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 00:23 On Jun 29, 9:51 pm, herbzet <herb...(a)gmail.com> wrote: > "Jesse F. Hughes" wrote: > > Of course, I'm really here for lower entertainment. I want posts about > > the Hammer, about how surrogate factoring moves the stock market, about > > the most influential mathematicians on the planet. But still I pretend > > to care about arguments, if only for appearance's sake. > > Well, I didn't come here for the low comedy -- that's simply not an idea > that had occurred to me. But, it being made explicitly an option -- I > suppose it's a valid choice, even one of some value. > > If I have any reservation, it would just be that I don't think that > encouraging crankery for its entertainment value crank: "an unbalanced person who is overzealous in the advocacy of a private cause" And what are signs of being unbalanced? How about unsubstantiated beliefs i.e. delusions? And what is overzealous? How about lying about what a reference contains and making personal attacks on people? And is any cause truly private - isn't it just a matter of degree, of how many people believe it? C-B > is such a great idea, > all things considered. Not that you're suggesting that. > > Nevertheless, thanks for enlarging the realm of possibilities. > > -- > hz
From: Charlie-Boo on 1 Jul 2010 00:34 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.) C-B > -- > I can't go on, I'll go on.
From: George Greene on 1 Jul 2010 01:12 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.
From: billh04 on 1 Jul 2010 01:20 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?
From: Charlie-Boo on 1 Jul 2010 01:31
On Jun 30, 5:43 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 29, 11:18 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > How about telling me the title of a book or article in which PA is > > proved consistent using only ZFC? > > Peter Hinman, 'Fundamentals Of Mathematical Logic' pg. 557. Theorem > 6.6.9. Hinman doesn't carry out any of his arguments in ZFC - he never mentions any of the axioms of ZFC at all - maybe I missed somthing - do you see any references to any ZFC axioms in his proof (other than perhaps Peano's Axioms in the form of the axiom of infinity but I don't even see that)? He actually basically declares that he's not going to express anything in ZFC (I guess as an answer to me - and others like me.) And of course ALL of it must be in ZFC. Do you know what a proof in ZFC is? It uses the ZFC axioms to prove something. Liar = The Liat Paradox? No, Liar =: "Liar liar pants on fire." (Above is the bat to extinguish the fire.) C-B > MoeBlee |