Re: what follows from denying an axiom

On Sat, 17 Jan 2009 10:44:17 -0800 (PST), george <greeneg@xxxxxxxxxxxxx>
said:
On Jan 16, 8:39 pm, Chris Menzel <cmen...@xxxxxxxxxxxxxxxxxxxx> wrote:
I'm not sure I see why one is any better (or worse) off using SOL.
Aw, C'mon: one is CLEARLY *worse* off; that's why it's not done. The
consequence relation is not r.e., so it is of no pragmatic value.

True, but not really relevant to my point.

The purpose of logic and proof is to allow you to EXTEND the level of
confidence that you already had in the axioms and the rules of
inference, TO the conclusions/theorems. If you can't have any
confidence that a possible consequence ACTUALLY IS a necessary one,
then the whole purpose is defeated. But THAT is NOT the point:

We agree.

THAT is the DOWN side of the trade-off. The upside is that YOU KNOW
WHAT YOU ARE TALKING ABOUT.

And my claim was that, because of Henkin's "general" semantics for
second-order languages, you don't know any better what you are talking
about when you use a second-order language and stipulate that it is to
be interpreted by the standard model theory (rather than Henkin's) than
when you use a first-order language and stipulate an intended model.

Consequences that are not TM-verifiable AS consequences (in 2OL)
become analogous to truths (in FOL) that were true in the standard
model but not TM- verifiable as consequences because they ARE NOT
(first-order) consequences, due to the existence of non-standard
models. My point is, you have unverified truths IN BOTH cases, but in
the 2nd-order case, at least you don't ALSO have to worry about
ontological contamination by things OTHER than those you INTENDED to
be talking about.

I don't know why there isn't as much a danger of "contamination" in the
second-order case due to the possibility that your language could be
interpreted a la Henkin. If you point out that there are nonstandard
models of PA in which the "PROOF" predicate is true of things that
aren't really codes of proofs in PA and argue that therefore the
predicate doesn't *mean* "(code of a) proof" in first-order logic, the
correct response is simply that nonstandard models are not the
interpretations I have in mind. In the same way, if I were to challenge
your claims about the fixity of meaning in second-order logic by
pointing out that there are nonstandard generalized models of
second-order PA, you'd (quite properly) bat it away with a similar
response. The two situations seem to me to be exactly on a part.

Second-order languages have a generalized model theory (from Henkin)
that is essentially first-order.

Yes and no.

Just Yes, I think.

The Henkin semantics IS NOT SECOND-ORDER ANYTHING, precisely BECAUSE
it is essentially first-order.

I never claimed Henkin's semantics was second-order anything. Quite the
opposite in fact if you'll have another look at what I wrote. The only
thing I referred to as second-order were the languages for which it is a
semantics. It is just a fact that languages with predicate quantifiers
are referred to as "second-order", and that's all I meant.

.

Relevant Pages

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  • Re: what follows from denying an axiom
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