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 28 Jun 2010 08:02 On Jun 28, 3:39 am, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-27, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > PA is not used but ZFC is? But ZFC invokes the Peano Axioms carte > > blanche to represent N > > ZFC does not invoke the Peano Axioms at all. In fact to say that a > formal theory "invokes" anything is at the very least somewhat odd. How do you prove Peano's Axioms in ZFC? A ZFC axiom gives them. C-B > Mathematicians working with ZFC may invoke axioms and inference rules > of ZFC to prove theorems such as the existence and uniqueness of a > given set they denote "N". They may then define operations such as > "S", "+" and "*" on its elements, and prove that these operations are > well-defined. Where in that do they "invoke the Peano Axioms carte > blanche"? > > - Tim
From: Charlie-Boo on 28 Jun 2010 08:19 On Jun 28, 5:57 am, Alan Smaill <sma...(a)SPAMinf.ed.ac.uk> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > On Jun 25, 5:19 am, Alan Smaill <sma...(a)SPAMinf.ed.ac.uk> wrote: > >> Aatu Koskensilta <aatu.koskensi...(a)uta.fi> writes: > >> > Charlie-Boo <shymath...(a)gmail.com> writes: > > >> >> Who has proved PA consistent using ZFC? If it were possible then I > >> >> assume someone would have done it. It certainly would be a very > >> >> educational exercise. > > Reference to a post with it? > MoeBlee's recent post in this thread recapitulates it for you. > He has posted that argument before. > Follow MoeBlee's post, and work it out for yourself; > that is the educationally worthwhile way to do it. > Follow MoeBlee's outline. Which post by MoeBlee says anything about how to prove PA consistent in ZFC? Here they are below. Can you copy that part to here? Thanks. C-B Copy to here: MoeBlee's posts: The common claim is that ZFC axiomatizes all (or virtually all) ordinary mathematics. It is not claimed that PA axiomatizes all (hor even virtually all) ordinary mathematics. What I've said on the subject is in my own words and is not properly simplified to "ZFC/PA is a good basis for all of our ordinary math", especially as I don't know what is supposed to be indicated by 'ZFC/ PA' in such slash notation. (PA is embeded in ZFC, of course.) For example, recently I said, "The common claim is that ZFC axiomatizes all (or virtually all) ordinary mathematics. " But I did not say that I personally make that common claim. I merely said what the common claim IS; I didn't say that it is also a claim that I make. And I didn't say anything about ZFC being a "good" basis. Good in what sense? ZFC has certain merits and (arguably) certain drawbacks. It may be a suitable theory in certain ways, but I did not claim that it is simply "good" as a basis. Also, I allowed that a reasonable view of the common claim may include that only VIRTUALLY all of ordinary mathematics may be axiomatized by ZFC. You claim that they are the same claim, though I noted specific differences. I don't need to argue whether they are the same, but only I note that I stand by my own wording and I don't obligate myself to defend your wording. ~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. 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. Right. I regard the claim as eminently plausible, from what I've seen; but there is a vast amount of ordinary mathematics that I have not checked for this claim. Come on, if you're going to engage my responses and my time, please don't waste my time with unfunny silliness. As far as what I say, you can just look at my posts. I don't see the point of your silliness above. Not as I recall. Rather Z-Infinity+~Infinity seems not to prove "there does not exist an infinite set". But ZF-Infinity+~Infinity DOES prove "there does not exist an infinite set". (Infinite defined as "not equinumerous with a natural number".) But, as I just mentioned, I don't know who says that is the case. Fine. And I didn't say you did. 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). You didn't read my remarks carefully. 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. 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). 'legerdemain' is a word in virtually any English dictionary. And I said nothing in my original post about Godel coding. PROVE IT. Look, in ordinary set theory such as Z set theory, we may prove that there is a certain object that is a certain theory and we define the rubric 'PA' to be that theory. But in ZF-Inf +~Inf we may prove that there does NOT exist any such object that meets the description of being a first order theory, let alone one having the specifics properties that, in Z, we ascribe to PA. So, in ZF-Inf +~Inf we cannot even define the rubric 'PA' in a way that PA is a theory. It can't be a very "meaningful" metatheory for PA since it proves that there IS NO theory that meets the description we give, in such as Z, to'PA'. Look, PA has a language that has an infinite set of variables. PA ITSELF is an INFINITE set of sentences. So, since ZF-Inf +~Inf proves there does NOT exist an infinite set, ZF-Inf +~Inf cannot define 'PA' in such a way that it is a theory. No, it's not a confusion. If we're talking about a FORMAL meta-theory - such as ZF-Inf +~Inf is a FORMAL theory - then when it refers to theories then we need to be able to prove IN THAT meta-thoery that there exists