How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
in [Math]
|
Prev: ? theoretically solved
Next: How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
From: Chris Menzel on 29 Jun 2010 00:26 On Mon, 28 Jun 2010 08:46:03 -0700 (PDT), billh04 <hale(a)tulane.edu> said: > 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? Well, Frederick said it, but so will I. > 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. You can formalize the metatheory for the language of PA in ZF and prove in ZF the existence of a model of PA and, hence, that PA is consistent, by the soundness theorem.
From: R. Srinivasan on 29 Jun 2010 03:09 On Jun 29, 3:48 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 28, 9:14 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 27, 10:49 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > Here is one possible approach. Consider the theory F that I had > > > defined in another post, where > > > F = ZF - Inf + D=0, > > > Where 0 is the null set and D is defined as > > > D = {x: An(x not in P_n(0))} > > I wonder why Srinivasan can't simply use the nth level of the > cumulative hierarchy, V_n, so that we don't have to worry > about whether P_n(0) is welldefined or not. > Being a finitist, I do not accept that there is any such thing as a cumulative hierarchy in the first place. There can be a doubt about whether P_n(0) is well defined only if you admit to having a doubt about what "finite" means in "n is finite". I do not have any such doubts. > > > Prove (and please state the formal language, logic, and axioms in > > which you conduct this proof) that there exists a D such that for all > > x, we have x in D iff An x not in P_n(0). > > Unless you can do that, then your 'D' is not well defined. > > OK, I see the problem here. In standard ZF, the object that Srinivasan > calls "D" would be too large to be a set. Thus, it's not evident that > the set D exists. > Here you are *assuming* that D=0 is not provable in the theory ZF-Inf. If that is the case, then indeed D can only exist as a proper class if we add the axiom Inf to ZF-Inf. I contend that if you go ahead and add D to the language of ZF-Inf, we have no option but to accept that D=0 is provable in ZF-Inf. 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 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. > > Rather than attempt to prove it, why can't we just let this be the > axiom itself -- so instead of writing D=0, the axiom becomes: > > ED (Ax (xeD <-> An (~xeV_n))) > > exactly the statement that MoeBlee asks Srinivasan to prove (except > inserting V_n). We can worry about whether such a set can equal 0 > down the road. > I do not want to allow any explicit references to infinite objects in the language of the theory ZF-Inf. We can and should deny the existence of infinite sets without explicitly defining any infinite object. Of course if we admit classes we may allow V_n as a class without committing that it is a set. > > > > Here P_n(0) is power set operation iterated n times on 0, and P_0(0) = > > > P(0), P_1(0)=P(P(0)), etc. Clearly only hereditarily finite sets can > > > exist in models of F. > > Prove it. > > OK, I see the problem here. Even if F has a model in which every set > is > HF, there might be non-isomorphic models of F with sets which aren't > HF at all. > Again this is subject to the assumption that such models are permitted. > > Still, I believe that Srinivasan is on his way to giving a theory that > will satisfy most finitists. > I believe that the logic NAFL is the ideal platform for launching such a theory F and not classical logic. In NAFL, we accept a meta-proof that the theory F proves 'D=0' and since F admits the notion of proper classes, this forces the non-existence of infinite sets in any model of F. Let me now address the huge disconnect between me and someone like MoeBlee, who accepts classical logic. This is well illustrated in this discussion of the theory Y=ZF-Inf+~Inf (or the equivalent that rules out the existence of any sets other than the hereditarily finite ones). It is very clear to me that MoeBlee would only accept an argument for the inconsistency of PA that first *presumes* the existence of infinite sets, defines the theory "PA" as an infinite object in terms of infinite sets and then attempts to prove "PA is inconsistent" by ruling out the existence of models for PA. But this is hopelessly circular, and any such "proof" is meaningless, for it would immediately be declared as a proof of a "false" statement in view of our having accepted the "true" existence of infinite sets in the first place. If this is "rigor", I do not want any part of it. This is where I get off the bus, and this is where someone like MoeBlee gets in. Instead, take a look at the theory Y. It does NOT define any explicit infinite object. Indeed, I would restrict the language of Y to ban *any* explicit reference to infinite sets, even for the purpose of denying their existence. So the theory Y cannot prove that "The theory PA does not exist" because such a proposition is not even legitimate in the language of Y. What Y does is to effectively define the theory PA *constructively* by displaying all its axioms and theorems in the language of Y (via a suitable interpretation of the theory PA in Y) and with no reference whatsoever to any infinite sets. Having done that, we also observe that Y admits Con(PA) as a legitimate proposition in its language. Thus Y does not commit itself to any notion of the theory PA as having been defined in terms of infinite sets but nevertheless is able to formalize "There exists a model of PA" (which is just Con(PA)). Lastly, observe that Y proves that "Infinite sets do not exist". Which means that, according to Y, there is no object corresponding to any model of PA in the universe of sets. Therefore I assert that Y proves ~Con(PA). And if one interprets Y in PA, this proof should translate to a proof of ~Con(PA) in PA. This is the *only* noncircular way in which a theory like Y can prove the inconsistency of PA. RS
From: Charlie-Boo on 29 Jun 2010 06:16 On Jun 28, 10:17 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Mon, 28 Jun 2010 04:56:30 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > > On Jun 27, 6:31 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > >> On Sun, 27 Jun 2010 10:50:16 -0700 (PDT), Charlie-Boo > >> <shymath...(a)gmail.com> said: > > >> > ... > >> > ZFC declares that there is a set that satisfies Peanos Axioms and > >> > defines N to be that set. > > >> You have it completely backwards. The set is defined first, independent > >> of any mention of PA. > > > So? > > So what you said was false. You said that ZFC defines N to be a set > that satisfies the Peano Axioms. False. > So, to complete the answer to your question: N is never *defined* to be > a set -- or, more exactly, the domain of a model -- in which the Peano > axioms are true. That *turns out* to be the case after the set and > corresponding model are defined. N is defined to be a set that meets the Peano Axioms whether they mention PA or not. You say it "turns out" to be the case. It turns out because of the answer to my question that you steadfastly refuse to mention: It is done with the ZFC Axiom of Infinity, which merely says that there is a set that models Peano's Axioms - which it calls N. ZFC Axiom of Infinty = PA Induction ZFC: Ex (0 e x ^ Ay e x (Y' e x) ) PA: 0 e N PA: Ay e N (Y' e N) Change x in ZFC to N and you get PA. That's just how the "proof" works. C-B > I realize none of this gets through your obdurate skull, but hopefully > an interested reader or two might benefit.
From: Charlie-Boo on 29 Jun 2010 06:21 On Jun 28, 10:25 pm, herbzet <herb...(a)gmail.com> wrote: > "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 If you lay with dogs, you are a dog. > *I* am proud of you, that you would make this startling announcement. > > -- > hz
From: Charlie-Boo on 29 Jun 2010 06:29
On Jun 28, 10:36 pm, herbzet <herb...(a)gmail.com> wrote: > billh04 wrote: > > Frederick Williams 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. > > I'm not an expert, bill, but I'm gonna guess that the correct > answer will be that the informal argument that there is a model > of PA in ZFC is itself formalizable in ZFC. The fact that a model implies consistency is what is missing from ZFC, so that is not a proof in ZFC. I don't think ZFC can prove PA consistent at all. PA can't and ZFC is just PA plus some shitty little axioms about what sets exist. That may be a way to avoid paradoxes, but it sure isn't a way to do mathematics. C-B > ZFC is certainly powerful enough to talk about itself as well > as to talk about PA, so it's not inherently contradictory for > ZFC to provide a formal proof of this or that about itself. > > I'm gonna guess that what needs to be assumed for such an > informal proof of PA's consistency relative to ZFC is not > anything especially exotic and does not exceed what is already > assumed axiomatically in ZFC. > > Hope to hear a reply to you from someone who actually knows > what he's talking about. > > -- > hz |