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

On Mar 21, 3:35 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Mar 20, 10:55 am, "R. Srinivasan" <sradh...@xxxxxxxxxx> wrote:
On Mar 20, 10:55 am, "R. Srinivasan" <sradh...@xxxxxxxxxx> wrote:

On Mar 19, 10:16 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:> On Mar 19, 12:09 am, "R. Srinivasan" <sradh...@xxxxxxxxxx> wrote:

Godel translated a proposition of the form P&~P->Q into a proposition
in the *object* language of arithmetic, say, PA.

What specific formula of the meta-language and its specific
translation into the language of PA are your referring to?

Why don't you ask this question to the Godelians? All I am interested
in at this point is that there is such a formula S in the language of
PA.

It would help if you were specific. If you wish not to be specific,
then so be it, but then it might happen that we reach an impasse when
we try to determine properties of something not specified.

Call this proposition
S. Note that S is a specific propositon involving only numbers.

What does "involving only numbers" mean?

Means it is a formula in the language of PA.

Okay. That's a lot more clear.

By your own assertion above you do not admit P&~P->Q in the object
language of arithmetic. So if you scan all proofs of PA, you will not
find a proof of P&~P-->Q (a proof must end with the propositon
proven). You will not find a refutation of P&~P --> Q either in the
list of PA-proofs. For if you did, then P&~P&~Q must hold in every
model of PA, for some specific (number-theoretic) propostions P and Q
and we know that is not possible.

No, the reason we don't have a refutation in PA is that '(P&~P) -> Q'
is not in the language of PA. It's in the meta-language.

Wrong. For PA to refute (P&~P)->Q, PA needs to prove (P&~P&~Q) for
*one* specific P and *one* specific Q, or, what amounts to the same
thing, PA needs to prove a contradiction.

You said, "Godel translated a proposition of the form P&~P->Q into a
proposition in the *object* language of arithmetic, say,
PA." [emphasis on 'object' YOURS]. Aside from the ambiguity of the
word 'proposition' (I'll take it here to be 'sentence'), what your
quote suggests is that there is a sentence of the FORM 'P&~P->Q' and
that that sentence is TRANSLATED into the language of PA. So that
sentence ITSELF is not a sentence of PA but rather a TRANSLATION of
that sentences is a sentence in the language of PA.

 E.g. if PA proves

GC & ~GC & ~TPC

then PA has refuted (P&~P)-->Q.

WHAT?! You're not even making sense in basic sentential logic.

If a theory proves:

(P & ~P) -> ~ Q,

then that does NOT entail that the theory has refuted:

(P & ~P) -> ~ Q.

Indeed ALL ordinary first order theories (both classical and
intuitionistic) have as theorems all sentences of both of the above
forms.

The refutation of

(P & ~P) -> ~Q

is not

(P & ~P) ->  Q

nor

(P & ~P) ->  ~~Q

but rather

~((P & ~P) -> ~Q)


Can you read what I have written above? Here it is again. P&~P->Q is
a *schema*. Right? If this schema is recursively enumerated, it would
include a sentence of the form

GC&~GC-> TPC. (*)

The negation of (*) is

GC&~GC&~TPC (**)

Note: you have apparently misread what I have written above. Do you
agree that (**) is the negation of (*)? If so then if PA proves (**),
do you agree that it has refuted the schema P&~P->Q, even though PA
cannot express

~((P&~P)->Q)

If not, I suggest you think about this. Do not read a book on logic.
There is no need to.



You can immediately see that PA would refute P&~P->Q if PA proves a
contradiction.

Of course any theory that proves a contradiction also refutes every
sentence in the language of a theory. But so what? (1) PA doesn't
prove a contradiction and (2) 'P&~P->Q' is not in the language of PA.

But PA does not have the expressive power to prove P&~P->Q. Of course, I can now
sense that you are going to jump up and down

You've got bigger problems than predicting whether I jump up and down.

insisting that PA has not refuted P&~P->Q even if PA is inconsistent.

You're going to have to be more specific what you mean by 'refute'.
Ordinarily, I take "PA refutes sentence F" to mean:

PA |- ~F

If you mean differently, then say so.

