How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: MoeBlee on 30 Jun 2010 11:52 On Jun 30, 10:43 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > "R. Srinivasan" wrote: > > Hmmm. I am not very conversant with classical model theory. So > > according to you, there is a "standard model for the LANGUAGE of PA" > > even if the theory PA is inconsistent. May I infer that you have used > > infinite sets to define this model? > > No sets, and a fortiori no infinite sets, are required (not for > first-order PA anyway): Perhaps in an informal sense of 'model'. My remarks are in the formal sense of a model, in which there is a set that is the universe for the model. > The above is a model of PA whether or not it and ZFC are consistent. Just to be clear, I was speaking of a model ('structure', if you prefer) FOR the LANGUAGE of PA, whether or not it is a model OF the THEORY PA. That there are models ('structures" if you prefer) for the language of a theory does not entail whether or not the theory is consistent. MoeBlee
From: Frederick Williams on 30 Jun 2010 11:57 MoeBlee wrote: > > On Jun 30, 10:43 am, Frederick Williams > <frederick.willia...(a)tesco.net> wrote: > > "R. Srinivasan" wrote: > > > Hmmm. I am not very conversant with classical model theory. So > > > according to you, there is a "standard model for the LANGUAGE of PA" > > > even if the theory PA is inconsistent. May I infer that you have used > > > infinite sets to define this model? > > > > No sets, and a fortiori no infinite sets, are required (not for > > first-order PA anyway): > > Perhaps in an informal sense of 'model'. My remarks are in the formal > sense of a model, in which there is a set that is the universe for the > model. Indeed so, I was thinking of what is _in_ the model modelling the language elements of PA: variables, constant and functions. Sorry if I misunderstood. And sorry to R. Srinivasan if I have added to his confusion. -- I can't go on, I'll go on.
From: MoeBlee on 30 Jun 2010 12:06 On Jun 30, 10:57 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > MoeBlee wrote: > > > On Jun 30, 10:43 am, Frederick Williams > > <frederick.willia...(a)tesco.net> wrote: > > > "R. Srinivasan" wrote: > > > > Hmmm. I am not very conversant with classical model theory. So > > > > according to you, there is a "standard model for the LANGUAGE of PA" > > > > even if the theory PA is inconsistent. May I infer that you have used > > > > infinite sets to define this model? > > > > No sets, and a fortiori no infinite sets, are required (not for > > > first-order PA anyway): > > > Perhaps in an informal sense of 'model'. My remarks are in the formal > > sense of a model, in which there is a set that is the universe for the > > model. > > Indeed so, I was thinking of what is _in_ the model modelling the > language elements of PA: variables, constant and functions. Sorry if I > misunderstood. And sorry to R. Srinivasan if I have added to his > confusion. But your point is also well taken in the broader sense of a model, not necessarily confined to the set theoretic "Tarski" sense. MoeBlee
From: MoeBlee on 30 Jun 2010 13:03 On Jun 30, 11:38 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > MoeBlee <jazzm...(a)hotmail.com> writes: > > By the way, would you recommend a good (hopefully, fairly easy) > > reference on the workaround you mentioned? > > The best reference is H jek and Pudl k's _Metamathematics of > First-Order Arithmetic_, where you will learn all you need to know about > doing recursion theory in fragments of first-order arithmetic. Alas, > it's neither cheap nor easy. Thanks, yeah, that one's been on my wish list. > But in fact you know enough already to puzzle it out on your own. You overestimate my knowledge and intelligence. > Recall > for example the following theorem of PA: > > (*) There is no consistent axiomatizable completion of PA. > > When we assert that (*) is provable in PA we obviously have something > more interesting in mind than just the triviality that we find in the > ontology of PA no infinitary objects, such as extensions of PA, at > all. What we mean is that PA proves: > > (*') Whenever i is an index of an r.e. set A such that all of the > axioms of PA are in A, A is either inconsistent or incomplete. Okay, I'm pretty much following. Pretty much in the same vein as Godel coding of such things as "Con(PA)", right? > Using similar devices we can in the language of PA quantify over sets > definable using arithmetical formulas of restricted quantifier > complexity, Okay, I'm hanging in here. Through coding, we can "talk about" PA formulas, thus, "talk about" the sets those formulas define, right? (I'm lost about the restricted quantifier complexity part, but nevermind for now, I'm just trying to get the big picture at this point). > and state results such as > > Every consistent Sigma-n set of sentences has a Delta-n+1 model. I know what a Sigma-n set of sentences is. But what's a Delta-n+1 model? > for a fixed n in the form > > Whenever A(x) is a Sigma-n formula that defines a consistent set of > sentences, there is a Delta-n+1 formula defining a model of the set of > sentences defined by A(x). > > and so on. Easy for YOU to say. Thanks for this. A followup question: Is there ANYTHING in mathematical logic you don't know about? MoeBlee
From: R. Srinivasan on 30 Jun 2010 13:36
