Re: Godel's comments about the "true reason" for incompleteness

On Mar 24, 11:00 pm, "R. Srinivasan" <sradh...@xxxxxxxxxx> wrote:
On Mar 24, 9:04 pm, "R. Srinivasan" <sradh...@xxxxxxxxxx> wrote:



On Mar 24, 5:12 pm, Aatu Koskensilta <aatu.koskensi...@xxxxxxxxx>
wrote:

On 2008-03-24, in sci.logic, R. Srinivasan wrote:

I intend the theory T to be the theory that Godel used as the
metatheory of PA, which formalizes the syntax of PA. As per what you
said, T is in substance equivalent to PA. So in the theory T is there
a  notion of "all sentences" of PA? Because if at all there is such a
notion, a sentence equivalent to (*) should be expressible in the
language of T ("All sentences follow from a contradiction").

What needs be provable in T is not "All sentences follow from a
contradiction" but "if PA proves a contradiction then, for all
sentences P, PA proves P".

I thought you implied (2) was something that "PRA proved of itself"
via Godel's translation procedure. What I am objecting to is what you
stated above, with PRA replaced by PA.

That PRA proves "if PRA proves 'P & not-P' for some 'P', then PRA proves
'A' for every sentence 'A'" does not mean that PRA proves

 (for all P)(for all Q)(P & not-P --> Q)

which is a claim about propositions not directly expressible in terms
of syntax at all.

Sure. But then why can't you express (*) in the language of the theory
T?

How would you express it? That is, assuming you have suitable
vocabulary to talk of wffs and proofs, taking the conjunction of a
wff, negating a wff, quantifying over wffs and so on, exactly how do
you propose to formalise (*)? Remember that variables ranging over
wffs are terms, not formulas.

I see the point you are making. Let us forget ex falso quodlibet and
consider a simpler example:

PA proves "For each P (PA |- Pv~P)"    (1)

Now clearly (1) is true by the translation procedure. Also consider

For each P, PA proves "((PA|-Pv~P) <-> Pv~P)"  (2)

Clearly (2) has to be a tautology because Pv~P is a tautology. Now why
can't we deduce from (1) and (2) that:

PA proves that "For each P (Pv~P)"   (3)

which is what you are denying? Note that I could just as well replace
(1) by

PA proves that "For each P (T0 |- Pv~P)"   (4)

where T0 is the theory corresponding to the null set of (extra-
logical) axioms.

Clearly (4) is also true by the translaiton procedure. Now it seems
very clear to me that (3) follows from (4).

Presumably the argument is that (2) consists of infinitely many PA-
proofs. I.e., PA proves (2) for each given P, but not for an arbitrary
P. This looks quite strange to me in light of the fact that the
formula in (4) is demonstrated for an arbitrary P. In the logic NAFL
where truth is identified with formal provability , this kind of a
paradoxical situation is avoided.  In NAFL, a situation in which
something is trure for each P, but not for an arbitrary P, is not
possible and is a metamathematical contradiction. It logically follows
that there is no such thing as a tautology in NAFL, whose theories
must have first-order languages.

Regards, RS



You have presumably avoided the contradiction by "banning" (3) from
being expressed, but in substance (1) (or even better, (4)) amounts to
(3). If your point is that (3) should not be demonstable by the
translation procedure, then neither (1) nor (4) should be demonstrable
either; at least that is what it looks like to me.

I also think the point of contention here is that you are assuming
(from the classical viewpoint) a certain distinction between truth and
provability that I am not assuming. But such a distinction between
truth and provability should really be a consequence of Godel's
theorems and should not be assumed while proving them. I would have to
think a little more about this, though.- Hide quoted text -


So to put it in a nutshell, we have the following situation:

PA proves that "For each P (T0 |- Pv~P)" (5)

For each P, PA proves "((T0|-Pv~P) <-> Pv~P)" (4)

From (4) and (5) does (3) follow, i.e.

PA proves that "For each P (Pv~P)" (3)

I have said in the other reply that (3) may not follow because
actually (4) ocnsists of infinitely many proofs not recognized by PA.
But there seems to be some ambiquity here. At first sight the
situation looks similar to the following. Suppose, for example

PA proves "For all n f(n) = n" (a)

For each n PA proves g(n)=f(n) (b)

From (a) and (b) can we conclude that

PA proves "For all n g(n)=n". (c)

The answer is presumably negative and there must exist a non-standard
model for PA in which g(rho) is not equal to rho, where rho is a non-
stnadard integer. However, if you look at (4) in analogy to (b), is
it possible that there can exist a non-standard model for PA in which
the equivalence mentioned in (4) can fail? Unless one admits non-
standard sentences P (e.g. of non-standard length), it does not seem
possible. So it is still not clear to me whether (3) follows from (4)
and (5). Looks ambiguous.

Regards, RS
.

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