How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: billh04 on 1 Jul 2010 14:24 On Jul 1, 10:47 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jul 1, 2:18 am, billh04 <h...(a)tulane.edu> wrote: > > > > > On Jul 1, 1:03 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > On Jul 1, 1:20 am, billh04 <h...(a)tulane.edu> wrote: > > > > > On Jun 30, 11:34 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > > > On Jun 30, 5:23 am, Frederick Williams <frederick.willia...(a)tesco.net> > > > > > wrote: > > > > > > > Charlie-Boo wrote: > > > > > > > The best way to explain that a ZFC axiom is not used is to give the > > > > > > > proof without using any ZFC axioms - good luck! > > > > > > > > How would you prove the PA axioms in ZFC, then? You keep saying it > > > > > > > isn't from an axiom but can't say how it is done - so how do you know > > > > > > > it isn't? > > > > > > > As Tim Little says elsewhere in the thread, you define S, +, and * in > > > > > > the language of ZFC and then prove the counterparts of PA's axioms as > > > > > > theorems. It's done in Suppes (probably without C). > > > > > > Which requires the ZFC equivalent of Peano's Axioms (the axiom of > > > > > infinity.) > > > > > I'm not sure that I understand what you are saying. > > > > > Let me expand what you are saying a little bit: > > > > > === Your statement ======== > > > > To prove the counterparts of PA's axioms as theorems requires the ZFC > > > > equivalent of Peano's Axioms (the axiom of infinity.) > > > > ======================= > > > > > Assuming this is correct, let me expand it a little bit more: > > > > > === Your statement ======== > > > > To prove the counterparts of PA's axioms as theorems requires a ZFC > > > > axiom (or seveal ZFC axioms) equivalent to Peano's Axioms (the axiom > > > > of infinity.) > > > > ======================= > > > > > If this is what you mean, then why do you think so? > > > > > Here is our statement, similar to yours, but not as strong: > > > > > === Our statement ======== > > > > To prove the counterparts of PA's axioms as theorems requires a ZFC > > > > axiom (or seveal ZFC axioms) that imply Peano's Axioms. > > > > ======================= > > > > > For example, we need to prove the following theorem in ZFC: > > > > > Theorem. (all x in N)(all y in N)(if S x = S y then x = y) > > > > > Here N is a set (defined as mentioned before). > > > > Here x and y are sets since everything in ZFC is a set. > > > > Here S is a set (the successor function). > > > > Thus, the theorem is a statement about sets. > > > > ZFC can proves statements about sets using the ZFC axioms. > > > > > Why do you think this theorem cannot be proved using the ZFC axioms? > > > > If it could be done without the axiom of infinity, then the axiom of > > > infinity wouldn't be needed in ZFC. Maybe it is, but I doubt that. > > > But the axiom of infinity is a ZFC axiom. So, I am allowed to use it > > in proving a theorem in ZFC. > > > It seems that you are just saying that we are cheating when we say > > that we can prove statements in ZFC that correspond to the statements > > of the axioms of PA > > No no no. This whole thing got started when I said that ZFC just > consists of PA plus some dinky little axioms about what sets exist, > and that the latter does not contribute to the proof so ZFC can't > prove PA consistent any more than PA can. I don't understand how this relates to the points that I was trying to make in my post. > > Then those who like to carpet bomb attacked the question of how ZFC > manages to include PA, whether it actually "includes" PA, the > definition of "includes", etc. etc. > > My mistake was to respond. It is of no consequence. > > C-B > > > because the axioms of ZFC imply the statements of > > the axioms of PA. Whether we are cheating or not, we don't deny that, > > and in fact that is pretty much what we are saying. We are not trying > > to hide that the axioms of ZFC imply the statements of the axioms of > > PA. That is what we are claiming. > > > > C-B > > > > - Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -
From: Charlie-Boo on 1 Jul 2010 15:20 On Jul 1, 4:37 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > On Jun 30, 5:23 am, Frederick Williams <frederick.willia...(a)tesco.net> > > wrote: > > > Charlie-Boo wrote: > > > > The best way to explain that a ZFC axiom is not used is to give the > > > > proof without using any ZFC axioms - good luck! > > > > > How would you prove the PA axioms in ZFC, then? You keep saying it > > > > isn't from an axiom but can't say how it is done - so how do you know > > > > it isn't? > > > > As Tim Little says elsewhere in the thread, you define S, +, and * in > > > the language of ZFC and then prove the counterparts of PA's axioms as > > > theorems. It's done in Suppes (probably without C). > > > Which requires the ZFC equivalent of Peano's Axioms (the axiom of > > infinity.) > > You seem to be saying that the axiom of infinity, suitably translated, > is a PA axiom, or maybe even all of the PA axioms collectively. The axiom of infinity was added to give ZFC the capabilities of PA. Now, "to give" here refers to the state of mind of its author, so let's not be psychotic. Let me say that it seems that's all it gives you - that's all it's used for. But all of you ZFC-ites can say if it has been used for anything else. C-B > -- > I can't go on, I'll go on.- Hide quoted text - > > - Show quoted text -
From: Charlie-Boo on 1 Jul 2010 15:25 On Jul 1, 12:23 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > Of course infinity = PA = arithmetic. > > Of course. PA doesn't mention infinity. Just because these chaps > > 0, 0', 0'', 0''', ... > > exist, doesn't mean that this chap > > {0, 0', 0'', 0''', ...} > > exists. Which--speaking loosely--is why ZF needs an axiom of infinity > and PA doesn't. No, PA and ZFC both have Peano's Axioms. The only difference is the universal set, which is N in PA and sets in ZFC. That's why it is expressed differently - different alphabets as well. C-B > -- > I can't go on, I'll go on.
