Re: Nonfirstorderizability
- From: "Michael De" <mikejde@xxxxxxxxx>
- Date: 7 Aug 2005 02:56:05 -0700
Torkel Franzen wrote:
> "Michael De" <mikejde@xxxxxxxxx> writes:
>
> > To show in general the nonfirorderizability of an English sentence we
> > translate it into the langauge of (first or second-order) arithmetic
> > and show that it is true in every nonstandard model but false in the
> > standard model.
>
> I assume that by "nonfirstorderizable" you mean that the meaning of
> the English sentence cannot be captured in first order logic. Your
> description of the supposed procedure is odd in several ways. Where
> did you learn that "To show in general..."?
I may have misinterpreted it but I can't see how. It's from George
Boolos' "To be is to be a value of a variable", The Journal of
Philosophy, Vol. 81, No. 8, 430-449. However I just noticed that it is
also in (Boolos again) "Nominalist Platonism", The Philosophical
Review, Vol. 94, No. 3, 327-344 and the answer to my question is given
in a footnote on pp.327.
This is how I originally got my interpretation of the technique (from
"To be is..."): "The proof due to Dave Kaplan that (B) has no
first-order equivalent is simple and exhibits an important technique in
showing nonfirstorderizability: Substitute the formula (x=0 v x=y+1)
for Axy in (B), and, observe that the result...is a sentence that is
true in all nonstandard models of arithmetic but false in the standard
one" [pp.432-33].
The answer in the footnote in "Nominalist..." is: "Since nonstandard
models exist and since (trivially) every first-order sentence has the
same truth-value in any model of the set of all first-order truths of
arithmetic as in any other, no such correct symbolization can be a
first-order sentence" [pp.327]. So I see.
.
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