How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Charlie-Boo on 3 Jul 2010 07:53 On Jun 29, 8:58 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > How about when you said that Gentzen proved PA consistent using ZFC? > > No one ever proves anything using ZFC in the sense of producing formal > derivations. Rather, they prove mathematical theorems using mathematical > principles formalized in ZFC. By certain elementary considerations we > know the formalizations of such theorems are formally derivable in ZFC. How do you know that all of the logical argument can be formalized in ZFC? This is an example of what I call The Fallacy of the Famous Result. People will attribute special status to a result (rather than other similar results) whose only distinction is that it is famous. They are often famous because they were the first (seminal) result in a certain subject (e.g. a branch of Computer Science e.g. the Theory of Computation or Incompleteness in Logic.) But why does that argument not apply to the other results? Turings Halting Problem theorem proved for computers what Godels First Incompleteness theorem proved for logic. But these results were simply the first theorems discovered in their respective areas. It would be quite the coincidence if they were equivalent in some way that distinguishes them from other theorems, especially more general theorems discovered later (e.g. Rossers extension of Godels results.) Why is it Gödels theorem instead of Rossers theorem that is equivalent to Turings result? Because Godels is more famous and the author cant figure out Rossers. It has no simple This is unprovable. puzzle/joke/amusement to describe it that anyone can talk about. Gregory Chaitin never mentions Rossers stronger results. Its too complicated for him to understand. The recursion theorem proves/explains/formalizes Godels Incompleteness Theorem. Why doesnt it prove/explain/formalize Rossers theorem? Or does it explain EVERY theorem? And here we have the assertion that ZFC formalizes a proof of consistency of PA. There are a good dozen axiomatizations of set theory. Does ZFC2 (read: NF, NFU, CST, CZF, IZF etc.) formalize this argument? How about ZFC3 that has only one short axiom!? How do you know when an axiomatization of set theory has enough axioms to do the job? Or the right axioms to do the job? Without showing this specific proof being carried out in ZFC, how can you show properties capabilities of ZFC that are sure to include what is needed in this particular proof? The answer is, you cant. Unfortunately, even people like you who know the names of lots of theorems also tow the party line despite the fact that in all references given (Gentzen by Frederick Williams and Hinman by MoeBlee) there is no mention of any of ZFCs axioms anywhere in their proofs (including earlier results on which it relies) at all. Not just not all but none at all. C-B > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Frederick Williams on 3 Jul 2010 08:48 Charlie-Boo wrote: > > On Jun 29, 10:16 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > Charlie-Boo <shymath...(a)gmail.com> writes: > > > Ok. What is the reference to the proof in ZFC of PA consistency doing > > > it that way? > > > > It's in Shelah's _Cardinal Arithmetic_ p.3245 - 4325238532. Basically, > > you just do a triple-fold transfinite recursion over a coherent extender > > sequence to obtain a suitable premouse, and iterate the upward Mostowski > > collapsing lemma a few times. To remove the extendible cardinal > > introduce some Aronszjan trees using Sacks forcing. > > Where's the part about how ZFC is used to do it? Woosh. -- I can't go on, I'll go on.
From: William Hale on 3 Jul 2010 10:39 In article <6edea512-8a83-4c88-834e-a4ab9f77e9bc(a)z8g2000yqz.googlegroups.com>, Charlie-Boo <shymathguy(a)gmail.com> wrote: > On Jun 29, 10:22�am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > Charlie-Boo <shymath...(a)gmail.com> writes: > > > What is that formal expression? > > > > To find out you need to read a logic book. > > Then why don't you (anyone) name one that has it? You can't name a > book that has the proof or even the theorem spelled out, and you can't > give it yourself, or even an outline. Yet you claim it can be done. > Is that Mathematics? I thought statements had to be proven in > Mathematics. I thought outlines were given, but you then asked for more detail, even going as far to ask for it to be written completely formal, starting from the axioms of ZFC without using any previous theorems. > > C-B > > > It appears the generous > > explanations various people have provided for your benefit in news are > > not sufficient. > > > > -- > > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > > �- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: R. Srinivasan on 3 Jul 2010 10:42 On Jul 2, 7:06 am, "K_h" <KHol...(a)SX729.com> wrote: > "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message > > news:90ce086b-c089-4b90-866d-c7391e95bbd6(a)j8g2000yqd.googlegroups.com... > On Jul 1, 3:15 am, "K_h" <KHol...(a)SX729.com> wrote: > > > > > "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message > > >news:46d58d89-34b1-40a9-a5a8-1ee250ba57e3(a)e5g2000yqn.googlegroups.com... > > 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: > > >> > > 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. > > >> Why do you think the axiom of infinity is false? What is > >> the basis for your belief in ~Inf? To me it is > >> self-evident that all the naturals exist. > > > First of all I happen to work in a logic (NAFL) where I have a *proof* > > of ~Inf. Essentially, if you define truth (as provability) such that > > all vestiges of Platonism are thrown out, infinite sets will not > > There are several problems with this. > Let me address the four issues you have raised. > > First, you have not explained how or why > you think that mathematical Platonism is false. > > In NAFL, Platonism is refuted from fundamental logical considerations. Consider a NAFL theory T, and let P be a proposition that is undecidable in T. The classical law of non-contradiction must fail for P and ~(P&~P) is not a theorem of T. To see this, consider the law of non-contradiction in the equivalent form P -> ~~P Classically, this tautology expresses "If P is the case then ~P is not the case". But according to the NAFL truth definition, this fails. In NAFL, "If P is the case...." becomes an assertion of axiomatic truth of P with respect to T, meaning "In the theory T+P, ~~P is provable". Similarly, ~P -> ~P may be classically interpreted as "If ~P is the case, then P cannot be the case". In NAFL, this has the meaning "In the theory T+~P, ~P is provable". To put it in a nutshell, ~(P&~P) can only be a theorem of those NAFL theories which either prove P or prove ~P. If P is undecidable in T, there is no proof of ~(P&~P) in T. What does "If P....." *mean* anyway? I interpret this classically as a hypothesis of some Platonic truth, i.e., "If P is true (in some world),....". But in NAFL, there is no such Platonic truth. Any argument that starts with "If P.....", made in the context of a theory T, cannot result in a proof of that argument in the theory T. For "If P..." is to be interpreted in NAFL as "If we add P as an axiom (to T).....". It is from these fundamental logical considerations that NAFL rejects Platonism. I find the NAFL argument totally convincing. I also find the classical argument for ~(P&~P) as a theorem of a theory T in which P is undecidable to be totally unconvincing and is in fact the source of Platonism in classical/intuitionistic logics. From the NAFL argument, one can refute the existence of infinite sets and I have done this in my paper. > > >Second, allowing N to exist as a > proper class is platonic (see my remarks below). > Your remarks below appeal to the diagonal argument. But this argument cannot be stated in NAFL, which does not allow infinite sets and also does permit quantification over proper classes. As I said, real analysis in NAFL is done via a translation of Euclidean geometry into the NAFL theory of finite sets F (which proves the existence of proper classes). In NAFL, Euclidean geometry in its diagrammatic form is taken as fundamental and there cannot be a consistent NAFL theory of non-Euclidean geometry. The square root of two does exist geometrically and it can be translated into the NAFL theory F (as a sequence of rationals, which is a proper class). > >Third, Cantor's theorem > suggests that N should be a set since |N|<|P(N)| cannot hold true for > a proper class. > > Cantor's theorem does not hold in NAFL, which does not permit quantification over infinite (proper) classes and does not permit an arbitrary proper class. In NAFL, N is a proper class and there is no such thing as P(N). You cannot quantify over all the infinite sub- classes of N to get P(N), because quantification over proper classes is not permitted. > > > Fourth, there is no reason why truth should be defined as provability, > especially in light of the incompleteness theorems. > > Again, you are arguing backwards. In NAFL, truth as provability is a fundamental logical requirement (as I have illustrated above while discussing the law of non-contradiction). Therefore the incompleteness theorems do not hold for NAFL theories (e.g. F, the theory of finite sets). RS > > > > Yet if > > infinitely many natural numbers are exceeded *within* N, it seems that > > the only way out is that N must contain an infinitely large number. > > Suppose that were the case then it would still be possible to get what is usually > called N by putting just the finite members into their own set. > > > This is precisely the intuition that leads to nonstandard models of > > arithmetic, where there are nonstandard integers that exceed every > > "standard" natural. To call such numbers "finite" is grotesque, to say > > the least. Yet that is the only way to save the consistency of > > classical Peano Arithmetic. We have to sacrifice our well-known and > > well-accepted intuition of what "finite" means, which is something I > > am not willing to do. > > No, the consistency of PA is not in any danger here. In the non-standard models > "finite" means something different. To retain the usual meaning of "finite" we > need to go beyond first-order logic to second-order logic. Once that is done > then it is possible to distinguish between standard and non-standard models and > PA is consistent in all of them. > > > If the above considerations do not already leave a bad taste in the > > mouth, consider the definition of N as a set. It is an essentially > > impredicative definition. Here I am talking about the simple basic > > definition of N, which uses universal quantifiers in an essential way. > > These quantifiers quantify over an universe that already contain N. > > That such a definition is "harmless" is a commonly stated assertion. > > If you think carefully, such a defense of circularity is based on > > Platonism, namely, that N "really" exists, and our attempted > > definition only tries to access something that is already "out there" > > in the universe of sets. Note that we do not have this problem with > > finite sets, even if these are defined using quantifiers. Because we > > can always define them predicatively by listing their elements. > > But you have already claimed that N really exists out there as a proper class. > And so, by your own logic, you have engaged in circular reasoning. The insight > underlying the diagonal argument shows that |N|<|P(N)| holds true for N and so N > cannot be a proper class. Therefore, inductive sets are needed to make sense of > things like square roots. There are many algorithms that calculate square roots, > for example root(2). So the founding axioms should guarantee a set, N, that > makes possible a list of every numeral in root (2): > > 0 <--> 1 > 1 <--> . > 2 <--> 4 > 3 <--> 1 > 4 <--> 4 > 5 <--> 2 > 6 <--> 1 > 7 <--> 3 > 8 <--> 5 > 9 <--> 6 > ... > > Unless you deny the existence of root(2) and deny the insight underlying the > diagonal argument, inductive sets have to really exists out there in the universe > of sets. That, together with the existence of algorithms that enumerate all > digits of root(2), provides a rock solid justification for an axiom stating the > existence of inductive sets. > > > Here is a post (by Brian Hart) in the FOM newsgroup that says > > Platonism is essential to defend the impredicative methods used in > > modern logic, physics, mathematics: > > >http://www.cs.nyu.edu/pipermail/fom/2010-May/014713.html > > > \begin{quote} > > If one axiomatizes the logical universe (the one containing strictly > > logical objects such as proper and hyper-classes) impredicativity is a > > requirement as these objects cannot be defined non-circularly. > > \end{quote} > > So when you write "However, there can and do exist infinite classes, like N, the > class of all natural numbers" you are being platonic. There is nothing wrong > with that unless you are denying Platonism. >
From: R. Srinivasan on 3 Jul 2010 10:46
On Jul 3, 7:42 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jul 2, 7:06 am, "K_h" <KHol...(a)SX729.com> wrote: > > > "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message > > >news:90ce086b-c089-4b90-866d-c7391e95bbd6(a)j8g2000yqd.googlegroups.com... > > On Jul 1, 3:15 am, "K_h" <KHol...(a)SX729.com> wrote: > > > > "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message > > > >news:46d58d89-34b1-40a9-a5a8-1ee250ba57e3(a)e5g2000yqn.googlegroups.com.... > > > 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: > > > >> > > 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. > > > >> Why do you think the axiom of infinity is false? What is > > >> the basis for your belief in ~Inf? To me it is > > >> self-evident that all the naturals exist. > > > > First of all I happen to work in a logic (NAFL) where I have a *proof* > > > of ~Inf. Essentially, if you define truth (as provability) such that > > > all vestiges of Platonism are thrown out, infinite sets will not > > > There are several problems with this. > > Let me address the four issues you have raised. > > > First, you have not explained how or why > > you think that mathematical Platonism is false. > > In NAFL, Platonism is refuted from fundamental logical considerations. > > Consider a NAFL theory T, and let P be a proposition that is > undecidable in T. The classical law of non-contradiction must fail for > P and ~(P&~P) is not a theorem of T. To see this, consider the law of > non-contradiction in the equivalent form > > P -> ~~P > > Classically, this tautology expresses "If P is the case then ~P is not > the case". > > But according to the NAFL truth definition, this fails. In NAFL, "If P > is the case...." becomes an assertion of axiomatic truth of P with > respect to T, meaning "In the theory T+P, ~~P is provable". > Similarly, > > ~P -> ~P > > may be classically interpreted as "If ~P is the case, then P cannot be > the case". In NAFL, this has the meaning "In the theory T+~P, ~P is > provable". > > To put it in a nutshell, ~(P&~P) can only be a theorem of those NAFL > theories which either prove P or prove ~P. If P is undecidable in T, > there is no proof of ~(P&~P) in T. > > What does "If P....." *mean* anyway? I interpret this classically as a > hypothesis of some Platonic truth, i.e., "If P is true (in some > world),....". But in NAFL, there is no such Platonic truth. Any > argument that starts with "If P.....", made in the context of a theory > T, cannot result in a proof of that argument in the theory T. For "If > P..." is to be interpreted in NAFL as "If we add P as an axiom (to > T).....". > > It is from these fundamental logical considerations that NAFL rejects > Platonism. I find the NAFL argument totally convincing. I also find > the classical argument for ~(P&~P) as a theorem of a theory T in which > P is undecidable to be totally unconvincing and is in fact the source > of Platonism in classical/intuitionistic logics. From the NAFL > argument, one can refute the existence of infinite sets and I have > done this in my paper. > > > > >Second, allowing N to exist as a > > proper class is platonic (see my remarks below). > > Your remarks below appeal to the diagonal argument. But this argument > cannot be stated in NAFL, which does not allow infinite sets and also > does permit quantification over proper classes. > > Sorry, I meant "... does NOT permit quantification over proper classes". RS > > >As I said, real > analysis in NAFL is done via a translation of Euclidean geometry into > the NAFL theory of finite sets F (which proves the existence of proper > classes). In NAFL, Euclidean geometry in its diagrammatic form is > taken as fundamental and there cannot be a consistent NAFL theory of > non-Euclidean geometry. The square root of two does exist > geometrically and it can be translated into the NAFL theory F (as a > sequence of rationals, which is a proper class). > > >Third, Cantor's theorem > > suggests that N should be a set since |N|<|P(N)| cannot hold true for > a proper class. > > Cantor's theorem does not hold in NAFL, which does not permit > quantification over infinite (proper) classes and does not permit an > arbitrary proper class. In NAFL, N is a proper class and there is no > such thing as P(N). You cannot quantify over all the infinite sub- > classes of N to get P(N), because quantification over proper classes > is not permitted. > > > Fourth, there is no reason why truth should be defined as provability, > > especially in light of the incompleteness theorems. > > Again, you are arguing backwards. In NAFL, truth as provability is a > fundamental logical requirement (as I have illustrated above while > discussing the law of non-contradiction). Therefore the incompleteness > theorems do not hold for NAFL theories (e.g. F, the theory of finite > sets). > > RS > > > > > > Yet if > > > infinitely many natural numbers are exceeded *within* N, it seems that > > > the only way out is that N must contain an infinitely large number. > > > Suppose that were the case then it would still be possible to get what is usually > > called N by putting just the finite members into their own set. > > > > This is precisely the intuition that leads to nonstandard models of > > > arithmetic, where there are nonstandard integers that exceed every > > > "standard" natural. To call such numbers "finite" is grotesque, to say > > > the least. Yet that is the only way to save the consistency of > > > classical Peano Arithmetic. We have to sacrifice our well-known and > > > well-accepted intuition of what "finite" means, which is something I > > > am not willing to do. > > > No, the consistency of PA is not in any danger here. In the non-standard models > > "finite" means something different. To retain the usual meaning of "finite" we > > need to go beyond first-order logic to second-order logic. Once that is done > > then it is possible to distinguish between standard and non-standard models and > > PA is consistent in all of them. > > > > If the above considerations do not already leave a bad taste in the > > > mouth, consider the definition of N as a set. It is an essentially > > > impredicative definition. Here I am talking about the simple basic > > > definition of N, which uses universal quantifiers in an essential way.. > > > These quantifiers quantify over an universe that already contain N. > > > That such a definition is "harmless" is a commonly stated assertion. > > > If you think carefully, such a defense of circularity is based on > > > Platonism, namely, that N "really" exists, and our attempted > > > definition only tries to access something that is already "out there" > > > in the universe of sets. Note that we do not have this problem with > > > finite sets, even if these are defined using quantifiers. Because we > > > can always define them predicatively by listing their elements. > > > But you have already claimed that N really exists out there as a proper class. > > And so, by your own logic, you have engaged in circular reasoning. The insight > > underlying the diagonal argument shows that |N|<|P(N)| holds true for N and so N > > cannot be a proper class. Therefore, inductive sets are needed to make sense of > > things like square roots. There are many algorithms that calculate square roots, > > for example root(2). So the founding axioms should guarantee a set, N, that > > makes possible a list of every numeral in root (2): > > > 0 <--> 1 > > 1 <--> . > > 2 <--> 4 > > 3 <--> 1 > > 4 <--> 4 > > 5 <--> 2 > > 6 <--> 1 > > 7 <--> 3 > > 8 <--> 5 > > 9 <--> 6 > > ... > > > Unless you deny the existence of root(2) and deny the insight underlying the > > diagonal argument, inductive sets have to really exists out there in the universe > > of sets. That, together with the existence of algorithms that enumerate all > > digits of root(2), provides a rock solid justification for an axiom stating the > > existence of inductive sets. > > > > Here is a post (by Brian Hart) in the FOM newsgroup that says > > > Platonism is essential to defend the impredicative methods used in > > > modern logic, physics, mathematics: > > > >http://www.cs.nyu.edu/pipermail/fom/2010-May/014713.html > > > > \begin{quote} > > > If one axiomatizes the logical universe (the one containing strictly > > > logical objects such as proper and hyper-classes) impredicativity is a > > > requirement as these objects cannot be defined non-circularly. > > > \end{quote} > > > So when you write "However, there can and do exist infinite classes, like N, the > > class of all natural numbers" you are being platonic. There is nothing wrong > > with that unless you are denying Platonism. |