So PA does not refute 'P&~P->Q' because 'P&~P->Q' is not even IN THE
LANGUAGE of PA and also PA refutes no sentence of the FORM 'P&~P->Q' ,
since PA is consistent. Or, if you claim that there is a sentence F of
the form ''P&~P->Q'  such that

PA |- ~F

then please PROVE that.

I said IF PA proves a contradiction (see (**) above THEN PA would have
refuted the schema P&~P->Q because to refute a schema, you need to
refute just one instance of it, whereas to prove a schema you need to
prove every instance of it. Why is this so hard for you to grasp?
Again do not read a book on logic, simply think about what I have
said.


Sure, if  P&~P->Q is not in the language of PA, its formal negation is
also not in the language of PA. But all I am saying here is that PA
can prove itself inconsistent by proving a contradiction.

If a theory proves a contradiction, then the theory is inconsistent.

Sure it is. The schema P&~P->Q expresses that PA is consistent. PA can
refute that schema by proving a contradiciton. Note: I am not saying
that PA proves a contradiciton. I am saying that PA can *in principle*
refute the schema P&~P->Q by proving the negation of one instance of
that schema. In order for that to happen, PA would have to prove a
contradiction and would therefore have to be inconsistent.

You don't
have to encode the notion of consistency into PA in order for PA to
prove itself inconsistent. In which case PA would have refuted P&~P->Q
in substance.

But PA doesn't prove a contradiction, moreover, now you have the
undefined "refute in substance".

Also, in case there is any doubt about this, there is a difference
between:

Theory T proves a contradiction, therefore theory T is inconsistent

and

Theory T proves theory T is inconsistent.

To see an analogy, it could be that FLT is not expressible in the
language of some theory T of arithmetic if you don't permit
exponentiation. But T could still prove ~FLT, by construction, if
there exist specific positive integers x, y, z and n>2 such
that....... Do you get it?

There are all kinds of things about intrepretations between theories
and expressibility and representability and arithmetical
representability. But you've not shown anything about them that
entails that PA is inconsistent.

This is what you could reasonably say:

There is a sentence S in the language of PA such that there is a
sentence F in the metatheory such that the metatheory proves "S is
true in the standard model of PA iff F", but S is not a theorem of
PA.

But that is not a contradiction.

And if there is ANY sentence in the language of PA that PA does not
prove, then PA is consistent.

Moreover, PA DOES prove every sentence IN ITS LANGUAGE of the form
'(P&~P)->Q'. And PA does NOT prove ANY sentence of the form '~((P&~P)-

Q)'.
You can also see that P&~P->Q *is* actually a consistency statement
for PA.

Please state exactly what YOU mean by "a consistency statement for PA"
and why you think 'P&~P->Q' is one.

Assuming you are a staunch and diehard Godelian,

What is a "staunch and diehard Godelian"? Presumably, it would be one
who staunchly advocates Godel's philosophy. So why in the world would
you assume that I am such a person? I haven't advocated Godel's
realist philosophy. Indeed in posts elsewhere I've said that I have no
inclination toward adopting Godel's realist philosophy. As to
mathematical results proven by Godel, that I recognize that he proved
such and such theorems doesn't make me a "Godelian" any more than my
recoginzing that, say, Kripke proved certain theorems would make me a
"Kripkean".

you should
now be able to get an intuitive feel for why if 'P&~P->Q' is
translated into a proposition S in the language of PA, then S has to
be undecidable in PA.

No, it's nonsense.

If a "translation" of 'P&~P->Q'  into the langauge of PA is a good
enough translation to preserve the tautologicalness of 'P&~P->Q' ,
then it IS a theorem of PA.

For that matter even ~(P&~P) or Pv~P, etc are
consistency statements for PA which are not expressible in the
language of PA and which cannot be proven in any first-order theory.

Okay, now you're just being silly for the sake of it, right?

(1) ALL sentences of the form ~(P&~P) or Pv~P that are in the language
of a first order theory are theorems of that first order theory (the
first in both classical and intuitionistic first order theories, and
the second in classical first order theories).

You are one confused guy. Just now we agreed that P&~P->Q or similar
schema are not within the language of first-order theories. P&~P->Q is
a theorem scheme, i.e., there are infinitely many theorems of the form
P&~P->Q in first order languages, and every first-order theory must
prove all these theorems. But no first-order theory can prove 'P&~P-
Q', where P and Q are *sentential variables*, because such a sentence
is not within the scope of a first-order language. I am saying that
sentences like these (which effectively involve quantification over
the P's and the Q's) are as good as consistency sentences for a first-
order theory like PA. Because PA would have to prove a contradiction
to refute such sentences. You can assume I have done some "coding" or
"translation" to arrive at this conclusion.

(2) ~(P&~P) or Pv~P are tautologies. In what sense are they
"consistency statements"?

It seems you've confused proving ~(P&~P) with proving "theory T does
not prove '~(P & ~P)'".

Yes, if '~(P & ~P)' is in the language of theory T, then "'theory T
does not prove '~(P & ~P)'" is equivalent (in the meta-theory) to
"theory T is consistent". Also PA has a sentence G that is true in the
standard model of PA (and ordinarily we simpify 'true in the standard
model' as 'true') iff PA is consistent. And PA does not prove G
(neither does PA refute G). But such a sentence G is NOT a tautology,
and specifically not a tautology of the form 'P&~P->Q'  and not the
negation of a tautology, and specifically not the negation of a
tautology of the form just mentioned.

Now Godel translated the meta-theoretical propostion P&~P->Q into a
specific number-theoretic propostion S in the language of PA. Note
that S is not a variable; it is a constant and we have a construction
for S.

So I take it 'S' is a defined constant in the meta-language.

 Since we just now argued that P&~P-->Q is not either provable
or refutable in PA,

Because it's not even in the language of PA.

it follows that S must also be undecidable in PA.

No, because S is in the language of PA. 'S' is a defined constant of
the meta-language. But S is a formula of the object language PA.

But this means that there must exist a model of PA in which S is
false, and such a model cannot exist, for the same reason argued
earlier. This is a contradiction.

Wow! What a completely amateurish confusion you have! You're competely
mixed up in use/mention and meta-language/object language. If a
formula is not in the language of theory, then it's not a theorem of
the theory, but that doesn't mean there are models for the language in
which the formula is false. Only formulas of the LANGUAGE have values
of 'true' or 'false' in models for the language. You're blatantly
conflated S with '(P&~P) -> Q'. They're in different languages. That
'(P&~P) -> Q' is not a theorem of PA doesn't entail that S is not a
theorem of PA, even though S is a "translation of '(P&~P) -> Q' .

Of course I know what you have said above, and I have explicitly
stated it in a subsequent post. I have already said that it is a
*requirement* of first-order logic that a sentence like P&~P->Q should
not be expressible in any first-order language, which will not allow
quantification over formulas.*That* is why we do not require that a
theory like PA should have a model in which P&~P->Q is false (even
though it is not provable in PA) because such a formula is not
admitted in the language of PA.

Now think of the following situation. Let us add P&~P-->Q (and only
this single sentence, where P and Q are arbitrary) into the language
of PA,

That makes NO SENSE! You don't add just a SENTENCE to a language.

Of course you can. If you do you would have gotten an inconsistency.
And that is what Godel indirectly did. He managed to add a sentence of
the form P&~P->Q to a first-order language (via an indirect
transalation procedure) but did not change the logic or its principles
of proof.



So that I can communicate at all with you, please tell me what basic
textbooks in mathematical logic you ordinarily consult. Because a
notion such as adding just a sentence to a language is so whack that I
need to be able to refer to the appropriate sections of a textbook to
get your straightened out about such basics before we can productively
continue.

I said if you did that, you would get an inconsistency. What I was
trying to convey to you is that such a thing is not possible without
changing the logic. But Godel did that through the back-door. Don't
read any books on logic. Just think about what I have said here.

