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 28 Jun 2010 09:43 On Jun 27, 1:54 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 26, 7:41 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > > Charlie-Boo <shymath...(a)gmail.com> writes: > > > "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. > > And the month of June continues with more and more posters > coming out of the woodwork to challenge ZFC. Challenge ZFC how? ZFC has never been used to prove anything non- trivial or not already known. ZFC was designed to avoid paradoxes by making explicit what can be a set. It doesn't do anything else except what the Peano Axioms give it. And it is known to not be able to decide the fundamental questions of set theory. What's to replace? What is needed is honest answers to questions where the standard answer refers to ZFC outside of avoiding paradoxes and PA. C-B > This thread > marks the return of Charlie-Boo and Srinivasan. > > Surprisingly, Charlie-Boo is one ZFC challenger whom I > have yet to defend, even though I know that he's been > posting here for years. I still remember several years > back when he once compared ZFC to noodle soup. (Of course, > I don't know whether Charlie-Boo still considers ZFC to be > like soup anymore.) > > In this thread, Charlie-Boo is criticized for lumping > together ZFC/PA as if they were interchangeable. I must > point out that I myself lump them together all the time, > but not because I consider them interchangeable -- we know > that ZFC is a much stronger theory than PA. (Okay, okay, I > mean that _a_suitable_extension_of_ZFC_ is a much stronger > theory than PA....) > > Nonetheless, I lump ZFC and PA together as the two main > standard _theories_ (not "theorists"). In particular, I > often make statements such as, "Those who use standard > theories such as ZFC/PA are much less likely to be called > five-letter insults than those who use other theories." In > this case, I'm not saying that ZFC and PA are equivalent > to each other, but only that either theory is a suitable > theory to use if one wants to avoid five-letter insults. > > Srinivasan, meanwhile, is trying to come up with NAFL, > which is supposed to be an alternative _logic_ to FOL. If > I remember correctly, in NAFL, it's possible for some > statement to be similarly true _and_ false, unlike in FOL. > > Srinivasan was once fascinated by Ed Nelson's set theory, > called Internal Set Theory or IST. One axiom schema of IST > is called the Transfer Principle. I came up with my > current username right in the middle of a discussion about > IST with Srinivasan. > > Both Srinivasan and IST's creator appear to be sympathetic > to finitism. Srivinasan discusses ZF-Infinity+~Infinity, a > theory which can be used by finitists, while Nelson is > working on a proof that PA is inconsistent. And of course, > if Nelson's proof goes through, it would also prove that > ZFC is inconsistent, since, as so many were quick to tell > Charlie-Boo, ZFC proves that PA is consistent. > > Hughes will undoubtedly disagree with me, but I find the > arrival of all these opponents of ZFC at the same time > simply hilarious...
From: Charlie-Boo on 28 Jun 2010 09:45 On Jun 28, 8:58 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > YOU CANNOT PROVE PA CONSISTENT USING ZFC. > > Yes, you can: take Gentzen's proof (or Ackermann's etc) and formalize it > in ZFC. Give the slightest bit of details. > > The reason is that ZFC is just PA plus some dinkly little axioms about > > which sets exist. Consistency of PA has nothing to do with which sets > > exist. > > It has everything to do with V_omega. That's not a ZFC axiom. C-B > -- > I can't go on, I'll go on.
From: Charlie-Boo on 28 Jun 2010 09:50 On Jun 28, 8:58 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > YOU CANNOT PROVE PA CONSISTENT USING ZFC. > > Yes, you can: take Gentzen's proof (or Ackermann's etc) and formalize it > in ZFC. Who has done it? You lied the last time I asked and said Gentzen did it. Why didn't he? Who really has? NOBODY!!!!!!!!!!!! No false references - especially at over $400.00!!! Any references - give a summary of how ZFC was used for the whole thing. No more lies, no more wild goose chases, no more $400 BS references!! If no one has done it, and, "If it could be done it would have been." that you agreed to, then it can't be, by your words. C-B > > The reason is that ZFC is just PA plus some dinkly little axioms about > > which sets exist. Consistency of PA has nothing to do with which sets > > exist. > > It has everything to do with V_omega. > > -- > I can't go on, I'll go on.
From: billh04 on 28 Jun 2010 11:46 On Jun 28, 7:58 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > YOU CANNOT PROVE PA CONSISTENT USING ZFC. > > Yes, you can: take Gentzen's proof (or Ackermann's etc) and formalize it > in ZFC. Are you saying that it is a theorem of ZFC that PA is consistent? I thought that the proof of consistency of PA relative to ZFC by showing that there is a model of PA in ZFC was a metatheorem, not a theorem stated in ZFC and proved using the axioms of ZFC. > > > The reason is that ZFC is just PA plus some dinkly little axioms about > > which sets exist. Consistency of PA has nothing to do with which sets > > exist. > > It has everything to do with V_omega. > > -- > I can't go on, I'll go on.
From: MoeBlee on 28 Jun 2010 11:53
On Jun 27, 4:32 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 28, 2:11 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 27, 1:25 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > On Jun 28, 12:44 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > The proof in the theory ZF-Inf > > > +~Inf that infinite sets do not exist encodes (among other things) > > > that models of PA cannot exist. > > > PROVE IT. > > Trivial. If it's trivial, then it shouldn't take you but a minute to prove it. Please do. > > Look, in ordinary set theory such as Z set theory, we may prove that > > there is a certain object that is a certain theory and we define the > > rubric 'PA' to be that theory. > > Sure. If we want a metatheory of PA in which it *is* possible to > construct a model of PA (like Z set theory) then I do agree that we > need all the machinery you describe. You're not listening. Theory Y (AF-I+~I) does not provide for constructing the THEORY PA ITSELF. Why do you keep skipping this?: Y proves there are no infinite sets. PA is an infinite set of sentences, moreover axiomatized by an infinite set of axioms, moreover with a language that has an infinite set of variables. So Y proves THERE IS NO SUCH theory. Please do not continue to argue PAST that very simple state of affairs. > >But in ZF-Inf +~Inf we may prove that > > there does NOT exist any such object that meets the description of > > being a first order theory, let alone one having the specifics > > properties that, in Z, we ascribe to PA. So, in ZF-Inf +~Inf we cannot > > even define the rubric 'PA' in a way that PA is a theory. > > But now we are looking at a metatheory that only needs to *deny* that > models of PA exist. Same as I just wrote above. > So it does not need all the machinery that you > describe to accomplish this. It is enough for this metatheory to deny > the existence of infinite sets for it to rule out the existence of > objects that can be interpreted as models of PA. Same as I just wrote above. > > > That is all I need to calll ZF-Inf > > > +~Inf as a metatheory of PA. > > > It can't be a very "meaningful" metatheory for PA since it proves that > > there IS NO theory that meets the description we give, in such as Z, > > to'PA'. > > This and what you state below is again your own confusion. You've shown no confusion on my part. Saying "your confusion" and "trivial" when asked for proofs of things is not substantive argument. Anyway, just go back to this: Y proves there are no infinite sets. PA is an infinite set of sentences, moreover axiomatized by an infinite set of axioms, moreover with a language that has an infinite set of variables. So Y proves THERE IS NO SUCH theory. Y proves not only is there such a theory, but there is not even such a LANGUAGE for such a theory. MoeBlee |