How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Chris Menzel on 27 Jun 2010 18:31 On Sun, 27 Jun 2010 10:50:16 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > ... > ZFC declares that there is a set that satisfies Peano's Axioms and > defines N to be that set. You have it completely backwards. The set is defined first, independent of any mention of PA. One can then prove (in ZF) that that set (and indeed, infinitely many others) can serve as the domain of a model of PA.
From: Charlie-Boo on 27 Jun 2010 20:14 On Jun 27, 5:41 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Sat, 26 Jun 2010 23:58:02 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > > On Jun 25, 9:58 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > >> On Sat, 26 Jun 2010 01:17:17 +0000 (UTC), Chris Menzel > >> <cmen...(a)remove-this.tamu.edu> said: > > >> > On Fri, 25 Jun 2010 14:29:26 -0700 (PDT), George Greene > > > > > Entirely true of course, but (as you explain quite clearly) that > > > > doesn't make PA a *subset* of ZF. It only shows that there is a > > > > natural embedding * of the language of PA into the language of ZF > > > > such that PA |- A iff ZF |- A*. > > > PA is not a subset of ZF, you say? > > Correct. > > > > Whoops, that needs to be "such that, if PA |- A, then ZF |- A*". > > Note this is only removing the right-to-left direction of the claim > above, which was false but irrelevant to the point. > > > if PA |- A then ZF |- A* you say? > > Indeed I do. Under an appropriate embedding *, the _translation_ A* of > any theorem A of PA into the language of ZF is a theorem of ZF. > > > PA => ZF > > Ah, there you go with the mystery symbols again. I'm guessing => > denotes one of the many undefined, magickal operators of CBF. > > > Oh, PA is a subset of ZF after all! > > Goodness me. PA would be a subset of ZF if it were the case that, if PA > |- A, then ZF |- A. But that's false. > > > > for example, proves Con(PA) which, of course, PA does not (assuming > > > its consistency). > > > You have to assume PA is consistent? > > You have to ask? Yes or no, please. C-B
From: Chris Menzel on 27 Jun 2010 21:40 On Sun, 27 Jun 2010 17:14:02 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: >> > > ZF, for example, proves Con(PA) which, of course, PA does not >> > > (assuming its consistency). >> > > > You have to assume PA is consistent? >> > > You have to ask? > > Yes or no, please. Yes. (You *really* had to ask?)
From: R. Srinivasan on 27 Jun 2010 23:49 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 theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > > > > exist"). > > > > ~Inf is not (1) "there does not exist an infinite set", but rather > > > ~Inf is (2) "there does not exist a successor inductive set". I don't > > > know that there is a proof in Z set theory of the equivalence of (1) > > > and (2). However, in ZF-Inf we can prove that equivalence, but it does > > > take a bit of argument. > > > > > This proof obviously implies that "There does not exist a > > > > model for PA", > > > > You get credit for skillful legerdemain here, but nothing more. > > > > Let's call ZF-Inf+~Inf by the name 'Y'. Now Y proves that there does > > > not exist ANY theory (in the ordinary sense in which first order PA is > > > a theory). Moreover, Y proves that there does not even exist any first > > > order LANGUAGE to be the language of a theory. This is all obvious > > > (once you think about it for a moment and are not distracted too much > > > by your logical legerdemain): A a first order language has an infinite > > > set of variables, and a theory is a certain kind of infinite set of > > > sentences, so if there are no infinite sets, then there are no first > > > order languages and no theories (in the sense in which first order PA > > > is a theory). Your meta-theory Y proves that there IS NO object that > > > satisfies the description we provide the rubric 'PA'. > > > > Any argument you base on theory Y proving anything about PA (with 'PA' > > > defined in some ordinary way) is just an exercise in vacuous > > > reasoning. Not only does Y prove that anything that fulfills the > > > description we give to PA does not have a model, but Y also proves > > > that anything that fulfills the description we give to PA DOES have a > > > model, since Y proves that THERE IS NO OBJECT that fulfills