Let us suppose that everything you say is true. This still does nothing to address two facts: (1) the set of true formulas is not arithmetically definable, but the set of provable formulas is, whence the two must be distinct; — Nagase
Correct me if i'm wrong, but aren't we both referring to the fact that the negation of PA's provability predicate doesn't actually enumerate what isn't derivable in PA?
(2) truth is not conservative over PA, whence it can't be redundant. I sketched that argument in my first post here precisely so we did not get entangled in fruitless discussions about how we can know that G is true or about the Kirby-Paris theorem. — Nagase
If by truth not being redundant, you are merely referring to the difference between
PA |-\- S (i.e. the event that PA doesn't derive S) versus PA |-- ~Prov('S')
then I think we're in agreement. But in this event, the notion of truth is redundant in the sense that it doesn't transcend the notion of derivability within PA, which is what I took the OP's point to be.
As for G's "truth" value, I can only recognise it as having undefined truth value, both in the sense of it's unknown status as a derivable theorem of PA, and also in the sense of it's decoded arithmetical interpretation pertaining to the undecidable solutions of Diophantine equations. As an aside, i don't recognize the truth of Goodstein's Theorem, due to it's proof relying on non-constructive notions pertaining to transfinite induction.
Obviously, that particular argument assumes the soundness of PA, which you have disputed (this is a minority position, but one that I respect, if only because in the case of Nelson it generated some interesting mathematics). But this is not necessary for the argument to go through: one can start with Q and argue that any recursively axiomatized theory that extends Q will fall into the same problem, namely truth will be arithmetically undefinable and theoremhood will be arithmetically definable. Since no one that I know of doubts the soundness of Q (not even Nelson), the argument should go through. — Nagase
I'm not so much of the view that PA will turn out to be inconsistent within my life time, but rather of the view that talk of the consistency of PA is meaningless in both demanding infinite evidence and also in being logically inexpressible, as demonstrated formally by Godel's second incompleteness theorem.
By the way, if your ii.c) is correct, then PA is inconsistent. In any case, that is not a valid substitution instance of ii.a): ii.a) says merely that (assuming soundness) PA |- S iff PA |- Prov('S'), not that PA |- S <-> Prov('S') (the latter is a reflection principle and is actually not provable in PA). — Nagase
ii a) doesn't express soundness in the sense of begging an external notion of arithmetic truth, rather it expresses the ability of Prov to enumerate what is derivable within PA:
PA |-- S <---> PA |-- Prov('S')
this is a provable equivalence, because Prov(''S') merely enumerates what
can be derived in PA and it doesn't lead to false derivations, because it directly encodes PA's axioms and rules of inference.
ii c): By itself ii c doesn't imply inconsistency, because it is specifically stated with reference to the Godel sentence G that diagonalizes the negative provability predicate:
PA |-- ~Prov('G') <--> Prov('G')
Only if PA |-- G is derivable does ii c lead to inconsistency.
ii c) is equivalent to Tarksi's undefineability of truth applied to PA, in which PA's derivability predicate, Prov, is taken to be it's "truth predicate" . It also implies the constructive content of Godel's first incompleteness theorem, namely that ~Prov is a misnomer in not being able to enumerate what PA
cannot derive.
That's why i said i agreed with Mad fool, for reasons you hinted at. Truth, as far as mathematical logic is concerned, IS derivability , for mathematical logic cannot be partitioned into a metalanguage talking about an independent object language, for both languages are isomorphic to one another and suffer from the same undecidability.