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From: James Burns on 29 Jun 2010 12:20
Charlie-Boo wrote:
> On Jun 29, 10:34 am, Chris Menzel
> <cmen...(a)remove-this.tamu.edu> wrote:
>>On Tue, 29 Jun 2010 03:34:12 -0700 (PDT), Charlie-Boo
>><shymath...(a)gmail.com> said:
>>>On Jun 29, 12:18 am, Chris Menzel
>>><cmen...(a)remove-this.tamu.edu> wrote:

>>>>The best known approach uses a mapping that Ackermann
>>>>defined from the hereditarily finite sets into N
>>>
>>>There are too many sets to map them 1-to-1 with
>>>the natural numbers.
>>
>>Apparently you have yet to master the semantic role
>>of adjectives.
>
> Ok, then tell me. What is the semantic role of adjectives?

Part of the semantic role of adjectives is to cause
"hereditarily finite sets" to mean something different
from "sets". Now that I've got you started, why don't
you go ahead and see if you can figure the rest of it out.

> http://blog.mrm.org/wp-content/uploads/2007/09/wizardofoz.jpg

Who are you supposed to be? The Wizard? Dorothy? Toto?

It can't be Toto, because Toto would have at least tried
to figure out what the semantic role of adjectives was.

Jim Burns

From: herbzet on 29 Jun 2010 22:33


Aatu Koskensilta wrote:
> Frederick Williams writes:
>
> > Yes, you can: take Gentzen's proof (or Ackermann's etc) and formalize
> > it in ZFC.
>
> This is a pretty silly way of proving the consistency of PA in set
> theory. That PA is consistent is a triviality. The interest in Gentzen's
> proof lies elsewhere.

The axioms of PA are supposed to be reasonably self-evident truths
about the naturals. But ...

I read somewhere that seeing the truth of the infinite number of
induction axioms of PA is in some sense equivalent to transfinite
induction up to epsilon_0.

That was a rather mysterious but intriguing remark to me. I wonder
if it has anything to with what you're hinting at here.

--
hz
From: herbzet on 29 Jun 2010 23:21


Aatu Koskensilta wrote:
> herbzet writes:
>
> > Hope to hear a reply to you from someone who actually knows
> > what he's talking about.
>
> How did you like my reply?

Fine, so far as it went.

billh04 distinguished in his post between the consistency of
PA being proved in ZFC, and a (meta) proof of the relative
consistency of PA with regard to ZFC.

You did not address the distinction drawn.

* * * * * *

Also, I was hoping for a reply from someone who knew
what he was talking about.

Hyuk, hyuk.

* * * * * *
As far as I'm concerned, if in theory A you have as theorems
the axioms of theory B, then the axioms of B are true as far
as theory A is concerned, and hence consistent, again as far
as A is concerned.

It's a bonus if in addition A proves "B is consistent".

But: if A also proves "A is inconsistent", then you punt.

--
hz
From: Chris Menzel on 30 Jun 2010 00:18
On Tue, 29 Jun 2010 23:21:49 -0400, herbzet <herbzet(a)gmail.com> said:
> Aatu Koskensilta wrote:
>> herbzet writes:
>>
>> > Hope to hear a reply to you from someone who actually knows
>> > what he's talking about.
>>
>> How did you like my reply?
>
> Fine, so far as it went.
>
> billh04 distinguished in his post between the consistency of
> PA being proved in ZFC, and a (meta) proof of the relative
> consistency of PA with regard to ZFC.
>
> You did not address the distinction drawn.

What distinction do you have in mind beyond the fact that they are
simply two rather different proofs?

From: Aatu Koskensilta on 30 Jun 2010 12:38
MoeBlee <jazzmobe(a)hotmail.com> writes:

> By the way, would you recommend a good (hopefully, fairly easy)
> reference on the workaround you mentioned?

The best reference is H�jek and Pudl�k's _Metamathematics of
First-Order Arithmetic_, where you will learn all you need to know about
doing recursion theory in fragments of first-order arithmetic. Alas,
it's neither cheap nor easy.

But in fact you know enough already to puzzle it out on your own. Recall
for example the following theorem of PA:

(*) There is no consistent axiomatizable completion of PA.

When we assert that (*) is provable in PA we obviously have something
more interesting in mind than just the triviality that we find in the
ontology of PA no infinitary objects, such as extensions of PA, at
all. What we mean is that PA proves:

(*') Whenever i is an index of an r.e. set A such that all of the
axioms of PA are in A, A is either inconsistent or incomplete.

Using similar devices we can in the language of PA quantify over sets
definable using arithmetical formulas of restricted quantifier
complexity, and state results such as

Every consistent Sigma-n set of sentences has a Delta-n+1 model.

for a fixed n in the form

Whenever A(x) is a Sigma-n formula that defines a consistent set of
sentences, there is a Delta-n+1 formula defining a model of the set of
sentences defined by A(x).

and so on.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
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