How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Charlie-Boo on 26 Jun 2010 21:09 On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy to > > prove that PA is consistent (its axioms and rules preserve truth) yet > > (by Godel-2) PA can't do such a simple proof as that. > > So what? ZFC can prove it. > ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove > that PA is consistent. This is why you can't say that "ZFC/PA > doesn't prove PA is consistent." "ZFC/PA" is just a meaningless > locution in any case. Since ZFC/PA is meaningles, how does ZFC show that we can't say that "ZFC/PA doesn't prove PA is consistent."? How can a system of logic show that an expression that contains meaningless syntax can't be said (is not truthful)? > ZFC is one thing. PA is another. And CBL is still another. However, CBL proves theorems with proofs that are about 1% the size of those published, while ZFC and PA take about 10 times the size published. So which is best? C-B
From: Charlie-Boo on 26 Jun 2010 21:28 On Jun 25, 5:19 am, Alan Smaill <sma...(a)SPAMinf.ed.ac.uk> wrote: > Aatu Koskensilta <aatu.koskensi...(a)uta.fi> writes: > > Charlie-Boo <shymath...(a)gmail.com> writes: > > >> Who has proved PA consistent using ZFC? If it were possible then I > >> assume someone would have done it. It certainly would be a very > >> educational exercise. > > > So why not have a try at it? You'll find all the details you need in any > > decent text. > > Not to mention that it has been outlined several times in sci.logic. Reference to a post with it? What of ZFC's set theoretic axioms is necessary - especially not bookkeeping ones like sets existing that are used throughout PA and are not needed in every proof? That is, I see little added by ZFC's axioms over PA's which are stolen anyway in the form of "definitions" that N has certain properties i.e. satisfies the PA axioms. So the question is what does ZFC provide that is needed that PA does not (implicitly in that it does mathematics at worst)? > It is of course more educational to work this out for oneself. Well, the first question is which proof to use. Then there is the question of how to formalize it. So it's at least 2 distinct steps. C-B > -- > Alan Smaill
From: Charlie-Boo on 26 Jun 2010 21:34 On Jun 25, 10:21 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > >> On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > >> > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy to > >> > prove that PA is consistent (its axioms and rules preserve truth) yet > >> > (by Godel-2) PA can't do such a simple proof as that. > > >> So what? ZFC can prove it. > >> ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove > >> that PA is consistent. This is why you can't say that "ZFC/PA > >> doesn't prove PA is consistent." "ZFC/PA" is just a meaningless > >> locution in any case. > >> ZFC is one thing. PA is another. > > > PA is a subset of ZFC, so I emphasize that by calling it ZFC/PA (it > > makes more sense to distinguish the two anyway.) This is besides the > > point. Who has proved PA consistent using ZFC? If it were possible > > then I assume someone would have done it. It certainly would be a > > very educational exercise. > > > In any case, it shows the weakness of PA. I added ZFC as that is so > > popular. > > No, I think you have a good point Thanks. > and an interesting new form of > argument. I'm gonna try it myself. > > People say that atoms are made up of subatomic particles. But you > can't make atoms up from protons, because they repeal each other. So > why would people think this? > > This is a great argument, because the class of protons is a subset of > the class of subatomic particles, just as the theorems of PA are a > subset of the theorems of ZFC (with suitable extension of the language > of ZFC). "with suitable extension of ZFC" Yikes! > I are as smart as Charlie. > > Final hint, Charlie: if someone says that ZFC suffices, and you show > that a subset of ZFC does *not* suffice, then you haven't refuted > their claim. my goodness C-B > -- > Jesse F. Hughes > "[M]oving towards development meetings for new release class viewer 5.0 > and since [I]'m the only developer, easy to schedule." > --James S. Harris tweets on code development- Hide quoted text - > > - Show quoted text -
From: Tim Little on 26 Jun 2010 22:51 On 2010-06-26, R. Srinivasan <sradhakr(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". > 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. > 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". - Tim
From: Charlie-Boo on 26 Jun 2010 22:59
On Jun 15, 3:18 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > George Greene <gree...(a)email.unc.edu> writes: > > PA doesn't know what an infinite set is. ZFC does. That is the main > > reason why ZFC can prove that PA is consistent (a model of PA *has* to > > be infinite, and PA can't prove that anything is infinite, since in > > its standard model, NOTHING IS). > > This doesn't make much sense; PA proves the consistency of many theories > that have only infinite models. Many => >1. Name two (w/ ref.s) C-B > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |