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 27 Jun 2010 01:02 On Jun 27, 12:21 am, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 25, 5:14 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > > So why not have a try at it? You'll find all the details you need in any > > decent text. > > This is a little sparse. > I really don't think that the model existence theorem is going to leap > out at him here. No no no. You have to prove it using ZFC's axioms and rules only. C-B
From: Transfer Principle on 27 Jun 2010 01:54 On Jun 26, 7:41 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > "with suitable extension of ZFC" > > Yikes! > Yes. The usual language of ZFC does not have a successor function > symbol, while the language of PA does. Thus, we must extend *the > language* of ZFC and also add a defining axiom for the successor > function. And the month of June continues with more and more posters coming out of the woodwork to challenge ZFC. This thread marks the return of Charlie-Boo and Srinivasan. Surprisingly, Charlie-Boo is one ZFC challenger whom I have yet to defend, even though I know that he's been posting here for years. I still remember several years back when he once compared ZFC to noodle soup. (Of course, I don't know whether Charlie-Boo still considers ZFC to be like soup anymore.) In this thread, Charlie-Boo is criticized for lumping together ZFC/PA as if they were interchangeable. I must point out that I myself lump them together all the time, but not because I consider them interchangeable -- we know that ZFC is a much stronger theory than PA. (Okay, okay, I mean that _a_suitable_extension_of_ZFC_ is a much stronger theory than PA....) Nonetheless, I lump ZFC and PA together as the two main standard _theories_ (not "theorists"). In particular, I often make statements such as, "Those who use standard theories such as ZFC/PA are much less likely to be called five-letter insults than those who use other theories." In this case, I'm not saying that ZFC and PA are equivalent to each other, but only that either theory is a suitable theory to use if one wants to avoid five-letter insults. Srinivasan, meanwhile, is trying to come up with NAFL, which is supposed to be an alternative _logic_ to FOL. If I remember correctly, in NAFL, it's possible for some statement to be similarly true _and_ false, unlike in FOL. Srinivasan was once fascinated by Ed Nelson's set theory, called Internal Set Theory or IST. One axiom schema of IST is called the Transfer Principle. I came up with my current username right in the middle of a discussion about IST with Srinivasan. Both Srinivasan and IST's creator appear to be sympathetic to finitism. Srivinasan discusses ZF-Infinity+~Infinity, a theory which can be used by finitists, while Nelson is working on a proof that PA is inconsistent. And of course, if Nelson's proof goes through, it would also prove that ZFC is inconsistent, since, as so many were quick to tell Charlie-Boo, ZFC proves that PA is consistent. Hughes will undoubtedly disagree with me, but I find the arrival of all these opponents of ZFC at the same time simply hilarious...
From: Charlie-Boo on 27 Jun 2010 02:01 On Jun 25, 9:17 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Fri, 25 Jun 2010 14:29:26 -0700 (PDT), George Greene > <gree...(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. If ZFC can't calculate what PA can, how can anyone say that ZFC is a good basis for doing mathematics - PA is used by lots of mathematicians. You are not even consistent in what you say. ZFC does just about everything. ZFC doesn't contain PA as a subset. ROTFALMAO! If ZFC doesn't contain PA, why the f*** don't they add it so it does? The truth is, Peano's axioms are defined when we represent N, so we have PA from the very beginning. > 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. The "details" are nothing - they just list Peano's axioms and define the set as meeting them (with no test of consistency) and then define N to be that set. What details? Its all smoke and mirrors - they "define" something (N) that is already defined - so they declare N to have the properties of the set that they say represents N. It would apply to defining the real numbers or prime numbers to prove they have the properties of N - because no properties of N are used to declare it to adhere to the properties they defined the set to have. How do you define "PA is a subset of ZF"? Can't ZF prove what PA can? How can you define it to be anything different than ZF proves what PA can? C-B > > > > > > > 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*.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Transfer Principle on 27 Jun 2010 02:24 On Jun 26, 7:51 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-26, R. Srinivasan <sradh...(a)in.ibm.com> wrote: > > The theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > > exist"). > Actually ~Inf does not assert "Infinite sets do not exist". It only > asserts "there does not exist a successor-closed set containing the > empty set". This has come up time and time again. I myself have claimed that the theory ZF-Infinity+~Infinity proves that every set is finite, and someone (usually MoeBlee or Rupert) points out that this theory only proves that there's no _successor-inductive_ set containing 0, not that there is no infinite set. And every time this comes up, I want to say _fine_ -- so if ZF-Inf+~Inf _doesn't_ prove that every set is finite, then there should exist a model M of ZF-Inf+~Inf in which "there is an infinite set" is true, even though "there exists a set containing 0 that is successor-inductive" is clearly false (assuming, of course, that ZF is itself consistent), just as the fact that ZFC doesn't prove CH implies that there is a model of ZFC in which CH is false (once again, assuming that ZF is itself consistent). Yet no one seems to accept the existence of this model M. Either this model M exists, or ZF-Inf+~Inf really does prove that every set is finite. There are no other possibilities. So let's settle this once and for all. Assuming that ZF is consistent, I ask: 1. Is there a proof in ZF-Inf+~Inf that every set is finite? 2. Does there exist a model M of ZF-Inf+~Inf in which "there is an infinite set" is true? Notice that exactly one of these questions has a "yes" answer and exactly one has a "no" answer. (Actually, come to think of it, since the base theory is ZF and not ZFC, it's possible that the answer to 1. is "yes" if by "finite" we mean one type of finite, say Dedekind finite, and "no" if we mean some other type of finite. In this case, I'd like to know which types of finite produce a "yes" answer.) If 1. is "yes," then I hope that I will never again see a post claiming that ZF-Inf+~Inf doesn't prove that every set is in fact finite. In fact, I'll go as far as to suggest that if 1. is "yes," then those who claim that ZF-Inf+~Inf doesn't prove that every set is finite deserve to be called five-letter insults -- if posters are going to call those who deny the proof of Cantor's Theorem by five-letter insults, then those who deny the proof of "every set is finite" in ZF-Inf+~Inf also ought to be called the same. If 2. is "yes," then what I'd like to know is how can I take _advantage_ of this fact? Suppose I want to consider a theory, based on ZF-Inf, which actually proves that an infinite set exists, yet also proves that no successor-inductive set containing 0 exists. In the current Tony Orlow thread, there is a discussion about whether TO is defining N+ to be a successor-inductive set. It is possible that the theory that I mentioned above might be useful to discussing TO's ideas. But of course, we can't proceed until we know, once and for all, whether ZF-Inf+~Inf proves every set to be finite or not.
From: Transfer Principle on 27 Jun 2010 02:29
On Jun 26, 6:09 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > > ZFC is one thing. PA is another. > And CBL is still another. However, CBL proves theorems with proofs > that are about 1% the size of those published, while ZFC and PA take > about 10 times the size published. So which is best? What's CBL? Is it "Charlie-Boo logic?" If so, then I'd like to learn more about this challenger to FOL. |