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

On Mar 19, 12:09 am, "R. Srinivasan" <sradh...@xxxxxxxxxx> wrote:

Unwillingness
on the part of your peers to discuss NAFL is obvious

Comments like that make you obnoxious. Lots of people on the threads
have discussed NAFL with you. Indeed many of them have been quite
generous with their time and patience.

A new logic like NAFL is self-evidently noteworthy

Oh, yes, surely self-evidently.

Right. Because people are not impressed with your work, it must be
that they are mass-produced clones. God forbid you countenance that
each individual is not impressed with your work based on that
individual having come to that conclusion himself.

You don't understand my work.

Because you don't properly explain it or because it's not coherent.

I do.

People tend to have some form of understanding of their own thoughts.

What are you talking about? Do you contend that the proof is not
formalizable in PRA or Robinsion arithmetic or PA or even Z set theory
- all first order theories.

The proof is formalizable with "coding".

No coding (other than from outside the object theory) needed to do it
in Z set theory.

I have clearly said that I am
questioning the legitimacy of the coding and you have failed to
understand that.

You've not shown anything not "legitimate".

Take the assertion that "From
a contradiction, an arbitrary propostion follows", or , say,

P&~P --> Q,    (*)

Let us take P to be fixed in (*), and consider this propositon for
arbitrary Q. Obviously there is an implied quantification over Q,

If 'Q' is a sentence letter, then it is not quantified over in the
language. (Of course, the meta-language may quantify over sentence
letters of the object language).

If 'Q' is a meta-variable in the meta-language, then it is not
quantified over in the meta-language (if the meta-language is first
order).

I object to this assertion that there can be a variable in a language
that is not quantified over.

Object all you like.

All variables range over some domain and
that *is* quantification.

Nope, there is quantification in the syntactical sense of a quantifier
and a variable followed by a formula. And there is the semantic sense
of the variables being interpreted.

This is the kind of confused thinking that
you are loudly accusing me of.

There's nothing confused about it at all. It is precise.

EIther Q is a constant (a fixed
propostion, which must be specified by construction) or else it is
quantified over.

False dichotomy.

If Q is a variable in the metalanguage, it is
quantified over in the metalanguage.
E..g when you say that P&P->Q in
the metalanguage, you are also asserting that this proposition holds
for an arbitrary P and arbitrary Q. That is quantification.

Sure. In the meta-language. Just as I said. But then the variables 'Q'
and 'P' are not (ordinarily) variables of the object language.

The point
is that the object (first-order) language does not contain any such
variables because actually propostions in the first-order language are
obtained after substitution of P and Q by specific formulas.

One may approach it that way. But ordinarily, we refer to the
recursive definition of 'formula'.

IF we admit it thus then it's a problem. But we DON'T. The assertion
that all sentences follow from a contradiction is not made in the
object language but rather in the meta-language.

Ah, so you do agree that there is quantification in the metalanguage.
Fine. Let us take it from there.

Of course I agree. I already mentioned it.

But with Godel's coding, I am alleging
that you are doing that tacitly and objectionably. And I have stated
my objection precisely. Here it is again.

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?

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

What does "involving only numbers" mean?

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.

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' .

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' .

MoeBlee
.

Relevant Pages

  • Re: Godels comments about the "true reason" for incompleteness
    ... arbitrary Q. Obviously there is an implied quantification over Q, ... letters of the object language). ... we want to convey that "all" propositions follow from a contradiction. ... refutable in PA and as confirmed by the supposed undecidability of S. ...
    (sci.logic)
  • Re: Godels comments about the "true reason" for incompleteness
    ... why not formally admit propositions ... arbitrary Q. Obviously there is an implied quantification over Q, ... letters of the object language). ... refutable in PA and as confirmed by the supposed undecidability of S. ...
    (sci.logic)
  • Re: Rules of deduction.
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    (talk.origins)
  • Re: free variables in FOL
    ... > the other to existential quantification? ... When the language is interpreted, ... convince yourself that the two inference rules preserve validity. ... If you don't want to bother with the variable/name distinction you ...
    (sci.logic)
  • Re: Godel Incompleteness Theorem
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    (sci.math)