Re: Continuum hypothesis

Nam D. Nguyen wrote:
Daryl McCullough wrote:
Nam D. Nguyen says...

That doesn't answer my question. My question has the clause:

"when one of the statement's hypothesis is a _subjunctive_ knowledge")

I didn't answer your question because I just wasn't in a subjunctive mood.

Anyway, it's a mathematical statement. There is no subjunctive:

T is consistent <-> T does not prove G_T

It - GIT - is actually a *meta* mathematical statement. Unlike inference
on 1st order which obeys the rules of inference and logical axioms, proofs
of meta levels, say in the context of GIT, requires the informal knowledge
or understanding of "arithmetic" (among other requirements of course),
therefore the conclusion must necessarily be subjunctive.

Now, Peter and you basically claimed the formality of the finitely
axiomatizable Robinson Arithmetic Q should be enough to define what we mean
by the phrase "as strong as arithmetic". But is it really enough?
What version of Q did Peter use? The version from

http://en.wikipedia.org/wiki/Robinson_arithmetic_Q

or the version from Shoenfield, both of which are of different language
signatures? To the extend that we have to agree what version of the-so-called
"Robinson Arithmetic" to be used for defining the phrase, then there - alone -
would make the phrase "as strong as arithmetic" already informal.

Can we use your version to define the phrase?

You wrote:
> Sure we do. For a theory T to be as "strong as arithmetic" just means
> that it is a recursively axiomatizable theory such that there is a
> recursive mapping M from formulas of arithmetic to formulas of T such
> that M(not Phi) = not M(Phi) and such that for any formula Phi in
> arithmetic, if Phi is a theorem of Robinson arithmetic (PA minus the
> induction axioms), then M(Phi) is a theorem of T.

The answer is "No": between your version and Shoenfield's version of Q
(his 'N' system), the formula:

(1) Axy [(x<y) \/ (x=y) \/ (y<x)]

is provable in one but not in the other; and *informally* (1) looks like
an "arithmetically" true formula. Besides, *within FOL framework alone*
(i.e. no transfinite arithmetic, higher order logic, etc...) none of these
systems are proven to be (syntactically) consistent. And if you have to depend

Should have read: "...none of these systems have been proven to be..."

on the "naturals" to *assume* these clearly defined formal systems be
consistent (so that the phrase "as strong as arithmetic" is sound),
the you end up with circularity!

In summary, Peter's and your's belief that the phrase "as strong as arithmetic"
can be precisely formally captured within the framework of FOL is incorrect.
(If nothing else, what Godel couldn't do, Peter and you, or anybody else for that
matter, won't be able to do!)


--
Daryl McCullough
Ithaca, NY


.

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