How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: R. Srinivasan on 29 Jun 2010 13:28 On Jun 29, 8:33 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 29, 2:09 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > 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. You don't even begin to understand what I am talking about here. But I have attempted to give an explanation further down. > > And I SHOWED you how to accomplish your axiom without resorting to D. > Yes, you did. But there are no issues with defining such a D and it is very much needed for my purposes in defining a theory of finite sets in the logic NAFL, as I was trying to explain. > > > But this would be controversial and there would > > still be some ambiguity on whether the existence of infinite sets is > > ruled out despite our forcing D=0. > > > Instead, what I have done in my work is to define D as a class and > > allow the proposition D=0 in a theory of finite sets F where F allows > > classes in its language. So according to the classical way of > > thinking, D would be a set (null set) if infinite sets do not exist > > and a proper class otherwise. > > But "THERE ARE NO proper classes" is a theorem of Y. (Why don't you > understand this?) Therefore, talk of proper classes in context of Y is > merely figurative, a figure of speech, for some other actual statement > in the meta-language. > Again you are showing utter inability to parse what I am saying. I was talking about a theory F which does permit classes, not about the theory Y. But now let us talk about the theory ZF-Inf. We *can* add D to the language of this theory and I claim that would entail the provability of D=0 in ZF-Inf. However despite this fact, there can arguably can exist models of ZF-Inf in which infinite sets can exist. D=0 would continue to be true in such models in the same way that in some set theories in which classes are not permitted, {x: x=x} is taken as the empty set. > > And in this case, you don't even NEED D. I showed you how. > Yes, and I agreed with you. But I do need D for my purposes, as I have explained below. > > > But in my logic NAFL, the undecidability > > of the proposition D=0 in the theory F is a contradiction, and so the > > conclusion is that D=0 must be provable in F. This would genuinely > > rule out the existence of infinite sets in the NAFL theory F. > > Whatever about "NAFL". > If you say so. > > > I do not want to allow any explicit references to infinite objects in > > the language of the theory ZF-Inf. > > What does that MEAN? What is your technical definition of "explict > reference to infinite sets in the language"? > It is a phrase in good old English. I am saying that no definitions of any infinite objects are allowed in the language of ZF-Inf and no symbol in the language of ZF-Inf represents any infinite set. For example "the set of PA-sentences" cannot even be defined in the language of ZF-Inf under this constraint. This is the best I can do with explanations. That such a language would be legitimate is obvious to me. > > > We can and should deny the > > existence of infinite sets without explicitly defining any infinite > > object. > > Yes you can! You ALREADY did it with theory Y: > > ZF-"Inf'+"~Inf" > > That theory entails that every object is finite. And there is no > definition of any infinite object possible in that theory. > OK. Here I want ~Inf to be stated in the form that you mentioned, that is, every set is hereditarily finite. > > > Of course if we admit classes we may allow V_n as a class > > without committing that it is a set. > > You CAN'T do that with ZF-Inf or ZF-Inf+"~Inf" or with ZF-Inf+"every > set is hereditarily finite" (i.e., your "D=0") and still have a > CONSISTENT theory. > > You have to maks some OTHER MODIFICATIONS. > Of course. That is understood. > [Big snip] > > > Therefore I assert that Y proves > > ~Con(PA). > > WRONG. WRONG. WRONG. I've explained already how you are mixed up about > this. You are now arguing just by RECLAIMING what you have not proven. > > Y proves "no x is infinite". > > In the language of Y there is a sentence S such that when S is > translated into the language of PA (under an interpretation of Y into > PA) as sentence S*, we have that S* is true in the standard model of > PA iff PA is inconsistent. > I am having trouble parsing your sentence. How can something be true "in the standard model of PA iff PA is inconsistent" ????? Are you saying that there is such a thing called the "standard model of PA" even if PA is inconsistent? Or are you saying that PA is "really" consistent and its standard model exists even if "PA is inconsistent" is true? Anyway, here is what I was claiming. The proof of "No x is infinite" in Y will translate to a proof of some sentence of PA when we interpret Y in PA. *That* proof ought to be a proof of the inconsistency of PA in PA. I do not have the foggiest idea whether *that* PA-sentence will turn out to be Godels ~Con(PA) or not. However, in light of what Aatu Koskensilta said, it seems that despite Y's proof of the non-existence of a model for PA, we cannot conclude that Y proves that PA is inconsistent, for the implied equivalence will be provable only in a theory with infinite sets permitted (like ZF). I have to think about this a little more. > > But you have NOT shown that S is a THEOREM of Y. > > Just because there is a sentence in the language of Y doesn't entail > that sentence is a THEOREM of Y. > I am referring to the THEOREM of Y that "every set is hereditarily finite". A proof of the translation of THAT sentence in PA ought to have been a proof of the inconsistency of PA in PA, unless Aatu's objection holds. RS
