Re: Beyond Self-Reference (II) by pc, sci.math 11 Aug 98


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

>	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!





Let us continue now.
Keeping firmly points 1.2.3.4.5. and the fact that
by the  definition by arithmetization of the complement of Pf
Goedel theorem cannot be accomplished, let us come to Lemma (4)
which is independent (it could be cut away), since it 
concerns only the diagonal lemma.   

>	There is a problem here...


Here there is really "the problem";
I was in doubt for many days  to include it
in BS-R  or not. Finally I decided for its inclusion
because  at the end what I want is a discussion at the
purpose to improve comprehension (""), 
and, as it seems to me, Lemma (4) is very important from
this theoretical point of view. 

You may come with me into the trouble now.

All my handbooks of logic have in their inside
a proposition like the following:

Proposition 2.8 (Mendelson): If B is a subformula of A, A' is
the result of replacing zero or more occurrences of B in A
by a wf C, and every free variable of B or C that is also a
bound variable of A in the list y1,...yk, then:
a) |- [(y1)...(yk) (B <-> C)] -> (A <-> A')  
b) If |- (B <-> C), then |-   (A <-> A')  
c) If |-  (B <-> C) and |- A, then |- A'.

a) is the Equivalence Theorem, b) is the Replacement Theorem.

On the other side by the diagonalization lemma (Lemma (1)  in BS-R)

1. |- G <-> phi('G')  

2. |- G <-> (x)Pf('G') 

3. |- G <-> (x)Rf('G') 

4. |- (x)Pf('G') <-> (x)Rf('G') 


After a little  reflection we can see that 
diagonalization lemma is only an Equivalence theorem.

What I add of new is that we are always able
to know the value of 4. as Pf and Rf are both
decidable and, as I already showed in  BS-R,
4 is only a contradiction.

So at the end diagonalization its only an equivalence theorem
which as the particularly to assert wrong equivalences.
In other words diagonalization it is only a statement that
being mistaken for an equivalence theorem simply introduce
contradictions in PA.   (Ilias, let me say you that for me 
is the same for diagonalization in ZF)


>	The statements you chose,  "every  x  is a proof(/refutation) of v",
>are in fact disprovable in PA, yes... each one of them.   Still, they do have
>fixed-points!   Here is an extreme example:  let  phi(v) be  " not v=v".  As
>"impossible" as they come, right?   Yet, deltas exist:  typically, delta would
>be a disprovable sentence:   3=7,...  its Goedel number, say, 98765,...  and
>phi(delta) would be  " not 98765=98765".   Sure enough,  PA |-  not delta,
>PA |- not phi(delta)...  and hence   PA |-   delta <-> phi(delta)  !


From what you say here the fact that
phi(v) be  " not v=v"
is acceptable because you accept the diagonal lemma.

By the decidability of both Pf and Rf 4. is a contradiction hence
I do not accept diagonalization as an equivalence (nor replacement)
theorem (s).

What can I say you: I think to a mathematic were contradiction
are called with their own name.
In the mean time you live in a mathematic were contradiction are
called fixed points and transfinite.  


By the way, let me add a topic I would like to
(polite) discuss further.


What it seems to me is that Lemma (4) restores the
biconditional in

 x = y -> (A(x,x) -> A(x,y)).

See also the use of extensionality in my Beyond Diagonalization.

And with this we are completely inside Frege  (...).


Paola Cattabriga.