Re: Enderton problem

Gc <Gcut667@xxxxxxxxxxx> writes:

On 11 joulu, 18:39, Alan Smaill <sma...@xxxxxxxxxxxxxxxx> wrote:
Gc <Gcut...@xxxxxxxxxxx> writes:
On 11 joulu, 17:00, Alan Smaill <sma...@xxxxxxxxxxxxxxxx> wrote:
Gc <Gcut...@xxxxxxxxxxx> writes:
You don`t understand. We don`t interpret < itself only the sign "<" of
our language. The < is in the structure what it is, because a
structure*IS* an interpration. You can interpret sign "<" to a > of
(M,>) where M contains a chain 0 >-1 > -2 > -3....and interpret each
"n" of our theory to - m in the model. So the theory says that it is
a theorem that there is a chain 0 < 1 < 2 < 3...

right; and that's an *ascending* chain, in the sense used in the
original problem, which is what others have been saying all along.

What do you mean? Are you saying that 0 < 1 < 2 < 3... is ascending?

Yes.
That's because in model theory, as you said,
"We don`t interpret < itself only the sign "<" of our language".

I
am talking about about 0 >-1 > -2 > -3.... If that`s descending I`m
right, because that`s the chain what corresponds to sentences "0" "<"
"1", "2" "<" "3" etc by the the weak theory in the model that I am
talking about by interperation "n" = - n and "<" = >.

The problem asked for a structure where, according to the interpretation
of the symbol "<", all the orginal statements about N with the usual
order are true, and also where there is an infinite descending chain
according to the interpretation of the symbol "<".

Hehe. I think you are moving the goalposts. The question was:
"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."

Now the exact conditions mentioned here are met by my proof.

Oh no they're not.

We all know that there are mathemetical structures with descending chains --
that's not at issue, as I hope you understand.

You need a sequence of distinct elements e1,e2,e3 such that the symbols

"x < y"

come out all true when x,y are interpreted successively as e2,e1
then e3,e2 then e4,e3 ...

You don't have that.

That`s true.

good.

But could you point me the exact words where these
conditions were asked?

"descending chain".
That's what it means, if you unfold the standard definitions from
model theory (and, hey, this is an exercise in model theory).

So it does not solve the original problem.

--
Alan Smaill


--
Alan Smaill
.

Relevant Pages

  • Re: Enderton problem
    ... So A appears to have a descending chain. ... It's also not elementarily equivalent to N, ... have language with usual truths about N expressible in this ... It depends on the interpretation of "<". ...
    (sci.logic)
  • Re: Enderton problem
    ... our language. ... and also where there is an infinite descending chain ... N is the naturals and < their usual order, ... Alan Smaill ...
    (sci.logic)
  • Re: Enderton SOLUTION
    ... given the language just had a binary predicate. ... Bridge's book "beginning model theory" is clear that the signature/type ... it needs no new constants to express a descending chain. ... interpretation of c in the structure that we've ...
    (sci.logic)
  • Re: Enderton SOLUTION
    ... given the language just had a binary predicate. ... Bridge's book "beginning model theory" is clear that the signature/type ... it needs no new constants to express a descending chain. ... interpretation of c in the structure that we've ...
    (sci.logic)
  • Re: Enderton SOLUTION
    ... given the language just had a binary predicate. ... Bridge's book "beginning model theory" is clear that the signature/type ... it needs no new constants to express a descending chain. ... interpretation of c in the structure that we've ...
    (sci.logic)