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 29 Jun 2010 10:34 On Tue, 29 Jun 2010 03:34:12 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > On Jun 29, 12:18 am, 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 > > There are too many sets to map them 1-to-1 with the natural numbers. Apparently you have yet to master the semantic role of adjectives. To say nothing of basic set theory.
From: Charlie-Boo on 29 Jun 2010 10:47 On Jun 29, 10:34 am, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Tue, 29 Jun 2010 03:34:12 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > > > > > > On Jun 29, 12:18 am, 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 > > > There are too many sets to map them 1-to-1 with the natural numbers. > > Apparently you have yet to master the semantic role of adjectives. Ok, then tell me. What is the semantic role of adjectives? C-B http://blog.mrm.org/wp-content/uploads/2007/09/wizardofoz.jpg > To > say nothing of basic set theory.- Hide quoted text - > > - Show quoted text -
From: MoeBlee on 29 Jun 2010 10:55 On Jun 28, 9:07 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jun 28, 12:24 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > And it's easy enough to see that if a theory has a model then that > > theory is consistent. > > Is that an axiomatic proof in ZFC? Plain Z-regularity proves that if a theory has a model then the theory is consistent. It's quite simple; you would come up with it yourself on just a moment's reflection. > 3. If a theory has a model then that theory is consistent. > Rule of Inference: It's easy enough to see. > Ha! Don't be a jerk (especially since I'm doing you the favor of providing information you've requested). I didn't say that just saying "it's easy to see" constitutes a proof. Of course, it should be formally proven, and saying "it's easy to see" is not a proof. But, my posts are not themselves a textbook of listed proofs one after another. Just think about the matter for a moment, and indeed you should see how a formal proof would be quite simple here. MoeBlee
From: MoeBlee on 29 Jun 2010 11:02 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? (Also, it occurred to me that I shouldn't have been so overconfident about the other direction (interpreting PA in Y). Does Ackermann do it also?) MoeBlee
From: Chris Menzel on 29 Jun 2010 11:10
On Tue, 29 Jun 2010 03:16:07 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > On Jun 28, 10:17 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> > wrote: >> On Mon, 28 Jun 2010 04:56:30 -0700 (PDT), Charlie-Boo >> <shymath...(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. > > > 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. > > N is defined to be a set that meets the Peano Axioms whether they > mention PA or not. You seem to use the word "define" in a way that no one else does. > You say it "turns out" to be the case. It turns out because of the > answer to my question that you steadfastly refuse to mention: It is > done with the ZFC Axiom of Infinity, which merely says that there is a > set that models Peano's Axioms. It simply says no such thing. > ZFC Axiom of Infinty = PA Induction > > ZFC: Ex (0 e x ^ Ay e x (Y' e x) ) I guess you mean Ex ({} e x ^ Ay e x (yU{y} e x) ) You use the constant '0' and a successor symbol but they are not part of the language of ZF. Nor are the empty set and the operation yU{y} just magically identical to the number zero and the successor function, respectively. Among the things that are proved when one builds the usual model of PA, in fact, are that the empty set and the operation in question can serve in these roles. It's part of what makes the matter interesting. Note more generally that there is no mention of Peano's axioms in the axiom of infinity nor is there any reference to the *true-in* relation between sentences and models. There would have to be if the axiom actually said "that there is a set that models Peano's axioms". Why you think otherwise is curious. > PA: 0 e N > PA: Ay e N (Y' e N) So that's what you think induction is, huh? That explains a lot. > Change x in ZFC to N and you get PA. That's just how the "proof" > works. Well, you've been pointed to many references that would be very helpful to you. |