How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Transfer Principle on 29 Jun 2010 16:44 On Jun 29, 10:28 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 29, 8:33 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:> On Jun 29, 2:09 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > 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. You don't even begin to understand > what I am talking about here. But I have attempted to give an > explanation further down. This big argument between MoeBlee and Srinivasan over what exactly this "D" is reminds me of the mathematician Goedel and his theory ZF+"V=L." So if MoeBlee is going to ask Srinivisan about "D," then maybe we should be asking about Goedel's "V" and "L." So what exactly are "V" and "L" anyway? V is supposed to be the universe of all sets. So V is too large to be a set. Thus, if MoeBlee is going to criticize Srinivasan for not proving that D is a set, maybe he should look at Goedel's V. What about "L"? L is supposed to be the constructible universe, and obviously if V=L and V is too large to be a set, then L must be too large to be a set. Even if ~V=L, I've heard that L contains all ordinals, and the ordinals are too large to be a set, and so L must be too large to be a set no matter what. On the contrary, if D=0 as asserted by Srinivasan, then D must be a set since 0 (the empty set) is evidently a set as well. Of course, we could call V and L "proper classes," but then again, MoeBlee has already pointed out that in ZF (unlike NBG) there are no proper classes. Thus, we can't talk about V and L, yet that didn't stop Goedel from writing the axiom "V=L." Therefore, Srinivasan's statement "D=0" is a valid axiom if and only if Goedel's statement "V=L" is a valid axiom. 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.
From: Frederick Williams on 29 Jun 2010 17:01 Transfer Principle wrote: > > Of course, we could call V and L "proper classes," but then > again, MoeBlee has already pointed out that in ZF (unlike NBG) > there are no proper classes. Thus, we can't talk about V and L, > yet that didn't stop Goedel from writing the axiom "V=L." Which set theory do you think G"odel worked with? Also, though it is true that there are no proper sets in ZF, there are formulae with one free variable. -- I can't go on, I'll go on.
From: MoeBlee on 29 Jun 2010 17:05 On Jun 29, 3:44 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 29, 10:28 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > On Jun 29, 8:33 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:> On Jun 29, 2:09 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > 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. You don't even begin to understand > > what I am talking about here. But I have attempted to give an > > explanation further down. > > This big argument between MoeBlee and Srinivasan over what > exactly this "D" is reminds me of the mathematician Goedel > and his theory ZF+"V=L." So if MoeBlee is going to ask > Srinivisan about "D," then maybe we should be asking about > Goedel's "V" and "L." Why? (1) Godel worked in NBG, in which there DO exist proper classes. (2) Even those who work in Z set theories often note that we mention proper classes such as V and L NOT as if the theory proves there are objects matching the definiens of 'V' and 'L' but rather as shortcut language in the meta-theory and that mention of such proper classes can be reduced to rubric without proper classes. > So what exactly are "V" and "L" anyway? V is supposed to be > the universe of all sets. So V is too large to be a set. Thus, if > MoeBlee is going to criticize Srinivasan for not proving that D > is a set, maybe he should look at Goedel's V. No, you don't know what you're talking about. IN Z-Inf, I just posted what happens with "D". Meanwhile, Godel was in NBG in which theory we DO prove the existence of certain proper classes. > What about "L"? L is supposed to be the constructible universe, > and obviously if V=L What do you mean "obviously"? There are very few set theorists to believe it the case. > and V is too large to be a set, then L must > be too large to be a set. So what? They're proper classes that are proven to exist in NBG. And in Z, the expression "V=L" is not one actually in the language of ZF but rather a nickname for an actual formula that is in the language of ZF. > Even if ~V=L, I've heard that L contains > all ordinals, and the ordinals are too large to be a set, and so L > must be too large to be a set no matter what. So what? It's a proper class, and (though I haven't personally worked through all the details), proven to exist in NBG. > On the contrary, > if D=0 as asserted by Srinivasan, then D must be a set since 0 > (the empty set) is evidently a set as well. I just posted a post that clarifies Srinivasan's defiens for 'D'. > Of course, we could call V and L "proper classes," but then > again, MoeBlee has already pointed out that in ZF (unlike NBG) > there are no proper classes. Thus, we can't talk about V and L, > yet that didn't stop Goedel from writing the axiom "V=L." Because Godel was in NBG!!! What the hell is your problem?!!! 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. > Therefore, Srinivasan's statement "D=0" is a valid axiom if and > only if Goedel's statement "V=L" is a valid axiom. No it is not. See my other post. > 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. Christ man, would you get HOLD of yourself? MoeBlee
