How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: MoeBlee on 27 Jun 2010 17:11 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 theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > > > exist"). > > > ~Inf is not (1) "there does not exist an infinite set", but rather > > ~Inf is (2) "there does not exist a successor inductive set". I don't > > know that there is a proof in Z set theory of the equivalence of (1) > > and (2). However, in ZF-Inf we can prove that equivalence, but it does > > take a bit of argument. > > > > This proof obviously implies that "There does not exist a > > > model for PA", > > > You get credit for skillful legerdemain here, but nothing more. > > > Let's call ZF-Inf+~Inf by the name 'Y'. Now Y proves that there does > > not exist ANY theory (in the ordinary sense in which first order PA is > > a theory). Moreover, Y proves that there does not even exist any first > > order LANGUAGE to be the language of a theory. This is all obvious > > (once you think about it for a moment and are not distracted too much > > by your logical legerdemain): A a first order language has an infinite > > set of variables, and a theory is a certain kind of infinite set of > > sentences, so if there are no infinite sets, then there are no first > > order languages and no theories (in the sense in which first order PA > > is a theory). Your meta-theory Y proves that there IS NO object that > > satisfies the description we provide the rubric 'PA'. > > > Any argument you base on theory Y proving anything about PA (with 'PA' > > defined in some ordinary way) is just an exercise in vacuous > > reasoning. Not only does Y prove that anything that fulfills the > > description we give to PA does not have a model, but Y also proves > > that anything that fulfills the description we give to PA DOES have a > > model, since Y proves that THERE IS NO OBJECT that fulfills the > > description we give to PA. > > > Of course, you could eschew vacuous reasoning as I just gave, but then > > you're NOT using Y as the meta-theory, since Y=ZF-Inf+~Inf deploys > > classical first order logic. If you wish to argue from some OTHER > > logic for the meta-theory then that meta-theory is not ZF-Inf+~Inf. > > > I expect that if you answer this point, you will do so with even more > > of your confusions about the basics of mathematical logic. In that > > case, I likely will not bother to serve as your nurse to clean up your > > mess. > > I was wondering why it took so long for you to make an appearance > here. My arguments are based entirely on FOL and I am not bringing in > any theory from NAFL as the metatheory of (classical) PA. Fine. And I didn't say you did. > So let us apply your example of "vacuous reasoning". The theory PA is > about numbers. That is its ordinary interpretation. However, where Z (or any extension) is used as a meta-theory about PA, then PA is "about" sets (as even numbers are sets). > It certainly does not know what "PA" is. Anything that > you claim about PA proving the consistency of PA is just utter > hogwash, You didn't read my remarks carefully. > since PA proves that "PA" does not exist, So Godel I didn't say PA proves that "PA" does not exist (because that's only an okay way to say it informally as long as we take "X exists" as short for "There exists an object that is as we defined with the symbol 'X', in this case, the symbol 'PA'). Go back to my post and see EXACTLY what I said. > What I outlined is simple straightforward stuff. By "metatheory" of > PA, I mean a theory in which PA is interpretable and therefore > automatically encodes sentences like con(PA). So certainly, when you > use suitable coding, the theory ZF-Inf+~Inf does know what PA is. A > second requirement is that the metatheory should suitably extend the > language of PA (under suitable "coding", of course) to either allow > for objects that can be interpreted as models of PA or deny the > existence of such objects. The theory ZF-Inf+~Inf takes the latter > route. You can encode all you like, but ZF-Inf+~Inf proves that there does not exist an object that has such things as an infinite set of variables (thus there is no object that is a language for a theory such as PA) and no object that is an infinite set of sentences (thus there is no object that is a theory such as PA). > It is all about *coding*. I am surprised that you failed to see this. > Or are you saying that "coding" was Godel's "logical > legerdemain" (whatever that means)? 'legerdemain' is a word in virtually any English dictionary. And I said nothing in my original post about Godel coding. > 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. 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. 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. > 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'. Look, PA has a language that has an infinite set of variables. PA ITSELF is an INFINITE set of sentences. So, since ZF-Inf +~Inf proves there does NOT exist an infinite set, ZF-Inf +~Inf cannot define 'PA' in such a way that it is a theory. > It is not necessary that a metatheory of > PA should treat PA as an explicit mathematical object in its language. > If you think so, that is your confusion. No, it's not a confusion. If we're talking about a FORMAL meta-theory - such as ZF-Inf +~Inf is a FORMAL theory - then when it refers to theories then we need to be able to prove IN THAT meta-thoery that there exists objects that are indeed theories. For example, Z set theory proves that there exist infinite sets that may serve as an infinite set of variables; and Z proves that there exist sets of sentences that are closed under entailment. So Z proves that there do exist THEORIES. In particular, we may specify a certain set of symbols and arity function so that that system is a language for a first order theory such as PA, then specify PA to be the theory that is the closure of the INFINITE set of axioms (the induction schema is an infinite set of axioms) that we specify. Can't do that in ZF-Inf +~Inf. So your argument relies on fooling us with a sloppy notion of meta- theory. ZF-Inf +~Inf is a FORMAL theory, and what it PROVES may be proven FORMALLY, but when we REALLY look at the formalizations, we see that ZF-Inf +~Inf proves that there does not exist a first order theory (as we may define, even in ZF-Inf +~Inf, a first order theory to be such and such a set of sentences, which is an infinite set of sentences), since ZF-Inf +~Inf proves there ARE NO infinite sets of ANY KIND, let alone certain kinds of infinite sets of sentences. I hope I will not sacrifice my time to explain this to you again. MoeBlee
