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 28 Jun 2010 09:45 On Jun 28, 8:58 am, Frederick Williams <frederick.willia...(a)tesco.net> 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. Give the slightest bit of details. > > The reason is that ZFC is just PA plus some dinkly little axioms about > > which sets exist. Consistency of PA has nothing to do with which sets > > exist. > > It has everything to do with V_omega. That's not a ZFC axiom. C-B > -- > I can't go on, I'll go on.
From: Charlie-Boo on 28 Jun 2010 09:50 On Jun 28, 8:58 am, Frederick Williams <frederick.willia...(a)tesco.net> 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. Who has done it? You lied the last time I asked and said Gentzen did it. Why didn't he? Who really has? NOBODY!!!!!!!!!!!! No false references - especially at over $400.00!!! Any references - give a summary of how ZFC was used for the whole thing. No more lies, no more wild goose chases, no more $400 BS references!! If no one has done it, and, "If it could be done it would have been." that you agreed to, then it can't be, by your words. C-B > > The reason is that ZFC is just PA plus some dinkly little axioms about > > which sets exist. Consistency of PA has nothing to do with which sets > > exist. > > It has everything to do with V_omega. > > -- > I can't go on, I'll go on.
From: billh04 on 28 Jun 2010 11:46 On Jun 28, 7:58 am, Frederick Williams <frederick.willia...(a)tesco.net> 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. > > > The reason is that ZFC is just PA plus some dinkly little axioms about > > which sets exist. Consistency of PA has nothing to do with which sets > > exist. > > It has everything to do with V_omega. > > -- > I can't go on, I'll go on.
From: MoeBlee on 28 Jun 2010 11:53 On Jun 27, 4:32 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 28, 2:11 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 27, 1:25 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > On Jun 28, 12:44 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > The proof in the theory ZF-Inf > > > +~Inf that infinite sets do not exist encodes (among other things) > > > that models of PA cannot exist. > > > PROVE IT. > > Trivial. If it's trivial, then it shouldn't take you but a minute to prove it. Please do. > > Look, in ordinary set theory such as Z set theory, we may prove that > > there is a certain object that is a certain theory and we define the > > rubric 'PA' to be that theory. > > Sure. If we want a metatheory of PA in which it *is* possible to > construct a model of PA (like Z set theory) then I do agree that we > need all the machinery you describe. You're not listening. Theory Y (AF-I+~I) does not provide for constructing the THEORY PA ITSELF. Why do you keep skipping this?: Y proves there are no infinite sets. PA is an infinite set of sentences, moreover axiomatized by an infinite set of axioms, moreover with a language that has an infinite set of variables. So Y proves THERE IS NO SUCH theory. Please do not continue to argue PAST that very simple state of affairs. > >But in ZF-Inf +~Inf we may prove that > > there does NOT exist any such object that meets the description of > > being a first order theory, let alone one having the specifics > > properties that, in Z, we ascribe to PA. So, in ZF-Inf +~Inf we cannot > > even define the rubric 'PA' in a way that PA is a theory. > > But now we are looking at a metatheory that only needs to *deny* that > models of PA exist. Same as I just wrote above. > So it does not need all the machinery that you > describe to accomplish this. It is enough for this metatheory to deny > the existence of infinite sets for it to rule out the existence of > objects that can be interpreted as models of PA. Same as I just wrote above. > > > That is all I need to calll ZF-Inf > > > +~Inf as a metatheory of PA. > > > It can't be a very "meaningful" metatheory for PA since it proves that > > there IS NO theory that meets the description we give, in such as Z, > > to'PA'. > > This and what you state below is again your own confusion. You've shown no confusion on my part. Saying "your confusion" and "trivial" when asked for proofs of things is not substantive argument. Anyway, just go back to this: Y proves there are no infinite sets. PA is an infinite set of sentences, moreover axiomatized by an infinite set of axioms, moreover with a language that has an infinite set of variables. So Y proves THERE IS NO SUCH theory. Y proves not only is there such a theory, but there is not even such a LANGUAGE for such a theory. MoeBlee
From: MoeBlee on 28 Jun 2010 12:14
On Jun 27, 10:49 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 28, 2:11 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > On Jun 27, 1:25 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > On Jun 28, 12:44 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > The proof in the theory ZF-Inf > > > +~Inf that infinite sets do not exist encodes (among other things) > > > that models of PA cannot exist. > > > PROVE IT. > > Here is one possible approach. Consider the theory F that I had > defined in another post, where And what is the meta-theory you are using for defining and proving things about this theory F? > F = ZF - Inf + D=0, > > Where 0 is the null set and D is defined as > > D = {x: An(x not in P_n(0))} Prove (and please state the formal language, logic, and axioms in which you conduct this proof) that there exists a D such that for all x, we have x in D iff An x not in P_n(0). Unless you can do that, then your 'D' is not well defined. So, instead, I'd state your axiom as: AxEn x in P_n(0) > Here P_n(0) is power set operation iterated n times on 0, and P_0(0) = > P(0), P_1(0)=P(P(0)), etc. Clearly only hereditarily finite sets can > exist in models of F. Prove it. (And please do not waste my time by typing a bunch of stuff that does not conform to some ordinary technical definition of 'structure (model) for a language' and 'model of a theory'.) > I assume that the theory F can be interpreted in PA. Assume this has > been done. Why do you assume that? It might be true, but still it deserves to be shown. > I have a question for you: > > Do you think that the sentence D=0 should map to a PA-sentence that is > equivalent to ~Con(PA) (since D=0 basically asserts that "infinite > sets do not exist")? (1) As I noted, your 'D' is not yet demonstrated to be properly defined. So I take the axiom as: AxEn x in P_n(0) (2) What do you mean by "map a sentence to"? What PARTICULAR FUNCTION do you have in mind? But wait, actually don't bother. I'm not interested in performing exercises for you. If you have a proof of something, then just state your formal system, rules, and axioms, and show me the proof. (3) I'm not even interested in whatever sense you might mean "basically states that" "infinite sets do not exist". If you have a proof of something, let's see it. MoeBlee |