Re: Beyond Self-Reference by pc, sci.math 10 Aug 98


In article  ilias kastanas 08-14-90, [email protected] writes:

>A careful review of what said results are based on 
>might lend focus to areas of agreement
>or disagreement.



I agree with you Ilias.
Please focus your attention to the following.
1.
All Godel incompleteness is based on the fact that
by  the _undecidability_ of its G = ((x) ~Pf(x,'G')) . 
given G we are not able to know if

               in PA |- G     or     in PA |- ~G.


By the _undecidability_ of G we can 
prove that not |- G and not  |- ~G.
2.
Any predicate P is recursive iff its characteristic 
function  is recursive. If the characteristic Cr_P is recursive
then Cr_P is computable, hence the predicate P is decidable.
I. e.
the function
                    0  if P is true
                  / 
             Cr_P
                  \
                    1  if P is false

is computable.
3.
Pf(x,v) is recursive. Rf(x,v) is recursive.
Hence Pf(x,v) and  Rf(x,v) are decidable.
4.
Thanks to Pf(x,v) we are able to decide if 
in PA |- G or not in PA |- G.
Thanks to Rf(x,v) we are able to decide if 
in PA |- ~G or not in PA |- ~G.

In fact  
                 Cr_Pf(m,'G')=0 iff Pf(m,'G')
and
                 Cr_Rf(m,'G')=0 iff Rf(m,'G')

and since Cr_Pf is computable we are able to know if
in PA |- G or not in PA |- G 
(indeed Cr_Pf(m,'G')=0 iff in PA |- G, 
Cr_Pf(m,'G')=1 iff not in PA |- G)
and since Cr_Rf is computable we are able to know if
in PA |- ~G or not in PA |- ~G 
(indeed Cr_Rf(m,'G')=0 iff in PA |- ~G, 
Cr_Rf(m,'G')=1 iff not in PA |- ~G).
---
Hence from the point of view of the pure arithmetization, i.e.
from the point of view of what is computable or not,
we are always able to know if in PA |- G or in PA |- ~G.
(To this purpose I want observe that Goedel never proposed this
point of view, and I confess I am still thinking around why having
Goedel in their hands the (wonderful) means to prove 
the completeness he finished to prove incompleteness. 
My idea/intuition/belief is that the reasons
are the same to those for Cantor: an unitary foundation 
of mathematics and hence an unitary foundation for two theories  
which cannot indeed be unified (except for to share the same 
language, fol) as they are substantially opposed/complementary, 
the theory of ordered numbers (induction-discrete) and 
the theory of numbers with no-order (nodenumerable-continuum) ... 
(see  also my "A proof for the axiom of choice"))
---
4. 
My Lemma (2) show only that


   (*)              Cr_Pf(m,'G')=0 iff Cr_Rf(m,'G')=1 

and this can be done simply by the arithmetization itself.

(This can be a quite elegant way to prove completeness don't you think so?
But G is closed here, and  we are not speaking of completeness ...
by the way arithmetization holds for any formula...)
---
Please completely forget Lemma (4) it is  useless now,
as we perfectly do not need it to prove that Goedel 
theorem cannot be accomplished 
(and as you correctly notice there is a problem there, 
indeed there is "the problem" there, and I will show you
later).
---
5.
Let us suppose to have 
PA |- G ,
then 
Pf(m,'G'),
and 
Cr_Pf(m,'G')=0.
Hence by (*)
Cr_Rf(m,'G')=1.

What do you need more now?

Am I obliged to say that if Cr_Rf(m,'G')=1 and Cr_Pf(m,'G')=0 then
by Cr_Pf  ~Pf(m,'G')=1    ?

Am I obliged to say that by Cr_Rf(m,'G')=1   m  cannot be the 
number of a refutation of G and that hence to speak of m as
a number for the unprovability of G has not sense?

Let us suppose to have 
PA |- ~G ,
then 
Rf(m,'G'),
and 
Cr_Rf(m,'G')=0.
Hence by (*)
Cr_Pf(m,'G')=1.

What should I add now? 

QED.

6.
......
Lemma (4) is completely independent to all the above.
I will  try to give some explanations around it
and I will try to give a coherent reply to

>	There is a problem here...
>
>	The Diag lemma does not assert  "PA |-  phi('delta')";   it asserts
>"PA |-   delta <-> phi('delta')".    And it delivers such a  delta...  no
>matter how "possible" or "impossible"  phi()  may be!
>

probably in a  next tread as now 
I am no longer able to give  you a coherent argumentation, and
I would like to precisely discuss around everything ...

and It seems to me this tread could blow up. 

So,              to be continued


Paola Cattabriga