From: R. Srinivasan on 27 Jun 2010 17:32 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 theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > > > > exist"). > > > > ~Inf is not (1) "there does not exist an infinite set", but rather > > > ~Inf is (2) "there does not exist a successor inductive set". I don't > > > know that there is a proof in Z set theory of the equivalence of (1) > > > and (2). However, in ZF-Inf we can prove that equivalence, but it does > > > take a bit of argument. > > > > > This proof obviously implies that "There does not exist a > > > > model for PA", > > > > You get credit for skillful legerdemain here, but nothing more. > > > > Let's call ZF-Inf+~Inf by the name 'Y'. Now Y proves that there does > > > not exist ANY theory (in the ordinary sense in which first order PA is > > > a theory). Moreover, Y proves that there does not even exist any first > > > order LANGUAGE to be the language of a theory. This is all obvious > > > (once you think about it for a moment and are not distracted too much > > > by your logical legerdemain): A a first order language has an infinite > > > set of variables, and a theory is a certain kind of infinite set of > > > sentences, so if there are no infinite sets, then there are no first > > > order languages and no theories (in the sense in which first order PA > > > is a theory). Your meta-theory Y proves that there IS NO object that > > > satisfies the description we provide the rubric 'PA'. > > > > Any argument you base on theory Y proving anything about PA (with 'PA' > > > defined in some ordinary way) is just an exercise in vacuous > > > reasoning. Not only does Y prove that anything that fulfills the > > > description we give to PA does not have a model, but Y also proves > > > that anything that fulfills the description we give to PA DOES have a > > > model, since Y proves that THERE IS NO OBJECT that fulfills the > > > description we give to PA. > > > > Of course, you could eschew vacuous reasoning as I just gave, but then > > > you're NOT using Y as the meta-theory, since Y=ZF-Inf+~Inf deploys > > > classical first order logic. If you wish to argue from some OTHER > > > logic for the meta-theory then that meta-theory is not ZF-Inf+~Inf. > > > > I expect that if you answer this point, you will do so with even more > > > of your confusions about the basics of mathematical logic. In that > > > case, I likely will not bother to serve as your nurse to clean up your > > > mess. > > > I was wondering why it took so long for you to make an appearance > > here. My arguments are based entirely on FOL and I am not bringing in > > any theory from NAFL as the metatheory of (classical) PA. > > Fine. And I didn't say you did. > > > So let us apply your example of "vacuous reasoning". The theory PA is > > about numbers. > > That is its ordinary interpretation. However, where Z (or any > extension) is used as a meta-theory about PA, then PA is "about" sets > (as even numbers are sets). > > > It certainly does not know what "PA" is. Anything that > > you claim about PA proving the consistency of PA is just utter > > hogwash, > > You didn't read my remarks carefully. > > > since PA proves that "PA" does not exist, So Godel > > I didn't say PA proves that "PA" does not exist (because that's only > an okay way to say it informally as long as we take "X exists" as > short for "There exists an object that is as we defined with the > symbol 'X', in this case, the symbol 'PA'). Go back to my post and see > EXACTLY what I said. > > > What I outlined is simple straightforward stuff. By "metatheory" of > > PA, I mean a theory in which PA is interpretable and therefore > > automatically encodes sentences like con(PA). So certainly, when you > > use suitable coding, the theory ZF-Inf+~Inf does know what PA is. A > > second requirement is that the metatheory should suitably extend the > > language of PA (under suitable "coding", of course) to either allow > > for objects that can be interpreted as models of PA or deny the > > existence of such objects. The theory ZF-Inf+~Inf takes the latter > > route. > > You can encode all you like, but ZF-Inf+~Inf proves that there does > not exist an object that has such things as an infinite set of > variables (thus there is no object that is a language for a theory > such as PA) and no object that is an infinite set of sentences (thus > there is no object that is a theory such as PA). > > > It is all about *coding*. I am surprised that you failed to see this. > > Or are you saying that "coding" was Godel's "logical > > legerdemain" (whatever that means)? > > 'legerdemain' is a word in virtually any English dictionary. > > And I said nothing in my original post about Godel coding. > > > 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. > > 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. > >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. 