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 29 Jun 2010 06:57 On Jun 27, 9:40 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Sun, 27 Jun 2010 17:14:02 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > >> > > ZF, for example, proves Con(PA) which, of course, PA does not > >> > > (assuming its consistency). > > > > > You have to assume PA is consistent? > > > > You have to ask? > > > Yes or no, please. > > Yes. (You *really* had to ask?) In Godel's proof that his system cannot prove its own consistency, he uses the caveat "assuming it is consistent". But he is not talking about PA at that point. He is talking about a system with a variable for the set of axioms, which includes consistent systems and inconsistent systems. Thus the "assuming it is consistent". Many unsophisticated people mistakenly refer to "assuming PA is consistent", not knowing that Godel was talking about sets of axioms outside of PA's. PA itself is consistent, as is easily proved based on the simple fact that the axioms and rules preserve truth. There is no "assuming PA is consistent", Chris Menzel - it is provably consistent. C-B
From: Charlie-Boo on 29 Jun 2010 08:22 On Jun 29, 7:57 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > "R. Srinivasan" <sradh...(a)in.ibm.com> writes: > > 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). > > Without the axiom of infinity we can't prove that if a theory has no > model it is inconsistent. Indeed, we can prove in ZF-Inf+~Inf that some > consistent theories have no model. Still waiting for the published proof of PA consistency using only ZFC. Godel's 1st: Truth does not equal provability. C-B's 137th: Truth does not equal claimability. qed C-B > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 29 Jun 2010 08:37 William Hale <hale(a)tulane.edu> writes: > 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. Most mathematicians don't use any formal theory in any apparent sense. Rather, they just go about their business of doing mathematics. That the mathematical principles and modes of reasoning they implicitly (or, rather less commonly, explicitly) invoke in this their business are formally captured in this or that formal theory is usually of no general mathematical interest. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Charlie-Boo on 29 Jun 2010 08:43 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. Just remembered another improvement to FOL - and this is a big (significant) one: 4. Bounded quantifiers are a big kludge (doesn't fit in and is added for one case rather than the general case - lots of programming languages do this - then later try to fix it.) The interest is that (allA<N)P(A,x) is recursive when P is, while (aA)P(A,x) is not recursive or even r.e. in general. So this kludge syntax and its rule are added. But we can prove that using the existing FOL syntax and some well- chosen rules of inference that have wide applicability (e.g. formalizing Rosser 1936 at a detailed level.) The expression is (aA) [LT(A,N)=>P(A,x)] where LT(a,b) means a<b. How do we prove it recursive given that P is? I mean formally with axioms and rules (everything I do is.) Another puzzle related to this: Give a function of x that always produces a set containing a finite number of elements but the relation between x and the elements of that x ("y is in the function of x") is not recursive. C-B
From: Charlie-Boo on 29 Jun 2010 08:58
On Jun 29, 8:23 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > What of ZFC's set theoretic axioms is necessary - especially not > > bookkeeping ones like sets existing that are used throughout PA and > > are not needed in every proof? > > Various bookkeeping axioms about set existence are used in the proof. Reference? You know, I asked for nonbookkeeping. If it's bookkeeping (not needed in PA), then the proof can be carried out in PA. Uh-Oh! > Of course, for the consistency of PA ZFC is a huge overkill. No excuses or obfuscations, plz. (I told you you've started acting like the stupid fucks.) Where's the beef (ZFC proof)? C-B > We can prove > PA is consistent in e.g. ACA, using a (definable) truth predicate for > arithmetical statements. > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |