How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Frederick Williams on 30 Jun 2010 05:23 Charlie-Boo wrote: > The best way to explain that a ZFC axiom is not used is to give the > proof without using any ZFC axioms - good luck! > > How would you prove the PA axioms in ZFC, then? You keep saying it > isn't from an axiom but can't say how it is done - so how do you know > it isn't? As Tim Little says elsewhere in the thread, you define S, +, and * in the language of ZFC and then prove the counterparts of PA's axioms as theorems. It's done in Suppes (probably without C). -- I can't go on, I'll go on.
From: Frederick Williams on 30 Jun 2010 09:16 "R. Srinivasan" wrote: > Looks to me like the cancer of infinity is so deeply ingrained in the > thinking of classical logicians that they are incapable of > appreciating any attempt to remove this cancer from logic. The "cancer of infinity" (to use your strange phrase) is one of the motives for the development of modern logic. -- I can't go on, I'll go on.
From: MoeBlee on 30 Jun 2010 10:42 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. > > > 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? > 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. MoeBlee
From: MoeBlee on 30 Jun 2010 10:46 On Jun 28, 11:18 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Mon, 28 Jun 2010 15:46:25 -0700 (PDT), MoeBlee <jazzm...(a)hotmail.com> > said: > > > One thing I don't know how to do is show the mutual-interpretability > > of PA and Y=ZF-"ax inf"+"~ax inf" > > > One direction seems not too difficult: interpreting PA in Y. > > > But how do we interpret Y in PA? Specifically, how do we define 'e' in > > PA and then prove, in PA, all the axioms of Y as interpreted in the > > language of PA? > > The best known approach uses a mapping that Ackermann defined from the > hereditarily finite sets into N that takes the empty set to 0 and, > recursively, {s_1,...s_i} to 2^(n_1) + ... + 2^(n_i), where n_i codes > s_i. For numbers n and m, let nEm iff the quotient of m/2^n is odd. > The relation E is obviously definable in PA. Ackermann showed that, by > defining the membership predicate as E, the axioms of Y are all theorems > of PA. Thanks. And if anyone is interested, it's mentioned in Kunen's 'Foundations Of Mathematics' too, but not with the full proof. Hopefully, one I'll work through it. MoeBlee
From: MoeBlee on 30 Jun 2010 10:55
On Jun 29, 8:17 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 29, 2:05 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 29, 3:44 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > > If MoeBlee is > > > going to insist that Srinivasan prove that D is actually a set, then > > > maybe MoeBlee should do the same for Goedel's V and L. > > You pontificate out of IGNORANCE. Godel worked in NBG where we prove > > that there do exist proper classes. > > Also, even if in ZF, we refer to V and L as "figures of speech" that > > must resolve back to actual formulas in the language of ZF. > > In that case, if Srinivasan were to work in NBG-Infinity instead of > ZF-Infinity, would he then be allowed to talk about his "D"? Of course. And he can talk about it using a Z set theory too, as long as it is understood that it is "figure of speech", and with care so that fallacious inferences are not drawn. Anyway, I showed him how to state is axiom without referrring to a proper class, just to be clear (and I think his axiom is equivalent in context to "all sets are hereditarily finite"?). > I > see no reason that D wouldn't be a class in NBG-Infinity As long as it is written along these lines: D = {x | x is a set & ~En x in P_n(0)}. But I don't know how you would derive (I suspect you cannot): EyAx(x in y <-> ~En x in P_n(0)). Suppose you could. Then V would be in y. Contradiction. Right? MoeBlee |