Re: what follows from denying an axiom
- From: Chris Menzel <cmenzel@xxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 17 Jan 2009 01:39:11 +0000 (UTC)
On Fri, 16 Jan 2009 15:55:31 -0800 (PST), george <greeneg@xxxxxxxxxxxxx>
said:
On Jan 16, 5:53 pm, Gc <Gcut...@xxxxxxxxxxx> wrote:
I maybe not should have said anything :) It is just so weird to think
that the completesness fails, it just feels it would bring down the
whole meaning of FOL.
Of course it would, but who cares? FOL is just some lame
approximation of both arithmetic AND set theory. FOL is fundamentally
inadequate to any realm that Godel's theorem applies to. IT IS
OBVIOUSLY INCOHERENT for there to exist a model of T + ~con(T), given
that, according to "the meaning of FOL", ~con(T) MEANS that models of
T do NOT exist!
I know, I'm starting to sound like the cranks; just because I think
something is counter-intuitive doesn't mean it's inconsistent. But
just because something has THE WRONG TRUTH-VALUE *does* mean that IT
DOES NOT MEAN whatever- it-is-allegedly-translating. Ditto for the
first-order take on the "Existence" (or lack thereof) of certain
bijections in the Lowenheim-Skolem theorem.
THE point is, IN ORDER to do this RIGHT, YOU HAVE to use SECOND- order
logic, if you are going to use "standard classical logic as we think
we know it" at all.
I'm not sure I see why one is any better (or worse) off using SOL.
Second-order languages have a generalized model theory (from Henkin)
that is essentially first-order. So writing something down in a
second-order language doesn't guarantee that it has a fixed meaning; you
have to stipulate that it is to be interpreted by standard second-order
model theory. But then how is that any better than using the language
of a first-order theory T (PA, say) to write down "~con(T)", say, and
stipulating that it is to be interpreted in the standard model of T? In
both cases one assumes an intended interpretation for the language in
order to express one's intended meaning.
.
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