How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
in [Math]
|
Prev: ? theoretically solved
Next: How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
From: Charlie-Boo on 27 Jun 2010 02:58 On Jun 25, 9:58 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Sat, 26 Jun 2010 01:17:17 +0000 (UTC), Chris Menzel > <cmen...(a)remove-this.tamu.edu> said: > > > On Fri, 25 Jun 2010 14:29:26 -0700 (PDT), George Greene > > Entirely true of course, but (as you explain quite clearly) that > > doesn't make PA a *subset* of ZF. It only shows that there is a > > natural embedding * of the language of PA into the language of ZF > > such that PA |- A iff ZF |- A*. PA is not a subset of ZF, you say? > Whoops, that needs to be "such that, if PA |- A, then ZF |- A*". ZF, if PA |- A then ZF |- A* you say? PA => ZF Oh, PA is a subset of ZF after all! > for example, proves Con(PA) which, of course, PA does not (assuming > its consistency). You have to assume PA is consistent? C-B The weaker (true) claim is all we need for the point: > > > Thus, we do have that {A* : PA |- A} is a subset of ZF. It is not > > difficult to prove this, but it is far from trivial. Hence, even if > > you understand the details (which I doubt very much Charlie does), to > > express this fact as "PA is a subset of ZF" is, at best, misleading. > > I believe the false claim above holds if we replace ZF with ZF-Inf+~Inf.
From: Charlie-Boo on 27 Jun 2010 03:46 On Jun 27, 1:54 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 26, 7:41 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > > Charlie-Boo <shymath...(a)gmail.com> writes: > posting here for years. I still remember several years > back when he once compared ZFC to noodle soup. (Of course, > I don't know whether Charlie-Boo still considers ZFC to be > like soup anymore.) You never got it? Any infinite r.e. set can replace ZFC's set of expressions as the "basis" of math, as that is the only property that is needed - the fact that it is an infinite listable set of expressions. > In this thread, Charlie-Boo is criticized for lumping > together ZFC/PA as if they were interchangeable. Who said they were interchangable? Quote it, you liar! It's not the Liar Paradox, it's the Paradox of the Liar. > I must > point out that I myself lump them together all the time, > but not because I consider them interchangeable -- we know > that ZFC is a much stronger theory than PA. The definition of N within ZFC simply "defines N" as having the Peano Axioms en masse. So now you "know" that ZFC is stronger - than something that it kidnaps under darkness of a definition? It dawned on you that if we can define N to have whatever properties we want - and simply declare Peano's Axioms to apply to N - them we can do what Peano's Axioms do? > (Okay, okay, I > mean that _a_suitable_extension_of_ZFC_ is a much stronger > theory than PA....) Oh just listen to all that machine talk! But is there something wrong with proving something two different ways? Must we translate one into the other? What if there were three? > Nonetheless, I lump ZFC and PA together as the two main > standard _theories_ (not "theorists"). In particular, I > often make statements such as, "Those who use standard > theories such as ZFC/PA are much less likely to be called > five-letter insults than those who use other theories." In > this case, I'm not saying that ZFC and PA are equivalent > to each other, but only that either theory is a suitable > theory to use if one wants to avoid five-letter insults. > > Srinivasan, meanwhile, is trying to come up with NAFL, > which is supposed to be an alternative _logic_ to FOL. If > I remember correctly, in NAFL, it's possible for some > statement to be similarly true _and_ false, unlike in FOL. > > Srinivasan was once fascinated by Ed Nelson's set theory, > called Internal Set Theory or IST. One axiom schema of IST > is called the Transfer Principle. I came up with my > current username right in the middle of a discussion about > IST with Srinivasan. > > Both Srinivasan and IST's creator appear to be sympathetic > to finitism. Srivinasan discusses ZF-Infinity+~Infinity, a > theory which can be used by finitists, while Nelson is > working on a proof that PA is inconsistent. And of course, > if Nelson's proof goes through, it would also prove that > ZFC is inconsistent, since, as so many were quick to tell > Charlie-Boo, ZFC proves that PA is consistent. > > Hughes will undoubtedly disagree with me, but I find the > arrival of all these opponents of ZFC at the same time > simply hilarious... Opponents of ZFC? You left out "those who lie about" between "of" and "ZFC". C-B
From: Charlie-Boo on 27 Jun 2010 04:02 On Jun 27, 2:29 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 26, 6:09 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > > > ZFC is one thing. PA is another. > > And CBL is still another. However, CBL proves theorems with proofs > > that are about 1% the size of those published, while ZFC and PA take > > about 10 times the size published. So which is best? > > What's CBL? Is it "Charlie-Boo logic?" If so, then I'd like to > learn more about this challenger to FOL. No, it's Computationally Based Logic, but you can call it Charlie-Boo Language or whatever you want. It is a system for representing proofs in Computer Science at a very high level. It axiomatizes Program Synthesis, Database Query Processing, Theory of Computation, Recursion Theory and Incompleteness in Logic. The primitive operators are: P/Q means (eM)M # P/Q M # P/Q means P=Q(M) This means that M is an object that represents relation P within Q. The simplest case is 1-place P and 2-place Q. Then P(x) <=> Q(M,x). For example, if Q(Turing Machine,Input Tape) iff the TM halts yes on the input tape, then P/Q means P is recursively enumerable and program M accepts set P. If Q(wff,substitute for free variables) means the wff with the substitution is provable, them P/Q means P is representable in our logic. PR=Provable, TW=True, DIS = Refutable sentences. ~PR/TW = Unprovability is expressible - the premise for Godel's 1st Theorem - intro to article. DIS/PR = Refutability is representable - the premis for Rosser's extension. Incompleteness axiom: -~P/P When P(x,y) is: Then -~P/P is. Turing Machine x halts yes on input y: The set of programs that do not halt yes on themselves is not r.e. Set x contains element y: There is no set of sets that do not contain themselves. Wff x with y substituted for its free variable is provable: The set of unprovable wffs is not representable. It shrinks proofs down to a tiny fraction of the size of published proofs. I have posted a good dozen proofs of various Incompleteness Theorems. I have an article on arxiv that gives its state about 10 years ago. C-B
From: William Hale on 27 Jun 2010 05:18 In article <ff54cc7d-b23f-4a45-9040-0459145ff6fc(a)j8g2000yqd.googlegroups.com>, Charlie-Boo <shymathguy(a)gmail.com> wrote: [cut] > If ZFC can't calculate what PA can, how can anyone say that ZFC is a > good basis for doing mathematics - PA is used by lots of > mathematicians. PA is not used by any mathematicians to do algebra, number theory, real analysis, complex analysis, topology, or differential geometry. These mathematicians represent most mathematicians. They use ZFC as their axiomatic system.
