Re: My talk about Godel to the post-grads.

On Jul 21, 4:50 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:

Whether Z is inconsistent or NOT, the incompleteness
theorem is provable in Z.

In that regard I'm referring to the first incompleteness theorem, and
in other places where I wrote 'the incompleteness theorems' I should
have specified that I mean the ordinarily discussed variants of the
first incompleteness theorem.

As far as I know (as far as I have gleaned from reading), the second
incompleteness theorem is also provable in Z set theory. However, I
have not gone through a proof of that in the detail that I have for
the first incompleteness theorems. The main outline is available in
Enderton (and expanded in the second edition), Boolos/Burgess/Jeffrey,
Smith, and other books. For a detail proof that includes many of the
details not given in ordinary textbooks, readers are usually referred
to Hilbert and Bernays 'Grundlagen der Mathematik', though that
classic has never been translated to English. (I wish someone would
translate it!)

MoeBlee




Really, Moeblee, since when could foundational issues of FOL reasoning be
independent of theory consistency?

I never claimed that the subject of first order reasoning is unrelated
to the notion of consistency.

(Don't you think that G(T) which is about
incompleteness, and CON(T) which is about consistency have some
 relationship?)

Of course there are relations among such subjects. I never claimed
otherwise. However, given the many relations among such subjects, it
is still the case that whether Z is inconsistent or NOT, the
incompleteness theorem is provable in Z.

MOREOVER, as I have read, as is widely recognized (though I haven't
personally confirmed the details), incompleteness is provable in a
system as weak as PRA (and, if I'm not mistaken, using only
constructive reasoning). Sure, you could even doubt the consistency of
constructive PRA (if 'constructive PRA' is not even a redundancy?),
but then what's the point? For then you might as well doubt the
consistency of ANYTHING. I mean, if you doubt the consistency of
constructive reasoing about finitistic mathematical events, then I
can't imagine anything you WOULDN"T doubt. And if you doubt everything
mathematical, then what is the point of singling out the
incompleteness theorems?

Give me axioms and rules of inference, then I'll believe theories such
as {Ax[x=a]}}, {Axy[x+y=0]}, ... are consistent.

Okay. I should have qualified to restrict to theories of some
foundational mathematical interest (and, of course, 'of some
foundational mathematical interest' is not intended to be a technical
term). So the question would be: If you doubt the consistency of PRA,
then what theory that even YOU consider to be of foundational interest
or importance would you not doubt its consistency?

So, you now can correct
your imagination: there are things - theory consistencies - I do *not*
doubt!

Okay, you don't doubt the consistency of certain trivial theories,
such as that there exists only one object. Any theories of more
interest that you don't doubt the consistency?

Oh, but I recall now. When we spoke about this a while back, indeed,
you wouldn't even assent to the most primitive reasoning about short,
finite strings of symbols. So, of course, with doubts as pervasive as
that, nothing could satisfy you.

Again, you're wrong: there are theory consistencies that I'm satisfied.

Again, any theories of more interest that you don't doubt the
consistency? Anyway, my immediate point above regarded doubt in
general - not just about inconsistency - as a while back you wouldn't
grant even the truth that certain very short finite strings match.

(Hint: perhaps you'd want to minimize the number of times you're wrong,
by refraining from asserting what *others* might know, think, doubt, ....!)

Rather, I should have qualified to restrict to mathematical theories
of some foundational interest.

You can verify that for YOURSELF just by following along each step in
textbook proofs. It is just willful ignorance on your part to keep
insisting that the incompleteness theorems are not provable from a
first order formal axiomatization and formal rules of inference.
What then is a meta theorem, to you?

I use the term 'meta-theorem' just as it is ordinarily used in
mathematical logic. There's no need for me to perform a ritualistic
definition for you, especially since I don't use it as TECHNICAL term..

That's the 1st clue here of your problems: 'meta-theorem' is a technical
term, and should be treated as such!

Specifically, a meta theorem is an *assertion* *about* formal reasoning of
a reasoning framework (e.g. FOL, SOL, etc...).

(1) But if that is a technical definition, then what are your
technical definitions of "about" and "formal reasoning" and
"framework"?

(2) I don't claim that one cannot devise a technical definition of
'meta-theorem'. It's just that I'm not presently aware of one, nor
have you shown one.

(3) Even technical terms can be used in non-technical senses. For
example, 'proof' has technical definitions and is also used in non-
technical senses (though, in this part of the discussion, I am using
'proof' in the technical sense)..

Where as, e.g. in FOL, axioms
are the originating points of a proof, provability of a meta theorem flows
from *knowledge* *about* the formal reasoning framework in question.

Surely, that's not part of your "technical" definition, right?

Examples of true meta theorems (or just meta theorems):

  - A FOL formula is finite in length.
  - GC is not an axiom of Q (as defined by, say, Shoenfield)
  - If ~GC is provable in Q then it's provable in any extension of Q.
Examples of false meta theorems (or just not meta theorems):

  - A FOL proof is not finite in length.
  - GC is an axiom of Q (as defined by, say, Shoenfield)
  - If Q is a consistent system, both GC and cGC are provable.
  ...

You don't need to provide such examples. I have a good sense of the
notion of a meta-theorem.

Can you give an example of a meta theorem that's *not* a
first order theorem?

Any theorem, whether a meta-theorem or not, that is stated in second
order language is not a first order theorem, though, of course, in
many cases, there may be a first order version too.

That's the 2nd clue here of your problems: I asked you for *a meta-theorem*,
and you came back with an answer that's twice nonsensical:

There's nothing nonsensical in my answer. Rather, your question is
aside the point. What possible bearing does anything I EXACTLY said
(or implied by anything I exactly said) depend on me barking out a
meta-theorem formulated in a second order theory have?

  - "Any theorem, *whether a meta-theorem or not*"!
  - Basically what you said is a 2nd order theorem is not a 1st order theorem!
    How does that become an example of a meta theorem?

No, I did not. However, it is true that a theorem formulated using
second order features that are not first order features is not a first
order theorem (whether meta-theorem or not) (in the trivial sense that
the exact formula is not a fist order formula). And I didn't say that
my remark "becomes an example of a meta-theorem".

You got to understand a question clearly, and answer straight to the point.
Of course you could have said "I don't know what a meta theorem is". But you
didn't.

No, I don't answer every question put to me as if I'm being
interrogated by the
Stasi. Your question is aside the point.

Anyway, what is the point of these questions?

Because the edifice of proof is different between a FOL theorem and
a meta theorem.

There's no requirement that first order theories may not prove meta-
theorems.

Wherever did you get the idea that there are not metatheorems
formulated in first order languages and proven in first order
theories?

And one has to understand the difference to appreciate why
GIT is a meta theorem. Naturally.

No, I understand that the incompleteness theorems are meta-theorems
because their subject matter is about theories themselves.

The fact that the
incompleteness theorems are provable in many ordinary formal first
order theories is not affected by my definition of 'meta-theorem' or
by the fact that there are also second order and higher order meta-
theorems too.

Sure. Z' above is an ordinary formal first theory in which incompleteness
theorems are provable. Which of course is true. But which is the only
thing you could say about?

First, Z' is not an ordinary theory (in the sense I meant 'ordinary',
which is one that is either ordinarily ...

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