Re: Model here, model there
- From: george <greeneg@xxxxxxxxxx>
- Date: Mon, 29 Oct 2007 09:18:01 -0700
On Oct 27, 11:30 am, Newberry <newberr...@xxxxxxxxx> wrote:
Does "Con(T)" express that T is consistent? "T is consistent" does. So
what is the difference in meaning between "Con(T)" and "T is
consistent"?
I'm almost flattered.
I came in with that very question over a decade ago.
You were privileged to get it answered by Peter Smith.
I had the far worse fortune to get Torkel Franzen.
The basic answer to your question is that
first-order theories, fundamentally, simply cannot
be ABOUT anything. You can INTERPRET the functions,
predicates, and constants of a first-order theory to refer to
any of many DIFFERENT things. The theory is simultaneously
about all AND none of these things, specifically.
This is usually a good thing.
Things like groups, fields, and rings occur in many places,
in many ways. Different examples of different kinds of mathematical
structure occur in different subfields, yet are recognizably the same
kind of structure. The fact that theories can have multiple
interpretations
IS USUALLY a GOOD thing.
An EVEN BETTER thing that theories can provide for you is
TheorEMS. A theorem is something that is provable from the axioms
of the theory. The important thing about theorems(because they're
provable)
is that they are true IN ALL MODELS of the theory. So basically, when
you have a theorem, YOU KNOW what it means. It is "always" true
and it always means "the same" thing (or the analogous thing) in EVERY
model into which you might re-interpret the theory.
Con(T), however, is NOT a theorem.
Despite the fact that T really is consistent, Con(T) is NOT provable
(from T, for sufficiently rich T) and therefore MUST be false in SOME
model of T.
The short answer to your question is, truly, therefore, that because
Con(T) is not a theorem, it simply does NOT HAVE a meaning, at least
not one that can be described without talking about a model as well
(and in that case, the meaning must reside at least partly in the
model, and not just in the statement Con(T)).
The generally accepted answer, however, is that T (and in this
case, T=PA) was invented FOR A REASON. There was a PARTICULAR
model, KNOWN IN ADVANCE, for which the axioms of T were KNOWN
to be true and from which the GOAL was to prove the REST of the truths
of T. If you interpret T *in*this*model*, THEN, Con(T) *does* mean "T
is consistent".
But completing/evaluating your interpretation of T in this model
requires the
employment of infinitary reasoning (or at least induction up to
epsilon_0).
T (or in this specific case, PA) has a STANDARD model.
In THAT model, all numbers are finite, so decoding numbers as proofs
is guaranteed to produce a (true) finite proof. In the non-standard
models
of PA, there are supernatural numbers, and because they are infinite,
they
can encode infinitary "proofs" of things that are disprovable
finitely.
So when Con(T) quantifies over "all numbers" (and therefore all
proofs),
it winds up being able to have "a proof" for any&everything, so
Con(T) is false.
But this doesn't change the fact that there are no "standard"(finite)
proofs
of anything that is standardly disprovable.
.
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