In fact my understanding is that PA proves S. If the meta-theoretical
sentence P&~P->Q does translate to S and if this coding is accepted as
legitimate, this is tantamount to the claim that PA proves P&~P-->Q,

NO! And that's the exact contrapositive of what I just mentioned. PA
does not prove  '(P&~P) -> Q' because '(P&~P) -> Q' is NOT IN THE
LANGUAGE of PA, but S IS in the langauge of PA, so PA may very well
prove S while OF COURSE it does not prove '(P&~P) -> Q' .

Sure, sure. I now see that you are the mother of all formalists.

Oh, shut the F up with grasping-at-any-straw insults such as "mother
of all formalists". (1) My remarks above reflect just ordinary
mathematical logic, no matter whether one's philosophy of mathematics
is formalist, realist, intuitionist, structuralist, or whateverrist.
(2) You don't know what formalism is. (3) Though there are formalist
considerations in my tentative views about this subject, my views are
not merely formalist, and definitely not "formalist" in the
sophomorically pejorative sense you mean.

So, please, don't degrade yourself even further with such sophomoric
cheap shots.

If
P&~P->Q has been translated to a formula S in the first-order language
of PA, and if PA proves S, are you denying that PA has *effectively*
proven itself consistent?

OF COURSE I deny that! It's ridiculous! A theory doesn't prove itself
consistent just for proving a particular TAUTOLOGY!

Read what I have written above. You dont even begin to understand what
I am talking about. A first-order theory does not prove *the*
tautology, P&~P->Q. There is no such thing as "THE" tautology. There
are only infinitely many senetences of this schema which a first order
theory proves. *I* am saying that "P&~P->Q", where P and Q are
sentential variables is a consistency sentence for PA or any first
order theory. Of course there is no first-order sentence with
variables in it, so no first-order theory can prove such sentences.

Even by your GODel,

Please grow up from such puerlity as spelling 'Godel' as "GODel'.

PA is inconsistent.

What?! Godel showed no such thing.

Now
I see that I have waved the red flag and you are going to charge with
further futile formlastic arguments for why that is not the case.

Ah, so your tack is to call me a 'formalist' over and over. Like a
political debate in which one candidate keeps calling the other
'liberal' over and over, as if by calling someone a 'liberal' you've
actually engaged the issues.

But I don't need Godel to make that argument. It is a requirement of
first-order logic that such a sentence cannot be translated into a
first-order language and we have started out by formulating the logic
that way.

What in the world are you TALKING about? There is no such
"requirement".

We have already fixed that while every instance of P&~P->Q
is provable in first-order theories, P&~P->Q itself is not provable
for arbitrary P and arbitrary Q.;

No, the OPPOSITE!. NOT EVERY instance is provable in the given theory;
only the instances that are IN THE LANGUAGE of the theory. And EVERY
such instance is provable in that given theory.

Admitting such a sentence into a
frist-order language, whether directly or by translation,

We dont' admit just individual SENTENCES into a language. And you've
not said what "admitting by translation" might mean.

will lead to
inconsistency because the logic would have been expanded beyond the
rules we ourselves have laid down. *If* such a sentence S (to which
P&~P->Q has been translated) exists in the language of PA, then S
*has* to be undecidable in PA in order to remain consistent with the
way we have formulated first-order logic. But such undecidability is
not possible.

Why are you protracting making a complete clod of yourself? There is
NO tautology in the langauge of a theory that is not a theorem of the
theory.

Sheesh, this is so very basic. NOW I DO understand why virtually
nothing you post makes sense.

Again, do NOT read a book on logic. You have missed the forest for the
trees. Here it is again. NO first-order theory proves any sentence of
the form P&~P->Q or ~(P&~P), etc. because these are sentences with
variables in them. You are right when you say that "There is no
tautology in the language of a theory that is not a theorem of the
theory". Because all tautologies are theorem schemes, not theorems,
and are metamathematically true, in the sense that first-order
theories cannot recognize the existence of a schema (or sentential
variables). Again: First-order theories can only prove all instances
of tautologies without ever recognizing any such thing as a tautology
(as a theorem scheme). Your failure to understand this has caused you
to get anguished. Think about this and then get back to me.

Regards, RS
.

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