objects that are indeed theories. For example, Z set theory proves that there exist infinite sets that may serve as an infinite set of variables; and Z proves that there exist sets of sentences that are closed under entailment. So Z proves that there do exist THEORIES. In particular, we may specify a certain set of symbols and arity function so that that system is a language for a first order theory such as PA, then specify PA to be the theory that is the closure of the INFINITE set of axioms (the induction schema is an infinite set of axioms) that we specify. Can't do that in ZF-Inf +~Inf. So your argument relies on fooling us with a sloppy notion of meta- theory. ZF-Inf +~Inf is a FORMAL theory, and what it PROVES may be proven FORMALLY, but when we REALLY look at the formalizations, we see that ZF-Inf +~Inf proves that there does not exist a first order theory (as we may define, even in ZF-Inf +~Inf, a first order theory to be such and such a set of sentences, which is an infinite set of sentences), since ZF-Inf +~Inf proves there ARE NO infinite sets of ANY KIND, let alone certain kinds of infinite sets of sentences. I hope I will not sacrifice my time to explain this to you again. > Alan Smaill
From: Charlie-Boo on 28 Jun 2010 08:31 On Jun 27, 2:25 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > The proof could be formalized in ZFC. I do not claim that Gentzen used > nothing but the *language* of ZFC. You certainly did. This whole thread is about the subject: "How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)". We are talking about the entire proof being carried out using ZFC's axioms and rules. What significance is there to merely mentioning ZFC during a proof? Any proof is still valid after adding a nonconsequential axiom from ZFC - that would be a reference to ZFC but have no significance as far as that proof goes. You also agreed that "If it can be done, someone would've done it." You still haven't cited anyone who has done it. And I think I know why. My original ideas were right. YOU CANNOT PROVE PA CONSISTENT USING ZFC. The reason is that ZFC is just PA plus some dinkly little axioms about which sets exist. Consistency of PA has nothing to do with which sets exist. And since PA can't prove PA consistent, ZFC cannot either. Here's more details: Describe any proof that PA is consistent in which any of ZFC's axioms outside of the Axiom of Infinity = Peano's Axioms has any bearing whatsoever. Without any such examples, we have the fact that: ZFC CAN PROVE PA CONSISTENCT iff PA CAN PROVE PA CONSISTENT. qed C-B > -- > I can't go on, I'll go on.
From: Frederick Williams on 28 Jun 2010 08:58 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. > The reason is that ZFC is just PA plus some dinkly little axioms about > which sets exist. Consistency of PA has nothing to do with which sets > exist. It has everything to do with V_omega. -- I can't go on, I'll go on.
From: Charlie-Boo on 28 Jun 2010 09:38
On Jun 27, 2:29 am, Transfer Principle <lwal...(a)lausd.net> wrote: > 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. It's not really a challenger to FOL. In fact, it uses FOL for what it does well: represent functions and relations over a single universal set - or subsets of it. CBL replaces FOL in certain contexts where we don't have a single universal set, because the domains of the components of functions and relations are not only different, they have diferent cardinalities! This is why the wffs for ZFC are so long, complex and sometimes debated as to validity. The primitives of something so primitive as set theory should be able to please Occam. And they can - in CBL. The ZFC axioms can be stated in a fraction of the size as using FOL. And how is the Theory of Computation formalized? How do we express its fundamental theorems and proofs? In CBL it is just a wff with a particular value of Q. Where does that occur? That is the definition of metamathematics. We are trying to draw relationships between sets of differing cardinality - e.g. to equate the aleph-1 set of wffs with the aleph-2 set of functions over wffs. In metamathematics, we want to know if a certain set can be represented within another set by substituting a constant for one component. We may also want to know and use this constant. So we relate a wff to a set. The universal sets having different cardinalities, we need a relation over sets of different cardinalities. In CBL, relation P being representable within relation Q this way is denoted P/Q. If M is a constant that meets the above definition, then we write M # P/Q and say that M "solves" or "characterizes" P in Q. Then P/Q is defined as (existsM)M#P/Q and M#P/Q is defined as P=Q(M). When Q is the set of halting programs, provable wffs, refutable wffs, true wffs, true English sentences, sets and their elements, etc. then we can express the basic assertions of metamathematics: P is recursively enumerable, representable, expressible, a set, etc. A proof in CBL is a theorem in each domain, when Q is each of the above sets. It formalizes much of Recursion Theory, and solves the problem of Program Synthesis. The conventional wisdom in Program Synthesis is that a system which has never been shown to produce anything solves the problem completely. I can give examples of all of this. Name your favorite theorem or result/problem/paradox in metamathematics. C-B |