From: Chris Menzel on 29 Jun 2010 14:39 On Tue, 29 Jun 2010 08:56:48 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > ... >> > ZFC Axiom of Infinty = PA Induction >> >> > ZFC: Ex (0 e x ^ Ay e x (Y' e x) ) >> >> I guess you mean >> >> Ex ({} e x ^ Ay e x (yU{y} e x) ) > > It's from Wolfram Mathworld http://mathworld.wolfram.com/AxiomofInfinity.html > Do you think that Wolfram is wrong? "Careless" might be a better or, at least, more charitiable, description. >> You use the constant '0' and a successor symbol but they are not part >> of the language of ZF. Nor are the empty set and the operation yU{y} >> just magically identical to the number zero and the successor >> function, respectively. > > Tell that to Stephen Wolfram. Oh, I'm sure he's an avid sci.logic reader. Unlike you, the MathWorld folks at least realize that a definition of the operator is needed, although they only state informally how the operator functions with respect to the ordinals, viz., 0=∅, 1=0'={0}, 2=1'={0,1}, etc. But note that doesn't say in general what the operator means when applied to arbitrary terms, in particular, variables, as in the axiom. The general definition is: x' =df xU{x} >> Among the things that are proved when one builds the >> usual model of PA, in fact, are that the empty set and the operation in >> question can serve in these roles. It's part of what makes the matter >> interesting. >> >> Note more generally that there is no mention of Peano's axioms in the >> axiom of infinity nor is there any reference to the *true-in* relation >> between sentences and models. There would have to be if the axiom >> actually said "that there is a set that models Peano's axioms". > > Where did I say that? I just said that N is defined to be a set that > satisfies Peano's Axioms. It has to satisfy Peano's Axioms, > otherwise the proof wouldn't work. But *that* N can serve as the domain of a model of PA is not part of its *definition*, contrary to what you seem to be asserting. That is something that is proved after the set is defined. >> > PA: 0 e N >> > PA: Ay e N (Y' e N) >> >> So that's what you think induction is, huh? > > Where did I say that? You boldly proclaimed "ZFC axiom of infinity = PA induction" at the top of your little demonstration. So since the two PA "axioms" were the only things remotely resembling induction, it seems pretty clear that they constitute your understanding of induction.
From: Chris Menzel on 29 Jun 2010 15:05 On Tue, 29 Jun 2010 09:18:59 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > On Jun 29, 11:56 am, MoeBlee <jazzm...(a)hotmail.com> wrote: >> On Jun 29, 8:06 am, Charlie-Boo <shymath...(a)gmail.com> wrote: >> >> > On Jun 29, 8:28 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> >> > wrote: >> >> > > George Greene <gree...(a)email.unc.edu> writes: >> > > > I really don't think that the model existence theorem is going >> > > > to leap out at him here. >> >> > > Pretty much any elementary text contains an account of the set >> > > theoretic construction of the various number systems. >> >> E.g., Enderton's 'Elements Of Set Theory'. >> >> > > It's a trivial exercise to verify the axioms of PA hold in the >> > > structure of naturals. >> >> The only slightly non-trivial part is the induction schema, but it >> falls right into place if one simply starts to do it. >> >> > >Combined with >> > > the soundness of first-order logic this immediately yields the >> > > consistency of PA. >> >> Soundness, e.g. demonstrated in Enderton's 'A Mathematical >> Introduction To Logic'. > > Enderton only makes a comment that ZFC can prove the consistency of PA > (pg. 270) with no details given. > > Is this why people are up in arms claiming that ZFC can prove PA > consistent, even with no references to it - because it is conventional > wisdom among academia? Then you really do have a mess - a claim that > nobody has ever demonstrated. It has been demonstrated in this very thread numerous times, just simply not to your satisfaction. It is a simple exercise to verify that the axioms of PA are true in the standard representation of the number structure in terms of finite von Neumann ordinals. It is a simple exercise to prove the existence of this representation in ZF. It is a simple exercise to code languages as sets and formalize basic first-order model theory in ZF. It is a simple exercise to prove the soundness theorem for first-order logic. Anyone with with a basic understanding elementary set theory and mathematical logic can now see that it follows immediately that ZF proves the consistency of PA. To work it out in complete, formal detail would be as tedious and frivolous an exercise as working out the proof that there is a prime number between 95 and 100 in Principia Mathematica.
From: MoeBlee on 29 Jun 2010 16:36 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. > 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. MoeBlee
From: MoeBlee on 29 Jun 2010 16:43
On Jun 29, 12:28 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 29, 8:33 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:> On Jun 29, > > > I do not want to allow any explicit references to infinite objects in > > > the language of the theory ZF-Inf. > > > What does that MEAN? What is your technical definition of "explict > > reference to infinite sets in the language"? > > It is a phrase in good old English. I am saying that no definitions of > any infinite objects are allowed in the language of ZF-Inf and no > symbol in the language of ZF-Inf represents any infinite set. For > example "the set of PA-sentences" cannot even be defined in the > language of ZF-Inf under this constraint. This is the best I can do > with explanations. That such a language would be legitimate is obvious > to me. That's ignorance and confusion (some of it CORRECT, but APPLIED incorrectly). Each day, like a new batch of tarballs on a Gulf coast beach. Just not enough time in the day to clean it all up. MoeBlee > > > That theory entails that every object is finite. And there is no > > definition of any infinite object possible in that theory. > |