How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Frederick Williams on 26 Jun 2010 11:10 "R. Srinivasan" wrote: > > On Jun 26, 1:58 am, Frederick Williams <frederick.willia...(a)tesco.net> > wrote: > > [...] > > You may wish to know that ZFC with the axiom of infinity replaced by its > > negation is a model of PA and vice versa. > > > > > There are two notions of consistency, namely the syntactic and model- > theoretic notions, which are supposed to be equivalent. > > Syntactically the consistency of PA is expressed by the sentence > Con(PA) which can be encoded in ZF and proven. Similarly ZF also > proves the model-theoretic consistency of PA ("There exists a model > for PA") by explicitly constructing a model for PA. > > So far so good. > > Now consider the theory ZF-Inf+~Inf (here Inf is the axiom of > infinity). This theory, being equivalent to PA, cannot prove Con(PA) > by Godel's incompleteness theorem. But does it prove ~Con(PA)? I claim > that it does. > > The theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > exist"). This proof obviously implies that "There does not exist a > model for PA", for a model of PA must have an infinite set as its > universe (according to the classical notion of consistency, which I am > going to dispute shortly). Not at all. If ZF-Inf+~Inf has models then the "points" (or whatever you want to call them) in that model aren't infinite sets. -- I can't go on, I'll go on.
From: R. Srinivasan on 26 Jun 2010 13:10 On Jun 26, 7:42 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > R. Srinivasan wrote: > > On Jun 26, 1:58 am, Frederick Williams <frederick.willia...(a)tesco.net> > > wrote: > > > [...] > >> You may wish to know that ZFC with the axiom of infinity replaced by its > >> negation is a model of PA and vice versa. > > > There are two notions of consistency, namely the syntactic and model- > > theoretic notions, which are supposed to be equivalent. > > Syntactically the consistency of PA is expressed by the sentence > > Con(PA) which can be encoded in ZF and proven. > > What does it mean for a formula A of L(T) to _syntactically signify_ the > (possible) consistency of T? > > By the way I do not agree that a formula of L(T) can express the consistency of T. As I have stated in my post later on, I strongly believe that the consistency of T is a metamathematical (or in this case metatheoretical) notion that cannot be expressed in the language of T. However, according the conventional wisdom, which is what I was stating above, a formula A of L(T) can represent the (code of the) assertion that "There does not exist a proof of '0=1' in the theory T", for theories T that can encode a certain amount of arithmetic. At least this is what Godel claimed. RS
From: Charlie-Boo on 26 Jun 2010 15:51 On Jun 25, 5:14 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Charlie-Boo <shymath...(a)gmail.com> writes: > > Who has proved PA consistent using ZFC? If it were possible then I > > assume someone would have done it. It certainly would be a very > > educational exercise. > > So why not have a try at it? You'll find all the details you need in any > decent text. Name one - including page numbers - that includes that proof. Just one. C-B > -- > 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 26 Jun 2010 15:56 On Jun 24, 6:02 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 24, 1:45 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > ZFC includes PA. > > I did NOT ASK for your ignorant OPINION, DUMBASS. > > I HAVE TWO RELEVANT DEGREES. > > You don't. > > I TOLD you that that ZFC does not "include" PA, > SO IT DOESN'T. Do you know how N is represented in ZFC? By constructing a representation of numbers, a set that meets Peano's Axioms, and then defining N to be that set. So they include Peano's Axioms with a "definition" that N satisifies them. See any text - e.g. Set Theory for the Working Mathematician, pg. 26-29. C-B > > More to the point, you were complaining that it couldn't prove PA > consistent. Well, IT CAN, but that is NOT because it INCLUDES > PA; rather, it is because it ENCODES PA, or can. > It ALLOWS things THAT SATISFY THE AXIOMS of PA to be > constructed by interpretation. > > PA has a ZERO. > ZFC *DOES*NOT*, so it does NOT INCLUDE > PA. > ZFC does, however, have AN EMPTY SET, so you may, > if you like, ENCODE the zero from PA as the empty set from ZFC. > Or as any other set you like, for that matter. > There is, however, a "natural" or "von Neumann" encoding in which > the successor of a set s (which we might write S(s), and which would > be defined for ANY set s, not JUST a PA natural number n) is > s U {S} , and the natural numbers are the successors of the empty > set, with zero encoded as the empty set. > You can then define addition via the cardinality of disjoint union and > multiplication via the cardinality of cross-product, with cardinality > being > defined bijectively. But all that IS A WHOLE LOT OF WORK, and > was definitely NOT just INCLUDED in ZFC to begin with!
From: MoeBlee on 26 Jun 2010 17:05
On Jun 26, 12:56 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > Do you know how N is represented in ZFC? By constructing a > representation of numbers, a set that meets Peano's Axioms, and then > defining N to be that set. Actually, more usually, we establish 0, w, S, +, and * (the empty set, omega, successor on omega, addition for omega, and multiplication for omega), and then show that taken as a system this is a model for the first order PA axioms (I'll just say 'PA' in this context). And doing THAT provides, in Z set theory, a proof that first order PA is consistent.. Only a moment's reflection is needed to see that if a theory has a model then that theory is consistent (and this is easily formalized in Z). Then we show in Z, that the above named system is a model of first order PA. Therefore, we've shown in Z that first order PA is consistent. So, since you seem to recognize that theory that has a model is a consistent theory, I don't know why you question that ZFC proves the consistency of PA. MoeBlee |