How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Chris Menzel on 28 Jun 2010 22:17 On Mon, 28 Jun 2010 04:56:30 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > On Jun 27, 6:31 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: >> On Sun, 27 Jun 2010 10:50:16 -0700 (PDT), Charlie-Boo >> <shymath...(a)gmail.com> said: >> >> > ... >> > ZFC declares that there is a set that satisfies Peano's Axioms and >> > defines N to be that set. >> >> You have it completely backwards. The set is defined first, independent >> of any mention of PA. > > So? So what you said was false. You said that ZFC defines N to be a set that satisfies the Peano Axioms. False. > Doesn't that set satisfy Peano's Axioms? No. Axioms are true in models and the set in question (assuming you mean the set of finite von Neumann ordinals) is not itself a model. Rather it is the domain of a model in which the axioms are true. > Whether it says it or not doesn't change the question. What question? >> One can then prove (in ZF) that that set (and indeed, infinitely many >> others) can serve as the domain of a model of PA. > > And then doesn't it define N to be that set? No. Again, a certain set is defined in ZF in purely set theoretic terms, e.g., the smallest set N containing the empty set and such that, if s is in N, so is sU{s}. A model can then be defined in ZF that interprets the primitives of the language of arithmetic in terms of N. (To do this all in pure ZF, constants, predicates, sentences etc are coded as sets.) It can then be shown (in ZF) that all of the Peano axioms are true in this model. So, to complete the answer to your question: N is never *defined* to be a set -- or, more exactly, the domain of a model -- in which the Peano axioms are true. That *turns out* to be the case after the set and corresponding model are defined. I realize none of this gets through your obdurate skull, but hopefully an interested reader or two might benefit.
From: Charlie-Boo on 28 Jun 2010 22:25 On Jun 28, 9:09 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Mon, 28 Jun 2010 05:02:20 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > > 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. > > False. Exactly why it is false has been explained to you numerous > times, but you have no interest in understanding. Or, perversely, like The best way to show it false is to give what you consider the correct method by which ZFC proves the Peano Axioms, wouldn't you say? How else do you prove something in ZFC if not by axioms? How does one guarantee the set representing N satisfies Peano's Axioms, if not with a ZFC axiom? C-B > a Tea Bagger politician, you are more interested in saying crazy stuff > for the sake of attention (in this quirky and insignificant little > corner of the net) than the facts.
From: Charlie-Boo on 28 Jun 2010 22:33 On Jun 28, 9:09 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Mon, 28 Jun 2010 05:02:20 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > > 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. > > False. Exactly why it is false has been explained to you numerous The best way to explain that a ZFC axiom is not used is to give the proof without using any ZFC axioms - good luck! How would you prove the PA axioms in ZFC, then? You keep saying it isn't from an axiom but can't say how it is done - so how do you know it isn't? C-B > times, but you have no interest in understanding. Or, perversely, like > a Tea Bagger politician, you are more interested in saying crazy stuff > for the sake of attention (in this quirky and insignificant little > corner of the net) than the facts.
From: herbzet on 28 Jun 2010 22:36 billh04 wrote: > Frederick Williams wrote: > > 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. > > Are you saying that it is a theorem of ZFC that PA is consistent? > > I thought that the proof of consistency of PA relative to ZFC by > showing that there is a model of PA in ZFC was a metatheorem, not a > theorem stated in ZFC and proved using the axioms of ZFC. I'm not an expert, bill, but I'm gonna guess that the correct answer will be that the informal argument that there is a model of PA in ZFC is itself formalizable in ZFC. ZFC is certainly powerful enough to talk about itself as well as to talk about PA, so it's not inherently contradictory for ZFC to provide a formal proof of this or that about itself. I'm gonna guess that what needs to be assumed for such an informal proof of PA's consistency relative to ZFC is not anything especially exotic and does not exceed what is already assumed axiomatically in ZFC. Hope to hear a reply to you from someone who actually knows what he's talking about. -- hz
From: Chris Menzel on 29 Jun 2010 00:18
On Mon, 28 Jun 2010 15:46:25 -0700 (PDT), MoeBlee <jazzmobe(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. |