From: MoeBlee on 29 Jun 2010 17:07 On Jun 29, 11:30 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jun 29, 12:13 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > On Jun 29, 10:25 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > On Jun 29, 10:55 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > 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. > > > > Sorry, but I don't know what proof you have in mind, so I can't > > > determine how the Axiom of Regularity would play a role. > > > It DOESN'T play a role, which is why I took it out. > > I asked what axiom is essential and would be needed to carry out the > proof in PA. I thought you were answering that. So what is the > answer? > > > In other words, we > > can prove in Z set theory even without the axiom of regularity > > (possibly without certain other axioms? but I've never done such > > detailed bookkeeping, as my only claim is that Z-R is SUFFICIENT). THAT is my answer. Damn, please READ my post if you're going to ask question about it. > > As to the proof. Would you just TRY to do it in your mind one time? > > I did. You can't just say the Axiom of Infinity provides a model as > you have to also prove that implies consistency. You claim to know > how, so be a mathematician and substantiate your claim. Of course, that is not the proof. Damn, I'm really using up my time with you. Look, in Z set theory we have: w (omega), 0, the successor function on w, the addition function for w, and the multiplication function for w. Now, take each axiom of PA and verify it is a theorem of Z as translated per the above and for all members of w. For example: 'Sn' in PA translates to 'nu{n}' in Z. PA axiom: Anm(Sn = Sm -> n=m) Then just verify that this is a theorem of Z: Anm((n in w & m in w & nu{n} = mu{m}) -> n=m). Then go on down the line verifying each axiom in PA as translated into Z. The induction schema for PA will be a bit of paperwork, but nothing conceptually too difficult. Then the model for PA will be: <w 0 S' +' *'> where S', +' and *' are the set theoretic operations on w corresponding to the operation sybmols for PA. (As, for example, I mentioned nu{n} corresponds to Sn). If you want to know more, then please just read up about the construction of the system of naturals in Z and about models in mathematical logic. I need to not type all day telling you things you can read for yourself. > > If > > you still can't see it, then, if I'm feeling generous, I'll outline it > > for you. As to showing an exact sequence of primitive formulas of the > > language of Z, no, that's just a chore. > > I don't know what you're referring to. I did ask for the statement of > the theorem in ZFC, but nobody has come up with that either. > > So in summary, > > 1. ZFC can prove PA consistent - it's easy and lots of people have > done it. > 2. Nobody can give a reference to its being done. It's done easily as an exercise, just as I gave you a start abovel. > 3. Nobody can describe the proof that has been done in ZFC. NO, I have DESCRIBED a proof. > 4. Nobody can give even the ZFC expression for the theorem itself. NO. If you paid me enough money, I'd do the labor of translating the informal proof to a perfectly formal one. > > > What is the proof and how is Regularity essential? > > > You're mixed up. Regularity is NOT needed. That's why I put Z- > > regularity, which means "Z without the axiom of regularity". Why don't you have the courtesy even to recognize that I've answered and corrected you? Even if not a matter of courtesy, but at least of communication so that I would know that you do recognize that a needed correction has been made to your misunderstanding. MoeBlee
From: MoeBlee on 29 Jun 2010 17:12
On Jun 29, 11:49 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > Didn't you just ask for a reference yourself a minute ago? So what? Aatu mentioned some material I'm not familiar with. I asked where I could read more about it. Meanwhile, you've been told that if you read a textbook in logic and one in set theory (even just the relevant portions, actually) then you'd see how to show that Z proves the consistency of PA as even just an added easy exercise, whether such books even make a point of the matter. And today I also gave you a basic outline and started you off with one of the "subsections" to complete. You are just a time/energy suck. Men of better discipline than I are lucky not to waste their time/energy on your juvenile silliness. MoeBlee |