Finitism / potential infinity requires the paraconsistent logic NAFL
Radhakrishnan Srinivasan
rk_srinivasan at yahoo.com
Thu Mar 9 20:03:26 EST 2023
Matthias said:
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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 ,,,
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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
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