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. > > > 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. RS > > Look, PA has a language that has an infinite set of variables. PA > ITSELF is an INFINITE set of sentences. So, since ZF-Inf +~Inf proves > there does NOT exist an infinite set, ZF-Inf +~Inf cannot define 'PA' > in such a way that it is a theory. > > > It is not necessary that a metatheory of > > PA should treat PA as an explicit mathematical object in its language. > > If you think so, that is your confusion. > > No, it's not a confusion. If we're talking about a FORMAL meta-theory > - such as ZF-Inf +~Inf is a FORMAL theory - then when it refers to > theories then we need to be able to prove IN THAT meta-thoery that > there exists objects that are indeed theories. > > For example, Z set theory proves that there exist infinite sets that > may serve as an infinite set of variables; and Z proves that there > exist sets of sentences that are closed under entailment. So Z proves > that there do exist THEORIES. In particular, we may specify a certain > set of symbols and arity function so that that system is a language > for a first order theory such as PA, then specify PA to be the theory > that is the closure of the INFINITE set of axioms (the induction > schema is an infinite set of axioms) that we specify. > > Can't do that in ZF-Inf +~Inf. > > So your argument relies on fooling us with a sloppy notion of meta- > theory. ZF-Inf +~Inf is a FORMAL theory, and what it PROVES may be > proven FORMALLY, but when we REALLY look at the formalizations, we see > that ZF-Inf +~Inf proves that there does not exist a first order > theory (as we may define, even in ZF-Inf +~Inf, a first order theory > to be such and such a set of sentences, which is an infinite set of > sentences), since ZF-Inf +~Inf proves there ARE NO infinite sets of > ANY KIND, let alone certain kinds of infinite sets of sentences. > > I hope I will not sacrifice my time to explain this to you again. > > MoeBlee
From: Chris Menzel on 27 Jun 2010 17:41 On Sat, 26 Jun 2010 23:58:02 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > On Jun 25, 9:58 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: >> On Sat, 26 Jun 2010 01:17:17 +0000 (UTC), Chris Menzel >> <cmen...(a)remove-this.tamu.edu> said: >> >> > On Fri, 25 Jun 2010 14:29:26 -0700 (PDT), George Greene > > > > Entirely true of course, but (as you explain quite clearly) that > > > doesn't make PA a *subset* of ZF. It only shows that there is a > > > natural embedding * of the language of PA into the language of ZF > > > such that PA |- A iff ZF |- A*. > > PA is not a subset of ZF, you say? Correct. > > Whoops, that needs to be "such that, if PA |- A, then ZF |- A*". Note this is only removing the right-to-left direction of the claim above, which was false but irrelevant to the point. > if PA |- A then ZF |- A* you say? Indeed I do. Under an appropriate embedding *, the _translation_ A* of any theorem A of PA into the language of ZF is a theorem of ZF. > PA => ZF Ah, there you go with the mystery symbols again. I'm guessing => denotes one of the many undefined, magickal operators of CBF. > Oh, PA is a subset of ZF after all! Goodness me. PA would be a subset of ZF if it were the case that, if PA |- A, then ZF |- A. But that's false. > > 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?
From: Chris Menzel on 27 Jun 2010 17:52 On Sun, 27 Jun 2010 00:46:44 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > ... > The definition of N within ZFC simply "defines N" as having the Peano > Axioms en masse. Oh my, you really don't have any idea what you're talking about, do you?
From: Chris Menzel on 27 Jun 2010 18:21
On Sat, 26 Jun 2010 23:01:37 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > ... > If ZFC can't calculate what PA can, how can anyone say that ZFC is a > good basis for doing mathematics - PA is used by lots of > mathematicians. By extending ZF with suitable (and very familiar) definitions of the primitives of the language of PA, all of PA's theorems can be "calculated", i.e., proved, in ZF. Actually, though, no one really claims that ZF is a good basis for *doing* any mathematics other than set theory. Most mathematicians simply work in theories tailored to their areas of interest, although some fragment of set theory is typically assumed as well, as most modern mathematical theories need to talk about sets containing mathematical objects of one stripe or another (e.g., in analysis, sets of reals). The claim about ZF you seem to have in mind -- that, under suitable definitions, all the theorems of classical mathematics are theorems of ZF -- is generally of interest only to those who are concerned about certain fairly arcane philosophical questions about the nature of mathematical objects and how it is we can know anything about them. Working mathematicians typically don't really care much about these questions. > You are not even consistent in what you say. ZFC does just about > everything. ZFC doesn't contain PA as a subset. > > ROTFALMAO! Alas, what you find amusing is simply a product of your own, sad ignorance. > If ZFC doesn't contain PA, why the f*** don't they add it so it does? > The truth is, Peano's axioms are defined when we represent N, so we > have PA from the very beginning. > >> It only shows that there is a natural >> embedding * of the language of PA into the language of ZF such that >> PA |- A iff ZF |- A*. Thus, we do have that {A* : PA |- A} is a subset >> of ZF. It is not difficult to prove this, but it is far from trivial. >> Hence, even if you understand the details (which I doubt very much >> Charlie does), to express this fact as "PA is a subset of ZF" is, >> at best, misleading. > > The "details" are nothing - they just list Peano's axioms and define > the set as meeting them (with no test of consistency) and then define > N to be that set. I'm sorry, but this is just terribly confused. It is so curious that you'd rather play the fool than take the time to educate yourself. |