From: R. Srinivasan on 27 Jun 2010 05:44
On Jun 27, 11:24 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On Jun 26, 7:51 pm, Tim Little <t...(a)little-possums.net> wrote: > > > On 2010-06-26, R. Srinivasan <sradh...(a)in.ibm.com> wrote: > > > The theory ZF-Inf+~Inf clearly proves ~Inf ("Infinite sets do not > > > exist"). > > Actually ~Inf does not assert "Infinite sets do not exist". It only > > asserts "there does not exist a successor-closed set containing the > > empty set". > > This has come up time and time again. I myself have claimed that > the theory ZF-Infinity+~Infinity proves that every set is finite, > and someone (usually MoeBlee or Rupert) points out that this > theory only proves that there's no _successor-inductive_ set > containing 0, not that there is no infinite set. > > There is a simple way to sidestep this controversy. Suitably extend the language of ZF-Inf to admit the set D, where D = {x: An (x not in P_n(0))} Here 0 is the null set, n ranges over the non-negative integers and P_n(0) is the power set operation iterated n times on 0 with P_0(0) = P(0). Note that by definition, D does not include any hereditarily finite set, but it will contain every other set. Consider the theory F = ZF-Inf+{D=0} It is clear that F will only admit models with hereditarily finite sets. Use the theory F instead of ZF-Inf+~Inf in my post. > > > And every time this comes up, I want to say _fine_ -- so if > ZF-Inf+~Inf _doesn't_ prove that every set is finite, then there > should exist a model M of ZF-Inf+~Inf in which "there is an > infinite set" is true, even though "there exists a set containing > 0 that is successor-inductive" is clearly false (assuming, of > course, that ZF is itself consistent), just as the fact that ZFC > doesn't prove CH implies that there is a model of ZFC in which > CH is false (once again, assuming that ZF is itself consistent). > > Yet no one seems to accept the existence of this model M. > > Either this model M exists, or ZF-Inf+~Inf really does prove that > every set is finite. There are no other possibilities. > > So let's settle this once and for all. Assuming that ZF is > consistent, I ask: > > 1. Is there a proof in ZF-Inf+~Inf that every set is finite? > 2. Does there exist a model M of ZF-Inf+~Inf in which "there > is an infinite set" is true? > > Notice that exactly one of these questions has a "yes" answer > and exactly one has a "no" answer. (Actually, come to think > of it, since the base theory is ZF and not ZFC, it's possible > that the answer to 1. is "yes" if by "finite" we mean one > type of finite, say Dedekind finite, and "no" if we mean some > other type of finite. In this case, I'd like to know which > types of finite produce a "yes" answer.) > > If 1. is "yes," then I hope that I will never again see a post > claiming that ZF-Inf+~Inf doesn't prove that every set is in > fact finite. In fact, I'll go as far as to suggest that if 1. > is "yes," then those who claim that ZF-Inf+~Inf doesn't prove > that every set is finite deserve to be called five-letter > insults -- if posters are going to call those who deny the > proof of Cantor's Theorem by five-letter insults, then those > who deny the proof of "every set is finite" in ZF-Inf+~Inf > also ought to be called the same. > > If 2. is "yes," then what I'd like to know is how can I take > _advantage_ of this fact? Suppose I want to consider a theory, > based on ZF-Inf, which actually proves that an infinite set > exists, yet also proves that no successor-inductive set > containing 0 exists. > > I am also very much interested in knowing the outcome of this controversy. A quick look at Wikipedia, http://en.wikipedia.org/wiki/Axiom_of_infinity , throws up the following; \begin{quote} Indeed, using the Von Neumann universe, we can make a model of the axioms [of ZF] where the axiom of infinity is replaced by its negation. It is V_\omega \!, the class of hereditarily finite sets, with the inherited element relation. \end{quote} > > > In the current Tony Orlow thread, there is a discussion about > whether TO is defining N+ to be a successor-inductive set. It > is possible that the theory that I mentioned above might be > useful to discussing TO's ideas. > > But of course, we can't proceed until we know, once and for > all, whether ZF-Inf+~Inf proves every set to be finite or not. |