# FOM: True = provable? Was: Goedel: truth and misinterpretations

V. Sazonov V.Sazonov at doc.mmu.ac.uk
Mon Nov 6 12:14:01 EST 2000

```Here are two simple (metamathematical) results related with my
attempts to give (one of) formalization to ``true, but not provable''.
They look as something opposite to this phrase. For definiteness,
consider that all the considerations below such as consistency of
theories or first order models or the concept of truth (probably
assumed implicitly) are given in the framework of ZFC.

As such a subject is not my current everyday mathematical work,
I would be especially grateful for any comments by specialists.
The results are rather simple, but are they known, etc.?

Let T be arbitrary sufficiently strong r.e. theory (say, PA
or PRA) and A, B, C, etc. are any sentences (having no free
variables) in the language of T. Consider the following axiom
schema

(*)    (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
of A.

Theorem 2. If T is consistent then it is consistent
T* = T + the above axiom schema (*).

Proof. Assume that T* has a contradiction. Then

T |-
(A & ~Pr[T](`A')) V (B & ~Pr[T](`B')) V ... V (C & ~Pr[T](`C'))

and by distributivity and de Morgan law we have

T |- ~(Pr[T](`A')) & Pr[T](`B')) & ... &  Pr[T](`C')) & ...

It follows that T |- con[T], and, therefore, due to Goedel,
T |- 0=1. QED.

Comment 1. Roughly speaking, Theorem 1 means that we can
reasonably assume that

every true sentence is provable.

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.

Arithmetical statement expressing strong consistency
could be denoted CON[T] in contrast with the ordinary
consistency con[T]. Evidently, CON[T] => con[T]
(and this is provable in PA).

Theorem 2. If T is strongly consistent then it is consistent
T** = T + the above axiom schema (**).

Proof. Assume that T** has a contradiction. Then

T |- ~[(A <=> Pr[T](`A')) & ... & (C <=> Pr[T](`C'))],

i.e.

T |- (~A <=> Pr[T](`A')) V ... V (~A <=> Pr[T](`A'))

what means that T is not strongly consistent (``pathological'').
QED.

Comment 2. Roughly speaking, Theorem 2 means that we can
reasonably assume that

true = provable.

Of course, here is no contradiction with the well known
(and non-deniable!) results on non-recursive enumerability
of arithmetical truth. (All of this except comments being
in precise (meta)mathematical sense. No philosophy!).

It is desirable to get analogous results for the case
of Feasible proofs (by using FPr, Fcon, FCON, etc.). But
it requires using some appropriate formalization of feasibility
concept. Thus, we could probably get that

true = feasible provable,

in the above sense. Otherwise, what does it mean provable in
practice? The previous results gives no guarantee for feasible
provability, only imaginary finite provability. Nevertheless,
they ``demonstrate'' the relation between ``true'' and ``provable''
in one of the ways discussed in FOM

Comments on the well-known reflection principle

Pr[T](`A') => A.

are probably also expectable.