How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: R. Srinivasan on 27 Jun 2010 06:50 On Jun 27, 7:51 am, 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". It may turn out to prove an equivalent statement under > the rest of the axioms, but ~Inf does not actually mean "infinite sets > do not exist". > > Please see my reply to Transfer Principle. We may sidestep this issue by replacing ZF-Inf+~Inf in my post with a theory F which will only admit models with hereditarily finite sets. > > > > 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 > > Even if your intepretation of ~Inf were correct, all it would prove is > that ZF-Inf+~Inf does not model PA. > Sure. But the point is that ZF-Inf+~Inf is *the* chosen metatheory (or model theory) of PA, in which models of PA, if they exist, can be constructed. I happen to have chosen a metatheory which will not admit any models of PA. In such a metatheory, PA is provably inconsistent because we have a proof that models of PA cannot exist. > > > > Now I am sure a lot of people are going to jump up and down and > > protest at this interpretation. But it is logical. > > No, it is not. It exhibits a fairly elementary failure of logic. The > statement "X models PA" implies "PA has a model". However, "X does > not model PA" does *not* imply "PA has no model". > > What you are effectively saying is that ZF-Inf+~Inf is the "wrong" metatheory because there *are* models of PA "out there" , meaning outside of our chosen model theory. This is just an assertion of Platonic existence. There is no particular reason for preferring ZF to ZF-Inf+~Inf as a model theory for PA. The fact that the latter theory yields an unpleasant result does not make it "wrong". RS
From: R. Srinivasan on 27 Jun 2010 08:17 On Jun 27, 2:42 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > R. Srinivasan wrote: > > On Jun 26, 7:42 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > >> R. Srinivasan wrote: > >>> 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. > >> What does it mean for a formula A of L(T) to _syntactically signify_ the > >> (possible) consistency of T? > > > By the way I do not agree that a formula of L(T) can express the > > consistency of T. > > So why did you use the phrases "syntactically" and "consistency of PA" > in your "Syntactically the consistency of PA is expressed by the sentence > Con(PA)"? > > As I said, I was expressing the conventional wisdom when I said that. I was arguing from the point of view of accepted classical logic. Basically I was playing along with the status quo to make a point later on. > > > > As I have stated in my post later on, I strongly > > believe that the consistency of T is a metamathematical (or in this > > case metatheoretical) notion that cannot be expressed in the language > > of T. > > So, again, why did you say "the consistency of PA is expressed by ... > Con(PA)", as below? > That is an accepted result of Godel from classical logic. Later on, when I was talking about NAFL, I disagreed with the classical result. NAFL is what I really believe in. RS > > >>> Syntactically the consistency of PA is expressed by the sentence > >>> Con(PA) which can be encoded in ZF and proven. > > > > > However, according the conventional wisdom, which is what I was > > stating above, a formula A of L(T) can represent the (code of the) > > assertion that "There does not exist a proof of '0=1' in the theory > > T", for theories T that can encode a certain amount of arithmetic. At > > least this is what Godel claimed. > > But all this is still syntactical, which you said above that "I do not > agree that a formula of L(T) can express the consistency of T". No?
From: Frederick Williams on 27 Jun 2010 08:21 Charlie-Boo wrote: > > On Jun 25, 4:58 pm, Frederick Williams <frederick.willia...(a)tesco.net> > wrote: > > Charlie-Boo wrote: > > > point. Who has proved PA consistent using ZFC? > > > Also, see Gentzen > > and Ackermann. Gentzen's proof used far less than full ZFC. > > References please. On-line?? Thanks! Gentzen: Mathematische Annalen, vol. 112, pp 493-565 and Forschungen zur Logik und zur Grundlegung der exakten Wissenschaften no. 4, pp 19-44. Ackermann: Mathematische Annalen, vol 117, pp 162-194. For Gentzen in English see his collected papers edited by Szabo. For an account of Ackermann's proof see Wang, Logic, Computers and Sets, Ch XIV. > > You may wish to know that ZFC with the axiom of infinity replaced > by > > its negation is a model of PA and vice versa. > > Wow, that sounds cool. I'll have to think anout that one. Where can > I read about it? I wish I could remember. Chris Menzel will tell us shortly. -- I can't go on, I'll go on.
From: Charlie-Boo on 27 Jun 2010 12:11 On Jun 25, 5:14 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > 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. > > So why not have a try at it? > You'll find all the details you need in any > decent text. What is the latest text with it - you know, modern treatment? Lots of new texts coming out all the time, aren't there? C-B > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Charlie-Boo on 27 Jun 2010 12:40
On Jun 27, 5:18 am, William Hale <h...(a)tulane.edu> wrote: > In article > <ff54cc7d-b23f-4a45-9040-0459145ff...(a)j8g2000yqd.googlegroups.com>, Charlie-Boo <shymath...(a)gmail.com> wrote: > > [cut] > > > 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. > > PA is not used by any mathematicians to do algebra, number theory, real > analysis, complex analysis, topology, or differential geometry. These > mathematicians represent most mathematicians. They use ZFC as their > axiomatic system. PA is not used but ZFC is? But ZFC invokes the Peano Axioms carte blanche to represent N - so PA is used by ZFC and thus by all of these Mathematicians. Good point! C-B |