How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Frederick Williams on 27 Jun 2010 14:37 Charlie-Boo wrote: > For the record, ZFC/PA is often described as �ZFC Arithmetic Proofs/ > Peano Arithmetic� and abbreviated ZAPPA. Google it and you�ll see > it�s very popular. Frank-ly, my dear, I don't give a damn -- I can't go on, I'll go on.
From: apoorv on 27 Jun 2010 14:42 On Jun 27, 3:50 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 27, 7:51 am, Tim Little <t...(a)little-possums.net> wrote:> On 2010-06-26, R. Srinivasan <sradh...(a)in.ibm.com> wrote: > > > > The theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > > > exist"). > > > Actually ~Inf does not assert "Infinite sets do not exist". It only > > asserts "there does not exist a successor-closed set containing the > > empty set". It may turn out to prove an equivalent statement under > > the rest of the axioms, but ~Inf does not actually mean "infinite sets > > do not exist". > > Please see my reply to Transfer Principle. We may sidestep this issue > by replacing ZF-Inf+~Inf in my post with a theory F which will only > admit models with hereditarily finite sets. > > > > This proof obviously implies that "There does not exist a model for > > > PA", for a model of PA must have an infinite set as its universe > > > Even if your intepretation of ~Inf were correct, all it would prove is > > that ZF-Inf+~Inf does not model PA. > > Sure. But the point is that ZF-Inf+~Inf is *the* chosen metatheory > (or model theory) of PA, in which models of PA, if they exist, can be > constructed. I happen to have chosen a metatheory which will not admit > any models of PA. In such a metatheory, PA is provably inconsistent > because we have a proof that models of PA cannot exist. > > > > Now I am sure a lot of people are going to jump up and down and > > > protest at this interpretation. But it is logical. > > > No, it is not. It exhibits a fairly elementary failure of logic. The > > statement "X models PA" implies "PA has a model". However, "X does > > not model PA" does *not* imply "PA has no model". > > What you are effectively saying is that ZF-Inf+~Inf is the "wrong" > metatheory because there *are* models of PA "out there" , meaning > outside of our chosen model theory. This is just an assertion of > Platonic existence. There is no particular reason for preferring ZF to > ZF-Inf+~Inf as a model theory for PA. The fact that the latter theory > yields an unpleasant result does not make it "wrong". > > RS Very well put. If we take it that PA captures the intuitive notion of numbers, then maybe, we should be able to interpret into PA that portion of any theory that purports to represent numbers . So, if we take the language of set theory, and the axioms Ey Ax ~x e y and Ax Ez Ay (y e z <->y e x or y=x), we have an interpretation into PA with e corresponding to <. If we want to preserve this interpretation, the axiom of infinity cannot be added to these two axioms, as it leads to a 'number' w other than 0 without a predecessor and PA does not admit of such a number.So, we have a choice between giving up infinity and and accepting PA as a 'gold standard' or accepting that induction extends across infinity to non-standard naturals. -apoorv
From: Charlie-Boo on 27 Jun 2010 14:44 On Jun 27, 1:59 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 24, 2:01 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > And the ZFC part is the DATA STRUCTURES of this "programming > > language". When programmers need to go beyond aleph-1 integers > > DAMN, you're stupid. > PROGRAMMERS NEVER need to go beyond aleph-1 integers. That's not true. Many computers print bills and they deal with decimal numbers - the number of dollars isn't always a whole number. > Indeed, they never even MAKE IT UP to aleph-ZERO integers! > COMPUTERS ARE FINITE! And so are you. > Jeezus. Good decision. C-B
From: Charlie-Boo on 27 Jun 2010 14:44 On Jun 27, 2:00 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 24, 5:01 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > Read the first 3 sentences of Godel's famous 1931 article (not famous > > enough, unfortunately.) > > YOU are TELLing US to READ something?? > I'm sorry, it doesn't work that way. > You ASK us a QUESTION about this, if you have read it. > We doubt this, frankly, since you obviously haven't understood it. How is that obvious? C-B
From: MoeBlee on 27 Jun 2010 15:44
On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > The theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > exist"). ~Inf is not (1) "there does not exist an infinite set", but rather ~Inf is (2) "there does not exist a successor inductive set". I don't know that there is a proof in Z set theory of the equivalence of (1) and (2). However, in ZF-Inf we can prove that equivalence, but it does take a bit of argument. > This proof obviously implies that "There does not exist a > model for PA", You get credit for skillful legerdemain here, but nothing more. Let's call ZF-Inf+~Inf by the name 'Y'. Now Y proves that there does not exist ANY theory (in the ordinary sense in which first order PA is a theory). Moreover, Y proves that there does not even exist any first order LANGUAGE to be the language of a theory. This is all obvious (once you think about it for a moment and are not distracted too much by your logical legerdemain): A a first order language has an infinite set of variables, and a theory is a certain kind of infinite set of sentences, so if there are no infinite sets, then there are no first order languages and no theories (in the sense in which first order PA is a theory). Your meta-theory Y proves that there IS NO object that satisfies the description we provide the rubric 'PA'. Any argument you base on theory Y proving anything about PA (with 'PA' defined in some ordinary way) is just an exercise in vacuous reasoning. Not only does Y prove that anything that fulfills the description we give to PA does not have a model, but Y also proves that anything that fulfills the description we give to PA DOES have a model, since Y proves that THERE IS NO OBJECT that fulfills the description we give to PA. Of course, you could eschew vacuous reasoning as I just gave, but then you're NOT using Y as the meta-theory, since Y=ZF-Inf+~Inf deploys classical first order logic. If you wish to argue from some OTHER logic for the meta-theory then that meta-theory is not ZF-Inf+~Inf. I expect that if you answer this point, you will do so with even more of your confusions about the basics of mathematical logic. In that case, I likely will not bother to serve as your nurse to clean up your mess. MoeBlee |