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 27 Jun 2010 02:34 On Jun 25, 2:16 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > On Thu, 24 Jun 2010 21:21:05 -0700 (PDT), Charlie-Boo > <shymath...(a)gmail.com> said: > > > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > >> On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > >> > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy > >> > to prove that PA is consistent (its axioms and rules preserve > >> > truth) yet (by Godel-2) PA can't do such a simple proof as that. > > >> So what? ZFC can prove it. > >> ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove > >> that PA is consistent. This is why you can't say that "ZFC/PA > >> doesn't prove PA is consistent." "ZFC/PA" is just a meaningless > >> locution in any case. > >> ZFC is one thing. PA is another. > > > PA is a subset of ZFC, > > No it isn't. (Curious that you continue to assert otherwise.) How do you define PA being a subset of ZFC and how do you know it fails that test? Peano's Axioms are part of the definition of N in ZFC. > > so I emphasize that by calling it ZFC/PA (it makes more sense to > > distinguish the two anyway.) This is besides the point. Who has > > proved PA consistent using ZFC? > > It is an elementary exercise. A proof can be found in almost any > reasonably thorough text in set theory. Ok, what's the latest one (easiest to get, modern syntax and terms)? C-B
From: Charlie-Boo on 27 Jun 2010 02:44 On Jun 25, 4:58 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > > PA is a subset of ZFC, > > No it isn't, but even if it were... How would you define PA being a subset of ZFC? (The above sounds stupid but the stuff later sounds smart!) > > so I emphasize that by calling it > > ... there would still be two things: something _and_ a subset of it > which would deserve two names. > > > ZFC/PA (it > > makes more sense to distinguish the two anyway. > > Well, yes, since they're different. No, I mean ZFC's set axioms vs. PA's axioms vs. ZFC/PA = both sets of axioms. Right now ZFC is said to have the PA axioms for free, but that's just trying to sneak them in. (They're claimed to be "definitions"!) > > ) This is besides the > > point. Who has proved PA consistent using ZFC? > > To "construct" a model of PA in ZFC is rather easy. You would think it not too big a deal. But what is strange is that I don't think what ZFC has that PA does not would help. So since PA can't I want to see how they used ZFC's stupid little set theoretic axioms or if they didn't how we got around Godel-2! But now I hear the (or a) proof uses the declaration of a model existing as being proof within ZFC of PA's consistency. No. Use ZFC axioms and rules only - that's the whole idea of ZFC. "That's the whole idea." - Barack Obama when asked if he inhaled. > Also, see Gentzen > and Ackermann. Gentzen's proof used far less than full ZFC. References please. On-line?? Thanks! > > If it were possible > > then I assume someone would have done it. > > You assume correctly. > > > It certainly would be a > > very educational exercise. > > > In any case, it shows the weakness of PA. > > PRA + induction up to epsilon_0 (which is what Gentzen used) is > incomparable with PA. Hmmm... you'll have to define some of that. (I assume you won't stoop to using the word "elementary" etc.) > > I added ZFC as that is so > > popular. > > So is watching telly (among a certain class of people at least). Yeah, I agree - I think most of the people here are at that level of intelligence - watch TV laughing at videos of children being injured in accidents. "America's Funniest Videos" > You may wish to know that ZFC with the axiom of infinity replaced by > its negation is a model of PA and vice versa. Wow, that sounds cool. I'll have to think anout that one. Where can I read about it? C-B > -- > I can't go on, I'll go on.
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 |