How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
in [Logic]
|
Prev: equivalence
Next: How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
From: MoeBlee on 30 Jun 2010 11:03 On Jun 29, 10:47 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Aatu Koskensilta wrote: > > Frederick Williams <frederick.willia...(a)tesco.net> writes: > > >> Yes, you can: take Gentzen's proof (or Ackermann's etc) and formalize > >> it in ZFC. > > > This is a pretty silly way of proving the consistency of PA in set > > theory. > > That PA is consistent is a triviality. > > In what formal system is this triviality in? It's a theory of Z-R, for example. Whether it's "trivial" to prove in Z-R depends on what strikes one as trivial. > (Iow, you didn't mean > it's a fact that PA is syntactically consistent, did you?) Consistent IS syntactically consistent. Here's one among equivalent definitions: DEFINITION OF CONSISTENT: A set of formulas S is in a language is consistent iff there is no formula P and the negation of P in S. PERIOD. That a set of FIRST order formulas is consistent iff that set of sentences is satisfiable is a RESULT we prove. And, of course, Aatu is claiming that PA is consistent. He's been saying it for at least about a decade. What don't you understand about that? MoeBlee
From: MoeBlee on 30 Jun 2010 11:05 On Jun 29, 11:09 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > R. Srinivasan wrote: > > On Jun 30, 1:36 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > >> On Jun 29, 12:28 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote > > May I infer that you have used > > infinite sets to define this model? How can you do that if the theory > > PA is inconsistent (which would make ZFC inconsistent as well)? > > The answer imho is simple: they, the "standard theorists" (and I use > the phrase in a respectful way), would assert they somehow "know" > the natural numbers and this "standard model for the LANGUAGE of PA" > is just the natural numbers, collectively! No, I use no such argument. MoeBlee
From: MoeBlee on 30 Jun 2010 11:13 On Jun 29, 11:35 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > the cancer of infinity But not to fear, chief oncologist, Dr. Srinivasan has just the right chemo...but watch for side effects worse than the disease. > is so deeply ingrained in the > thinking of classical logicians that they are incapable of > appreciating any attempt to remove this cancer from logic. You're unfamiliar with even some of the most famous literature in the subject. MoeBlee
From: MoeBlee on 30 Jun 2010 11:15 On Jun 29, 11:45 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Tue, 29 Jun 2010 08:02:06 -0700 (PDT), MoeBlee <jazzm...(a)hotmail.com> said: > > On Jun 28, 11:18 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> > > wrote: > >> On Mon, 28 Jun 2010 15:46:25 -0700 (PDT), MoeBlee <jazzm...(a)hotmail.com> > >> said: > > >> > One thing I don't know how to do is show the mutual-interpretability > >> > of PA and Y=ZF-"ax inf"+"~ax inf" > > >> > One direction seems not too difficult: interpreting PA in Y. > > >> > But how do we interpret Y in PA? Specifically, how do we define 'e' in > >> > PA and then prove, in PA, all the axioms of Y as interpreted in the > >> > language of PA? > > >> The best known approach uses a mapping that Ackermann defined from the > >> hereditarily finite sets into N that takes the empty set to 0 and, > >> recursively, {s_1,...s_i} to 2^(n_1) + ... + 2^(n_i), where n_i codes > >> s_i. For numbers n and m, let nEm iff the quotient of m/2^n is odd. > >> The relation E is obviously definable in PA. Ackermann showed that, by > >> defining the membership predicate as E, the axioms of Y are all theorems > >> of PA. > > > Thanks. Would you recommend a book (or site) where I can read it in > > all details? > > Hm, don't really know of any books. There's a recent article by Kaye > and Wang called "On Interpretations of Arithmetic and Set Theory" that > occurred in 2007 or 2008 in the Notre Dame Journal of Formal Logic that > discusses the mapping in detail. Great, thanks, I think that NDJFL is free online. I'll look. Meanwhile, I found a little bit also in Kunen's 'Foundations Of Mathematics'. MoeBlee
From: MoeBlee on 30 Jun 2010 11:17
On Jun 30, 2:44 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > The notion of provability in a theory is not formalizable in NAFL > theories. It must remain as a metamathematical notion. Meanwhile, metamathematical notions, including 'provability' are formalizable in such theories as Z set theory. You understand that, right? MoeBlee |