Re: Enderton problem

On 6 joulu, 02:56, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx> wrote:
On Wed, 5 Dec 2007 14:41:18 -0800 (PST), Gc <Gcut...@xxxxxxxxxxx>
wrote:



On 5 joulu, 14:09, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx> wrote:
On Tue, 4 Dec 2007 13:15:54 -0800 (PST), Gc <Gcut...@xxxxxxxxxxx>
wrote:

On 4 joulu, 13:13, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx> wrote:
On Tue, 04 Dec 2007 04:56:01 -0600, David C. Ullrich

<ullr...@xxxxxxxxxxxxxxxx> wrote:
On Mon, 3 Dec 2007 16:34:37 -0800 (PST), dpo...@xxxxxxxxxxxxx wrote:

If you have Enderton's A Mathematical Introduction to Logic handy,
this problem comes from number 7 in chapter 2, section 6.

Consider a language with a two-place function predicate symbol <, and
let N = (N; <) be the structure consisting of the natural numbers with
their usual ordering. Show that there is some A elementarily
equivalent to N such that <A has a descending chain.

Okay, here's what I'm thinking. We let the domain of A be the set {1/
n : n in the natural numbers} and define (m, n) is in <A iff (n, m) is
in <N.

This makes no sense - surely you meant (1/m, 1/n) is in <A iff (n, m)
is in <N? (Didn't notice at first because I didn't really read your
description of A, since it was clear that this is the wrong
way to go about it anyway.)

So A appears to have a descending chain. Now I need to show
that A and N are elementarily equivalant. I can do this by showing A
is a model for ThN. But...how do I do this?

Don't know.

Come to think of it, it's clear that your A is _not_ elementarily
equivalent to N. For example, "there is a smallest element" can
be expressed in the language.

How well that goes along with your construction?

Huh? It has nothing to do with the solution I gave.

OK. But your theory is not a conservative extension of the original
theory, that`s way its models can`t be elementarily equivalent to the
(N,<).

Sigh. Yes, it _is_ a conservative extension.

What's an example of a sentence in the second theory, in
the language of the first theory, which is not also in
the first theory?

Yes, you are right. I was confused about the meaning of the descending
chain. I thought that was just an ever decreasing chain.


.

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