How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Jesse F. Hughes on 27 Jun 2010 09:12 Charlie-Boo <shymathguy(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. In "good" cases, one can prove that the extension is conservative. Utterly standard. Wikipedia has pages on both "Conservative extensions" and "Extensions by definitions". -- "Witty adolescent banter relies highly on the use of 'whatever.' Anyone out of high school forced to watch more than an hour of 'Laguna Beach' might possibly feel the urge to beat themselves about the head with a large stick." -- NY Times on an MTV reality show
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 |