On Wed, Sep 4, 2013 at 1:07 AM, James Propp <jamespropp@gmail.com> wrote:
Sorry for being unclear, Andy. What I meant by "existence of a truth-value for P within PA" is "definability within PA of a number whose value is 1 if P and 0 if not-P". Now that I think about it, I suspect that for every P in PA, there's a way to define (within PA) a number n such that "n=1 iff P" and "n=0 iff not-P" are both provable in PA. But it's 1:00am, so I don't put much faith in what I think or suspect right now.
I'm still not sure what your definition of "defineable" is. If you use the one that I gave earlier, your statement is trivially true; such an n is defined by the formula "n=1 iff P and n= 0 iff not-P". Andy On Tue, Sep 3, 2013 at 5:05 PM, Andy Latto <andy.latto@pobox.com> wrote:
I'm not sure exactly what you mean by "defineable". I'm assuming what you mean is that there is a formula P(x) with one free variable, such that
ExP(x) ^ ((AxAy P(x) ^ P(y)) -> x=y)
is a theorem of PA,then the number x such that P(x) is defineable in PA.