the > > > description we give to PA. > > > > Of course, you could eschew vacuous reasoning as I just gave, but then > > > you're NOT using Y as the meta-theory, since Y=ZF-Inf+~Inf deploys > > > classical first order logic. If you wish to argue from some OTHER > > > logic for the meta-theory then that meta-theory is not ZF-Inf+~Inf. > > > > I expect that if you answer this point, you will do so with even more > > > of your confusions about the basics of mathematical logic. In that > > > case, I likely will not bother to serve as your nurse to clean up your > > > mess. > > > I was wondering why it took so long for you to make an appearance > > here. My arguments are based entirely on FOL and I am not bringing in > > any theory from NAFL as the metatheory of (classical) PA. > > Fine. And I didn't say you did. > > > So let us apply your example of "vacuous reasoning". The theory PA is > > about numbers. > > That is its ordinary interpretation. However, where Z (or any > extension) is used as a meta-theory about PA, then PA is "about" sets > (as even numbers are sets). > > > It certainly does not know what "PA" is. Anything that > > you claim about PA proving the consistency of PA is just utter > > hogwash, > > You didn't read my remarks carefully. > > > since PA proves that "PA" does not exist, So Godel > > I didn't say PA proves that "PA" does not exist (because that's only > an okay way to say it informally as long as we take "X exists" as > short for "There exists an object that is as we defined with the > symbol 'X', in this case, the symbol 'PA'). Go back to my post and see > EXACTLY what I said. > > > What I outlined is simple straightforward stuff. By "metatheory" of > > PA, I mean a theory in which PA is interpretable and therefore > > automatically encodes sentences like con(PA). So certainly, when you > > use suitable coding, the theory ZF-Inf+~Inf does know what PA is. A > > second requirement is that the metatheory should suitably extend the > > language of PA (under suitable "coding", of course) to either allow > > for objects that can be interpreted as models of PA or deny the > > existence of such objects. The theory ZF-Inf+~Inf takes the latter > > route. > > You can encode all you like, but ZF-Inf+~Inf proves that there does > not exist an object that has such things as an infinite set of > variables (thus there is no object that is a language for a theory > such as PA) and no object that is an infinite set of sentences (thus > there is no object that is a theory such as PA). > > > It is all about *coding*. I am surprised that you failed to see this. > > Or are you saying that "coding" was Godel's "logical > > legerdemain" (whatever that means)? > > 'legerdemain' is a word in virtually any English dictionary. > > And I said nothing in my original post about Godel coding. > > > 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. > 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))} 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. I assume that the theory F can be interpreted in PA. Assume this has been done. I have a question for you: Do you think that the sentence D=0 should map to a PA-sentence that is equivalent to ~Con(PA) (since D=0 basically asserts that "infinite sets do not exist")? I claim so, because, as I said earlier, a proof of D=0 in F rules out the possibility of constructing any models of PA in F. But I suspect you will disagree (based on what you have posted here). If by any chance you agree with me, you will have no problem in agreeing that that the proof of D=0 in F, when interpreted in PA, is a proof of ~Con(PA) in PA. RS
From: Tim Little on 28 Jun 2010 03:28
On 2010-06-27, Charlie-Boo <shymathguy(a)gmail.com> wrote: > How do you define "PA is a subset of ZF"? "Every sequence of symbols in the set of theorems of PA is a sequence of symbols in the set of theorems of ZF". This statement is obviously false, since there are symbols in the theorems of PA that are not even in the language of ZF. > Can't ZF prove what PA can? Strictly speaking, ZF itself cannot. In an informal sense, it can. You can extend ZF by definitions into a superset (let's call it ZF* following convention earlier in the thread) of PA, which does prove what PA can. In practice almost nobody uses ZF itself, but if you're going to use the terminology you should get it right. - Tim |