
>many 0,1,...), which is a finite number that is not fixed and can be
>arbitrarily large. The set of formulas in (*), roughly {c>0, c>1, ...},
>is thus not (actual) infinitely large. From that perspective, the
>nonstandard numbers are indefinitely large finite numbers ,,,
>***************************************
>
>My response:
>
>Here I assume that by "finite" you mean "standard finite", i.e., your nonstandard numbers do not exceed the completed infinity of standard natural numbers.
>
>Within first-order logic (FOL), you certainly have the freedom to deny the existence of infinite sets (in the classical sense) and replace them with your potentially infinite sets. But I do think that you have a similar freedom in FOL to deny the existence of nonstandard natural numbers (in the classical sense) and replace them with your indefinitely large finite numbers. In my previous post, I gave a well-known argument that uses the compactness theorem of FOL to establish the existence of nonstandard natural numbers. Here let me consider another route, namely, Goedel's first incompleteness theorem.
>
>To keep things simple, let us consider a universally quantified sentence "For all x phi(x)" that is true (in the standard model), but unprovable in the FOL theory PA. The completeness theorem implies that there must exist a nonstandard model of PA in which ~phi(c) is true, where c is a nonstandard natural number. Here it seems quite clear that you cannot replace c with an indefinitely large finite number that never exhausts the standard natural numbers. Because no matter how large such a c is, phi(c) would always be true. So how would your model theory (that uses only potentially infinite sets) establish the existence of a nonstandard c such that ~phi(c) holds?
>
>Regards,
>R. Srinivasan
```