FOM: True = provable? Was: Goedel: truth and misinterpretations
torkel at sm.luth.se
Tue Nov 7 03:33:23 EST 2000
Vladimir Sazonov says:
>(*) (A => Pr[T](`A')
>where Pr[T](x) is arithmetically definable predicate of
>provability in T of a formula with Goedel number x and
>`A' is numeral (term SSS...S0) representing Goedel number
>Theorem 2. If T is consistent then it is consistent
>T* = T + the above axiom schema (*).
As your proof essentially points out, (*) is consistent with T since
in fact the stronger statement "everything is provable in T" is consistent
>Comment 1. Roughly speaking, Theorem 1 means that we can
>reasonably assume that
> every true sentence is provable.
This is a very mysterious comment, however. The mere fact that (*) is
consistent with T does not imply that it is at all reasonable to assume
(*), any more than the consistency of "PA is inconsistent" with PA means
that we can reasonably assume that PA is inconsistent.
>Now consider a stronger schema
>(**) (A <=> Pr[T](`A')
>Let us call a theory T *strongly consistent* (PA is an example of
>such theory) if it does not proves ``pathologies'' as
> ~A <=> Pr[T](`A')
>or any their finite disjunctions.
This can't be quite what you mean, since PA is not "strongly
consistent" in this sense. ~A <=> Pr[T](`A') is provable in PA if
A is a Godel sentence for PA.
>Theorem 2. If T is strongly consistent then it is consistent
>T** = T + the above axiom schema (**).
Your proof of this is correct, but does not apply to PA, as noted above.
The schema (**) is incompatible with the diagonal lemma.
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