From: Charlie-Boo on 1 Jul 2010 15:30 On Jul 1, 12:32 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jul 1, 11:00 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > how about a high > > level summary of the proof and how ZFC axioms are needed to formalize > > that? > > That's been posted already! (No, I'm not gonna get post numbers for > you; It's enough that at least a few posters have been providing such > summaries all along.) > > As to the axioms used, the axioms of Z-R suffice. Whether there is a > proper subset of those axioms that suffice, I don't opine. The point is that ZFC would have to have an axiom other than Infinity (i.e. PA) that is necessary to prove PA consistent in order for it to be impossible in PA and possible in ZFC. But again you don't show that. And of course there are still no references to anyone doing it. Just 2 or 3 BS references that I exposed as being bogus - par for academia. C-B > MoeBlee
From: Charlie-Boo on 1 Jul 2010 15:45
On Jul 1, 1:01 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jul 1, 11:07 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > On Jul 1, 11:53 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Jun 30, 11:17 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > > The problem isn't to conclude that a model exists, using ZFC. The > > > > problem is to prove that PA is consistent, using ZFC. > > > > I've gone over this already with you. ZFC proves that if a theory has > > > a model then the theory is consistent. It's an extremely simple > > > exercise I proposed you might think about and complete in about two > > > minutes. > > > Then there's no reason for not giving it. Bravado is no substitute > > for Mathematics. > > The reason for giving it as an exercise is to get you STARTED > THINKING. My fault that you don't answer the question? Did you know that a common tactic of abusive people is to blame the abused for making the abuser abuse them? It's not a theorem until proven. I have nothing to do with that. C-B > WHAT bravado? > > Come on, really. > > A theory is a set of sentences (all in a language) closed under > entailment. If a theory T is inconsistent, then there is a sentence P > such that both P and ~P are in T. But P is true in model M iff ~P is > false in model M, and no sentence is both true and false in a given > model M (by the definition-by-recursion function that maps sentences > to true or (exclusive or) to false per a model). So if a theory is > inconsistent, then the theory has no model (lest there be a sentence P > that is both true and false in the model, which is impossible). > > Note: This does not preclude that there are models ('structures' if > you prefer) for the LANGUAGE of an inconsistent theory. For any > language, there are many models for that language. But if a theory is > inconsistent, then there is no model in which all of the sentences of > the theory are true. > > Now, I PROMISE myself. No more explanation of this for you. If you > don't understand or have some question or objection about it. Then > just study the matter. I suggest Enderton's book for this particular > matter. > > > Didn't you read my response? Hinman doesn't refer to ZFC's axioms at > > all in his proof. > > The axioms are used in the various steps leading up to the proof. > That's how mathematics works. A proof of a theorem may rely on > previously proven theorems. > > > He even admits that. > > What specific quote do you have in mind? > > He doesn't need ZFC. ZF is sufficient (less is sufficient too). > > One disclaimer: I've given you the Hinman reference since you've asked > for a reference. I have not scrutinized his particular proof, since > this is something I proved myself long before I got Hinman's book. > Nevertheless, it is a reference as you asked for one, and you may > elect or not to read all the steps in the book that lead to his > exposition of why ZF proves PA is consistent. > > MoeBlee > |