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 1 Jul 2010 11:41 On Jul 1, 2:04 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > To see it means to have actually created it, and its actual creation > > would answer the very interesting question of whether ZFC can prove > > that PA is consistent even though PA can't. > > We already know the answer. The axioms used in the proof are a > restricted form of comprehension, infinity, union and pairing. Why do we need each of these? Of course infinity = PA = arithmetic. But why are the others needed? > Spelling > out the formalization of "PA is consistent" in the language of set > theory would be a tedious, trivial Tedious AND trivial? Wow! So easy we need not do it. And so hard why should we spend so much time on THAT!? Like Goldilocks who only wanted a bed "just right" and got eaten by the bears because of it*. Another word of wisdom for children of all ages. > and pointless undertaking, of no > apparent mathematical interest whatsoever. Proving PA consistent in ZFC has no mathematical interest at all? Wow!! > > It would substantiate your assertions. Is that of value? > > Well, no. I can't think of any pressing reason I should want to convince > you of anything. Me? What do I have to do with it? We're talking about Mathematics, here. Have we degraded to the point of arguing against the merits of the substantiation of mathematical assertions? (The new Conservative Atta accustoms himself to the worn-out, discredited tactics of the past - refighting old battles.) C-B *I may be mixing up fairy tales. > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Charlie-Boo on 1 Jul 2010 11:47 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. 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 11:51 On Jul 1, 2:34 am, Ki Song <kiwisqu...(a)gmail.com> wrote: > Perhaps an analogy is in order! > > I feel like what Charlie-Boo is asking people to do is analogous to > asking someone to perform the addition: > > Sqrt{2}+Sqrt{3}, with a decimal precision of 10^(10^(10^100)) place. I thought I was asking for a reference to substantiate the claim that ZFC can prove PA consistent. By my reasoning, it can't. I gave my reasoning but other don't give their's for the opposing view. The next thing to actually do is to show how or why PA can't prove PA consistent in detail - take Godel's proof and "move the subroutines in- line" - see exactly what it says about the wffs, axioms, rules, theorems etc. of PA. Then apply that to ZFC. C-B > On Jul 1, 2:04 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > > > > Charlie-Boo <shymath...(a)gmail.com> writes: > > > To see it means to have actually created it, and its actual creation > > > would answer the very interesting question of whether ZFC can prove > > > that PA is consistent even though PA can't. > > > We already know the answer. The axioms used in the proof are a > > restricted form of comprehension, infinity, union and pairing. Spelling > > out the formalization of "PA is consistent" in the language of set > > theory would be a tedious, trivial and pointless undertaking, of no > > apparent mathematical interest whatsoever. > > > > It would substantiate your assertions. Is that of value? > > > Well, no. I can't think of any pressing reason I should want to convince > > you of anything. > > > -- > > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus- Hide quoted text - > > - Show quoted text -
From: MoeBlee on 1 Jul 2010 11:53 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. > 1. Reference doesn't have it. > 2. No reference. > 3. Talks about a proof of something else in ZFC. > 4. Isn't carried out in ZFC. I just gave you the reference to Hinman's book. (It won't do you any good, though, since even if you got it, you wouldn't go through the steps in the book leading up to said proof; also because the book is at a somewhat advanced level that would be difficult for someone, such as you, who has not first established an understanding of certain basics in symbolic logic.) MoeBlee
From: Charlie-Boo on 1 Jul 2010 11:53
On Jul 1, 2:45 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Ki Song <kiwisqu...(a)gmail.com> writes: > > Perhaps an analogy is in order! > > > I feel like what Charlie-Boo is asking people to do is analogous to > > asking someone to perform the addition: > > > Sqrt{2}+Sqrt{3}, with a decimal precision of 10^(10^(10^100)) place. > > The analogy is apt, but it's not really quite that bad. Based on nothing > at all I'd estimate that a formalization of "PA is consistent", fully > spelled-out in the language of set theory, fits in less than ten > pages. Ergo don't give as much as a hint of any detail as to how it would be constructed or any other aspect of it, aside from being certain that it can be done (and it is too complex to even think about)? C-B > But perhaps you were thinking of Bourbaki, whose term for the > number 1 already consists of 4,523,659,424,929 symbols and > 1,179,618,517,981 disambiguatory links? > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, dar ber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |