How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
in [Logic]
|
From: Jesse F. Hughes on 27 Jun 2010 15:22 Charlie-Boo <shymathguy(a)gmail.com> writes: > On Jun 27, 9:12 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Charlie-Boo <shymath...(a)gmail.com> writes: >> > On Jun 26, 10:41 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> >> Charlie-Boo <shymath...(a)gmail.com> writes: >> >> > On Jun 25, 10:21 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> >> >> 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! >> >> >> Yes. The usual language of ZFC does not have a successor function >> >> symbol, while the language of PA does. Thus, we must extend *the >> >> language* of ZFC and also add a defining axiom for the successor >> >> function. >> >> > "add an axiom" >> >> > Yikes! Yikes! >> > > You might want to learn about conservative extensions of a theory. > Any > > time you add a function symbol to a language, you must also add a > > defining axiom to the theory if you want the function to be > defined. > > You should already have that axiom as a theorem. > > You are adding Peano's Axioms, one at a time (as opposed to the > standard way of all at once when a set for N is defined.) As I said, you really should read about conservative extensions. And pick up a good introductory text on set theory. -- Jesse F. Hughes "I send papers to math journals and I damn well get a reply. Sure, they're polite rejections but they had better reply to me." -- James S. Harris, on influence.
From: Alan Smaill on 28 Jun 2010 05:57 Charlie-Boo <shymathguy(a)gmail.com> writes: > 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? MoeBlee's recent post in this thread recapitulates it for you. He has posted that argument before. > 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? Follow MoeBlee's post, and work it out for yourself; that is the educationally worthwhile way to do it. > 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. Then you have a puzzle to work out. > So the question is what does ZFC provide that is needed that PA does > not (implicitly in that it does mathematics at worst)? I'm sure you are smart enough to work this out for yourself. >> 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. Follow MoeBlee's outline. You can do it, can't you? > > C-B > >> -- >> Alan Smaill > -- Alan Smaill
From: herbzet on 28 Jun 2010 22:25 "Jesse F. Hughes" wrote: > But, Walker, you really have the wrong impression of me. I come to > sci.math mostly to read the cranks. I'm not proud of that fact *I* am proud of you, that you would make this startling announcement. -- hz
From: Aatu Koskensilta on 29 Jun 2010 07:57 "R. Srinivasan" <sradhakr(a)in.ibm.com> writes: > The theory ZF-Inf+~Inf clearly proves ~Inf ("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 (according to the classical notion of consistency, which I am > going to dispute shortly). Therefore we may take the proof of ~Inf in > ZF-Inf+~Inf as a model-theoretic proof of the inconsistency of PA, > which must be equivalent to its syntactic counterpart ~Con(PA). Without the axiom of infinity we can't prove that if a theory has no model it is inconsistent. Indeed, we can prove in ZF-Inf+~Inf that some consistent theories have no model. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 29 Jun 2010 08:23
Charlie-Boo <shymathguy(a)gmail.com> writes: > 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? Various bookkeeping axioms about set existence are used in the proof. Of course, for the consistency of PA ZFC is a huge overkill. We can prove PA is consistent in e.g. ACA, using a (definable) truth predicate for arithmetical statements. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |