How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: George Greene on 25 Jun 2010 17:14 On Jun 25, 12:21 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > PA is a subset of ZFC, NO, IT ISN'T. I am not going to belabor this further since I ALREADY EXPLAINED why. Typical that you did not reply at all to the explanation.
From: George Greene on 25 Jun 2010 17:29 > >> On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > >> > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy > >> > to prove that PA is consistent (its axioms and rules preserve > >> > truth) yet (by Godel-2) PA can't do such a simple proof as that. > > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > >> So what? ZFC can prove it. > >> ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove > >> that PA is consistent. This is why you can't say that "ZFC/PA > >> doesn't prove PA is consistent." "ZFC/PA" is just a meaningless > >> locution in any case. > >> ZFC is one thing. PA is another. > > > PA is a subset of ZFC, On Jun 25, 2:16 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > No it isn't. Thank you kindly for taking the simple side, for once. I was living in fear of again being attacked by irrelevant complexity from superior education, on the grounds that one can prove "a certain sort of equivalence" between PA and ZF withOUT the axiom of infinity. As I said when I first rebutted C-B, > A stronger theory can often prove the consistency of a weaker one. > ZFC *can* (and does) prove that PA is consistent: the set required > to exist by ZFC's Axiom of Infinity is (the domain of) a model for PA. In addition to being able to use the von Neumann encoding to interpret PA in ZF, one can also interpret natnums as finite sets in order to encode a theory of finite sets in PA (infinite sets would get encoded as non-standard hyper-finite numbers, in that case). But, of course, just because something CAN be defined under something else Does Not Make the something- defined into a "subset" of the larger something-else under which it WAS defined. The newer smaller thing would've needed to be present in the larger one FROM THE BEGINNING in order for it to be properly deemed a "subset" -- BUT IT WASN'T -- all the WORK OF DEFINITION AND ENCODING had to be ADDED to the original environment to produce the smaller thing within it. In any case, if you take this mutual encodability as connoting "some sort of" equivalence, then PA "is like" ZF without the axiom of infinity, so to the extent that the set of axioms (and theorems) without that one axiom is a subset of the one with, There Could Sort Of Be A Point to what C-B was trying to say. But it is not one that he would understand and certainly not one that would justify anyone's talking about "ZFC/PA".
From: Chris Menzel on 25 Jun 2010 21:17 On Fri, 25 Jun 2010 14:29:26 -0700 (PDT), George Greene <greeneg(a)email.unc.edu> said: >> >> On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: >> >> > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy >> >> > to prove that PA is consistent (its axioms and rules preserve >> >> > truth) yet (by Godel-2) PA can't do such a simple proof as that. > >> > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: >> >> So what? ZFC can prove it. >> >> ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove >> >> that PA is consistent. This is why you can't say that "ZFC/PA >> >> doesn't prove PA is consistent." "ZFC/PA" is just a meaningless >> >> locution in any case. >> >> ZFC is one thing. PA is another. >> >> > PA is a subset of ZFC, > > On Jun 25, 2:16 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: >> No it isn't. > > Thank you kindly for taking the simple side, for once. I was living > in fear of again being attacked by irrelevant complexity from superior > education, George, George, I never attack you, I only ever correct you. With love. > on the grounds that one can prove "a certain sort of equivalence" > between PA and ZF withOUT the axiom of infinity. 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*. Thus, we do have that {A* : PA |- A} is a subset of ZF. It is not difficult to prove this, but it is far from trivial. Hence, even if you understand the details (which I doubt very much Charlie does), to express this fact as "PA is a subset of ZF" is, at best, misleading. > As I said when I first rebutted C-B, >> A stronger theory can often prove the consistency of a weaker one. >> ZFC *can* (and does) prove that PA is consistent: the set required to >> exist by ZFC's Axiom of Infinity is (the domain of) a model for PA. > > In addition to being able to use the von Neumann encoding to interpret > PA in ZF, one can also interpret natnums as finite sets in order to > encode a theory of finite sets in PA (infinite sets would get encoded > as non-standard hyper-finite numbers, in that case). But, of course, > just because something CAN be defined under something else Does Not > Make the something- defined into a "subset" of the larger > something-else under which it WAS defined. The newer smaller thing > would've needed to be present in the larger one FROM THE BEGINNING in > order for it to be properly deemed a "subset" -- BUT IT WASN'T -- all > the WORK OF DEFINITION AND ENCODING had to be ADDED to the original > environment to produce the smaller thing within it. Right. It is in fact literally impossible to make PA a subset of ZF for the simple reason that the language of PA (as standardly understood) is disjoint from the language of ZF and, hence, no PA sentence is a ZF sentence. The best one could do is extend the language of ZF to include "0", "s", "+", and "x" and then form a new theory ZF* in this language whose axioms are those of ZF plus the usual definitions of the new expressions in terms of membership. One could then show that all of the usual axioms of PA are theorems of ZF* and hence that PA is a subset of ZF*.
From: Chris Menzel on 25 Jun 2010 21:58 On Sat, 26 Jun 2010 01:17:17 +0000 (UTC), Chris Menzel <cmenzel(a)remove-this.tamu.edu> said: > On Fri, 25 Jun 2010 14:29:26 -0700 (PDT), George Greene > <greeneg(a)email.unc.edu> said: > ... >> on the grounds that one can prove "a certain sort of equivalence" >> between PA and ZF withOUT the axiom of infinity. > > 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*. Whoops, that needs to be "such that, if PA |- A, then ZF |- A*". ZF, for example, proves Con(PA) which, of course, PA does not (assuming its consistency). The weaker (true) claim is all we need for the point: > Thus, we do have that {A* : PA |- A} is a subset of ZF. It is not > difficult to prove this, but it is far from trivial. Hence, even if > you understand the details (which I doubt very much Charlie does), to > express this fact as "PA is a subset of ZF" is, at best, misleading. I believe the false claim above holds if we replace ZF with ZF-Inf+~Inf.
From: R. Srinivasan on 26 Jun 2010 09:38
On Jun 26, 1:58 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: [...] > You may wish to know that ZFC with the axiom of infinity replaced by its > negation is a model of PA and vice versa. > > There are two notions of consistency, namely the syntactic and model- theoretic notions, which are supposed to be equivalent. Syntactically the consistency of PA is expressed by the sentence Con(PA) which can be encoded in ZF and proven. Similarly ZF also proves the model-theoretic consistency of PA ("There exists a model for PA") by explicitly constructing a model for PA. So far so good. Now consider the theory ZF-Inf+~Inf (here Inf is the axiom of infinity). This theory, being equivalent to PA, cannot prove Con(PA) by Godel's incompleteness theorem. But does it prove ~Con(PA)? I claim that it does. The theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not exist"). This proof obviously implies that "There does not exist a model for PA", for a model of PA must have an infinite set as its universe (according to the classical notion of consistency, which I am going to dispute shortly). Therefore we may take the proof of ~Inf in ZF-Inf+~Inf as a model-theoretic proof of the inconsistency of PA, which must be equivalent to its syntactic counterpart ~Con(PA). Now I am sure a lot of people are going to jump up and down and protest at this interpretation. But it is logical. In the first case, we added the axiom Inf to ZF-Inf, and exhibited an infinite construction to prove that a model exists for PA. We also claimed that this is a consistency proof of PA. In the second case, we added the axiom ~Inf to ZF-Inf and proved that no such infinite construction exists. If the first case is a model-theoretic proof of consistency of PA, the same yardstick should apply to the second case as an inconsistency proof. Note here that in the first case ZF is the metatheory for PA and in the second case, it is ZF-Inf+~Inf. So we now have two proofs, one of Con(PA) i n the theory ZF, and one of ~Con(PA) (at least its model-theoretic equivalent) in the theory ZF- Inf+~Inf. The second proof is obviously finitary. The first one is not. Which proof is to be trusted? The finitary one or the infinitary one? Enter Platonism, to save the day. Here is the Platonist justification of the first proof: \begin{Platonist argument} We *know* that PA is consistent, since its axioms are "true" to start with and the rules of inference preserve truth. Therefore Con(PA) must be "true". That being the case, it also follows that infinite sets must "really" exist. Therefore the theory ZF-Inf+~Inf is a "false" theory, one that proves "false" statements like ~Inf. \end We must reject this metaphysical (metalogical?) claim wholesale. Enter the logic NAFL, whose existence has been acknowledged by quantum physicists, via the paper: http://dx.doi.org/10.1142/S021974991000640X See <Arxiv: quant-ph/0504115> for a downloadable version of this paper. Logicians and philosophers, in contrast to physicists, seem to have a blind spot to this logic and determinedly fail to see its existence. Extraordinary, but that can be discussed later. In the logic NAFL, one may actually show that ~Inf must be provable in a theory of finite sets F, which is the NAFL counterpart of the theory ZF-Inf. So in NAFL, it is the finitary proof of the inconsistency of (the classical theory) PA that has weight, and the infinitary consistency proof in ZF is rejected outright. Note carefully that we are referring to the classical notion of "consistency" here. However, the NAFL notion of consistency is different from the classical one. The NAFL version of PA, namely, NPA, has the following properties: 0.Truth exists only with respect to axiomatic theories in NAFL. Look at the two classical theories ZF and ZF-Inf+~Inf and assume for the moment that these are legitimate in NAFL (which they are not). The proposition Inf would be true with respect to the first theory and false with respect to the second one according to the NAFL notion of truth. There is absolutely no reason to prefer one theory to another, since NAFL rejects Platonism outright. 1. The consistency of NPA (i.e., the existence of a model for NPA) is a metamathematical concept that cannot be formalized in NPA. 2. NPA proves the existence of the infinite proper class N of natural numbers, which is certainly the universe for a model of NPA, if it exists. So in a suitable metatheory which extends NPA, we may hopefully demonstrate the consistency of NPA without having to perform the completely arbitrary action of adding an Axiom of Infinity (not permitted in NAFL in any case). 3. Godel's theorems do not hold in NPA. Consistency of NPA *demands* that there are no NPA-undecidable propositions in the language of NPA. 4. The informal argument that the axioms of NPA are true and its rules of inference preserve truth can be sustained in NAFL, without having to entertain any metaphysical Platonist claims like the "true" existence of infinite sets. 5. Etc,, etc. Regards, RS |