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 01:52 On Jul 1, 12:17 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > If so, what is the formal expression in ZFC that PA is consistent? > > On any scheme for expressing such statements in the unextended language > of set theory it's a virtually incomprehensible string of symbols. Why would it be incomprehensible? It's well-defined. It may take a long time to read through it all - but we can get the idea by the kind author giving us a summary at the highest level of what axioms of ZFC would be used and how they would be used and the overall structure of the proof. How would that go, then? Is there anything at all that you can say about the nature of this elusive formal proof written in ZFC? Other than that you're quite sure that it exists. Is this Hilbert all over again??? > If you really want to have a look, go through a set theory text and > write a computer program to output it for you. If any part of a book is formally expressed, then it can be programmed, of course. But here we have something that nobody has ever shown formalized by any system - by ZFC or anything else. > Why would you want to > see it in the first place? 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. In fact, its construction would show an intimate link between the ZFC axioms and an aspect of PA that its axioms lack. It would give a true answer to the question of whether ZFC can prove PA consistent. It would substantiate your assertions. Is that of value? C-B You see, if you lie with dogs, you are a dog. And I know exactly how dogs bark. > -- > 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 02:03 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. C-B - Hide quoted text - > > - Show quoted text -
From: Charlie-Boo on 1 Jul 2010 02:05 On Jul 1, 1:12 am, George Greene <gree...(a)email.unc.edu> wrote: > On Jul 1, 12:34 am, Charlie-Boo <shymath...(a)gmail.com> wrote:> Which requires the ZFC equivalent of Peano's Axioms (the axiom of > > infinity.) > > You are not JUST wrong, you are the EXACT OPPOSITE of right. > It's like the right answer to the question was 3, but instead of > saying > 1, or 2, or 0, or 5, or 42, or 69, you said > NEGATIVE 3. Prove it. C-B
From: billh04 on 1 Jul 2010 02:18 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 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 -
From: Ki Song on 1 Jul 2010 02:34
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. 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 |