How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
in [Math]
|
Prev: ? theoretically solved
Next: How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
From: Transfer Principle on 29 Jun 2010 21:17 On Jun 29, 2:05 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 29, 3:44 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > If MoeBlee is > > going to insist that Srinivasan prove that D is actually a set, then > > maybe MoeBlee should do the same for Goedel's V and L. > You pontificate out of IGNORANCE. Godel worked in NBG where we prove > that there do exist proper classes. > Also, even if in ZF, we refer to V and L as "figures of speech" that > must resolve back to actual formulas in the language of ZF. In that case, if Srinivasan were to work in NBG-Infinity instead of ZF-Infinity, would he then be allowed to talk about his "D"? For I see no reason that D wouldn't be a class in NBG-Infinity (but then, whether this class is a _proper_ class or the empty set depends on the axiom that Srinivasan wishes to add to it).
From: herbzet on 29 Jun 2010 21:54 herbzet wrote: > ZFC is certainly powerful enough I think the phrase I wanted was "expressive 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.
From: R. Srinivasan on 29 Jun 2010 23:23 On Jun 30, 6:17 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 29, 2:05 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > On Jun 29, 3:44 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > > If MoeBlee is > > > going to insist that Srinivasan prove that D is actually a set, then > > > maybe MoeBlee should do the same for Goedel's V and L. > > You pontificate out of IGNORANCE. Godel worked in NBG where we prove > > that there do exist proper classes. > > Also, even if in ZF, we refer to V and L as "figures of speech" that > > must resolve back to actual formulas in the language of ZF. > > In that case, if Srinivasan were to work in NBG-Infinity instead of > ZF-Infinity, would he then be allowed to talk about his "D"? For I > see no reason that D wouldn't be a class in NBG-Infinity (but > then, whether this class is a _proper_ class or the empty set > depends on the axiom that Srinivasan wishes to add to it). > Exactly. If you see my paper on NAFL, you will find I *have* worked precisely in NBG-Infinity. So D is a proper class when infinite sets exist, and the proposition D=0 is undecidable in NBG-Infinity. However, in the corresponding NAFL theory F, such undecidability is a contradiction and we conclude infinite sets cannot exist in NAFL. If we do work in ZF-Infinity, we only need the *convention* that in models of ZF-Infinity where infinite sets exist, D=0 still holds (this would make D=0 *provable* in ZF-Infinity). As I explained, this is like the convention {x:x=x}=0 in some set theories with no classes. RS
From: R. Srinivasan on 29 Jun 2010 23:45 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 > > to much flotsam still for me to spend more time than I've already > spent. > > However: > > > > We PROVE from ZF-Inf that there IS NO SUCH object that you are calling > > '> D'. (or at least we have not before us a proof that there IS such an > > > object). Just adding a constant symbol 'D' and saying whategver you > > > want about it does not override. > > You do not have any such proof. > > I SAID, "or at least we have not before us a proof that there IS such > an object". > > But it's simple anyway: > > Therorem of ZF-I: > > Ex~En x in P_n(0) -> ~EyAz(zey <-> ~En z in P_n(0)) > > Proof: Toward a contradiction suppose Ex~En x in P_n(0) and > Az(zey <-> ~En x in P_n(0)). > Let ~En x in P_n(0). > Let j be arbitrary. > ~En xu{j} in P_n(0). > So Aj j in Uy. > > Theorem of ZF-I: > > ~Ex~En x in P_n(0) -> Ey(Az(zey <-> ~En z in P_n(0)) & y=0) > > Proof: Immediate. > > Then, as far as I know (which is pretty limited) it is not decided in > ZF-I whether Ex~En x in P_n(0). Someone may inform me further on that, > but I'm pretty sure that ZF-I doesn't tell us whether there are or are > not sets other than the hereditarily finite sets. > I think it is not known whether this proposition (That there are sets other than the hereditarily finite sets) is undecidable, refutable, or provable in ZF-I. Undecidability of this propostion is just an assumption as far as I know. > > > How can something be true > > "in the standard model of PA iff PA is inconsistent" ????? > > Typo of omission. I meant, "true in the standard model for the > LANGUAGE of PA", as I had posted in previous messages. > Hmmm. I am not very conversant with classical model theory. So according to you, there is a "standard model for the LANGUAGE of PA" even if the theory PA is inconsistent. 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)? Anyway, in NAFL there is no such thing as the "standard model for the LANGUAGE of PA". Truths are with respect to (consistent) axiomatic theories and there are no truths in just the language of a theory. RS
From: R. Srinivasan on 30 Jun 2010 00:35
On Jun 30, 9:09 am, 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 > > >> to much flotsam still for me to spend more time than I've already > >> spent. > > >> However: > > >>>> We PROVE from ZF-Inf that there IS NO SUCH object that you are calling > >>> '> D'. (or at least we have not before us a proof that there IS such an > >>>> object). Just adding a constant symbol 'D' and saying whategver you > >>>> want about it does not override. > >>> You do not have any such proof. > >> I SAID, "or at least we have not before us a proof that there IS such > >> an object". > > >> But it's simple anyway: > > >> Therorem of ZF-I: > > >> Ex~En x in P_n(0) -> ~EyAz(zey <-> ~En z in P_n(0)) > > >> Proof: Toward a contradiction suppose Ex~En x in P_n(0) and > >> Az(zey <-> ~En x in P_n(0)). > >> Let ~En x in P_n(0). > >> Let j be arbitrary. > >> ~En xu{j} in P_n(0). > >> So Aj j in Uy. > > >> Theorem of ZF-I: > > >> ~Ex~En x in P_n(0) -> Ey(Az(zey <-> ~En z in P_n(0)) & y=0) > > >> Proof: Immediate. > > >> Then, as far as I know (which is pretty limited) it is not decided in > >> ZF-I whether Ex~En x in P_n(0). Someone may inform me further on that, > >> but I'm pretty sure that ZF-I doesn't tell us whether there are or are > >> not sets other than the hereditarily finite sets. > > > I think it is not known whether this proposition (That there are sets > > other than the hereditarily finite sets) is undecidable, refutable, or > > provable in ZF-I. Undecidability of this propostion is just an > > assumption as far as I know. > >>> How can something be true > >>> "in the standard model of PA iff PA is inconsistent" ????? > >> Typo of omission. I meant, "true in the standard model for the > >> LANGUAGE of PA", as I had posted in previous messages. > > > Hmmm. I am not very conversant with classical model theory. So > > according to you, there is a "standard model for the LANGUAGE of PA" > > even if the theory PA is inconsistent. > > MoeBlee has to clarify but I'd think that's what he meant. > > > 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! > And from this lofty platform of "rigor", they denounce any dissent as "ignorant confusion". I must have been "ignorantly confused" when I asserted that it is perfectly valid to ban any explicit references to infinite sets in the language of a theory which *proves* that there are only hereditarily finite sets. Looks to me like the cancer of infinity is so deeply ingrained in the thinking of classical logicians that they are incapable of appreciating any attempt to remove this cancer from logic. > > > > Anyway, in NAFL there is no such thing as the "standard model for the > > LANGUAGE of PA". > > Truths are with respect to (consistent) axiomatic > > theories and there are no truths in just the language of a theory. > > I don't know much about NAFL but I'd agree with 1st half of this > statement. > One clarification. I meant: In NAFL, truhs for *formal propositions* are with respect to (consistent) axiomatic theories in whose language such propositions are legitimate. There are absolute (Platonic) truths in NAFL (e.g. a NAFL theory is either consistent or inconsistent) but such propositions are not formalizable in the language of a NAFL theory. These must remain as metamathematical truths. |