On Jun 30, 7:42 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 29, 10:45 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > On Jun 30, 1:36 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Jun 29, 12:28 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote > > > > to much flotsam still for me to spend more time than I've already > > > spent. > > > > However: > > > > > > We PROVE from ZF-Inf that there IS NO SUCH object that you are calling > > > > '> D'. (or at least we have not before us a proof that there IS such an > > > > > object). Just adding a constant symbol 'D' and saying whategver you > > > > > want about it does not override. > > > > You do not have any such proof. > > > > I SAID, "or at least we have not before us a proof that there IS such > > > an object". > > > > But it's simple anyway: > > > > Therorem of ZF-I: > > > > Ex~En x in P_n(0) -> ~EyAz(zey <-> ~En z in P_n(0)) > > > > Proof: Toward a contradiction suppose Ex~En x in P_n(0) and > > > Az(zey <-> ~En x in P_n(0)). > > > Let ~En x in P_n(0). > > > Let j be arbitrary. > > > ~En xu{j} in P_n(0). > > > So Aj j in Uy. > > > > Theorem of ZF-I: > > > > ~Ex~En x in P_n(0) -> Ey(Az(zey <-> ~En z in P_n(0)) & y=0) > > > > Proof: Immediate. > > > > Then, as far as I know (which is pretty limited) it is not decided in > > > ZF-I whether Ex~En x in P_n(0). Someone may inform me further on that, > > > but I'm pretty sure that ZF-I doesn't tell us whether there are or are > > > not sets other than the hereditarily finite sets. > > > I think it is not known whether this proposition (That there are sets > > other than the hereditarily finite sets) is undecidable, refutable, or > > provable in ZF-I. Undecidability of this propostion is just an > > assumption as far as I know. > > As far as you know. > Yes. Undecidability of that proposition is a mere assumption. > > > > > How can something be true > > > > "in the standard model of PA iff PA is inconsistent" ????? > > > > Typo of omission. I meant, "true in the standard model for the > > > LANGUAGE of PA", as I had posted in previous messages. > > > Hmmm. I am not very conversant with classical model theory. So > > according to you, there is a "standard model for the LANGUAGE of PA" > > even if the theory PA is inconsistent. > > No just according to me. > > I have no idea why you ask. This is basic mathematical logic. > > > May I infer that you have used > > infinite sets to define this model? > > Of course. > > > How can you do that if the theory > > PA is inconsistent (which would make ZFC inconsistent as well)? > > If ZFC is inconsistent, then we can prove ANYTHING in the langauge of > ZFC. There wouldn't be any "epistemological" value to such proves, but > still they exist as formal objects. > > And this "what if PA (or Z-R or whatever foundational theory) is > inconsistent?" is not special to the matter of Z-R proving that PA is > consistent. If PA is inconsistent, then vast amounts of ordinary > mathematics (including that used in everyday technology) become no > less "questionable" than the mathematics used to prove, in Z-R, that > PA is consistent. Please read Franzen's book on incompleteness. > > Have you ever read ANY book at all on the subject of set theory and/or > mathematical logic? > At my age (53) and given my day job, I have to be conservative about what I read. I do not want to fill my head with stuff that I do not believe in. Classical model theory fits that description perfectly. And in any case, these days I am reluctant to even attempt to read anything that does not appear in a minimum of 14-pt font (preferably 16 or 18) on a computer screen with full brightness. For what it is worth: Your paradigm is based on the philosophy called Platonism, which is the only way out for you to deny circularity. I am not going to waste time explaining this to you. It is enough for you to know that I reject Platonism wholesale and developed a logic NAFL which satisfies that requirement. And since I reject Platonism, I assert that what you have laid out above and what you call "basic mathematical logic" is circularity at its worst. Cut back to about 350 BC and look at the basic laws of logic laid out by Aristotle and others. Those are the *real* basic laws of classical logic, and my point of departure starts from there. So there is no real point in my going too deep into most modern books on classical logic. > > > Anyway, in NAFL there is no such thing as the "standard model for the > > LANGUAGE of PA". Truths are with respect to (consistent) axiomatic > > theories and there are no truths in just the language of a theory. > > I tried to follow what the hell about NAFL. I sincerely went over and > over your paper. I could not make sense of it. I sincerely asked you > for your explanations. I could not make sense of them, especially as > most it was waffle about how you were going to one day show this or > that. In particular, I never got from you a definition of "an NAFL > theory", at least not a coherent one. Sorry, but I've moved on now. > Whatever you say about NAFL, I really can't say much in reply. But if > your EVER wish to present to me a FORMAL theory and FORMAL semantics, > I'll probably be around. > I did spend my energy reading about a very interesting experiment in quantum physics, namely Afshar's experiment. Afshar claimed that this experiment demonstrates that a famous principle of quantum mechanics, namely Bohr's complementarity principle (BCP) is violated. I immediately recognized in this experiment a beautiful application of the logic NAFL. I wrote a paper on that and published it in a mainstream physics journal, with top-of-the line physicists in its editorial board. This paper explains why BCP is still intact, and also explains the logical fallacies inherent in Afshar's interpretation based on classical thinking. Physicists are much more open-minded that logicians or philosophers, and less tradition-bound. Which is why a bold theory like quantum mechanics got published in the first place. And now they have dared to publish a bold new interpretation of quantum mechanics based on the logic NAFL, that is tailor-made to explain the foundations of quantum mechanics (and indeed, all part of science that is worth doing, as far as I am concerned). If I had wasted my time trying to dig into the rubbish that you have laid out above, I would not have had much time or energy left to deal with the kind of stuff that *i* consider worth doing. Like I said, I am not privileged to have a day job in which I can spend my time working on NAFL. RS |