How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: MoeBlee on 28 Jun 2010 18:20 On Jun 28, 2:36 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > On Jun 28, 10:57 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > On Jun 28, 11:54 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > On Jun 28, 8:53 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Jun 27, 4:32 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > > On Jun 28, 2:11 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > > On Jun 27, 1:25 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > > > > On Jun 28, 12:44 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > > > > On Jun 26, 6:38 am, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > > > > > > The proof in the theory ZF-Inf > > > > > > > +~Inf that infinite sets do not exist encodes (among other things) > > > > > > > that models of PA cannot exist. > > > > > > > PROVE IT. > > > > > > Trivial. > > > > > If it's trivial, then it shouldn't take you but a minute to prove it. > > > > Please do. > > > NO ANSWER HERE. STILL NO ANSWER FROM YOU. (And your confused stuff that follows later does not answer here.) > > OF COURSE PA exists, as we prove formally in Z set theory, but NOT in > > Y. In Z set theory we may formally define 'first order theory', and > > all kinds of things, and then PROVE that there is a certain first > > order theory with certain properties, and we call that particular > > first order theory 'PA'. That happens in Z set theory or some other > > sufficiently strong formal theory (if we formalize) or just in > > informal set theoretic mathematical discussion (if we don't > > formalize). But it does NOT happen in Y. > > > > IS NOT EVEN A PROPOSITION in the language of Y. > > > YES IT IS, you ignoramus. Please recognize here that I have properly corrected your incorrect assertion, as below (though I should have added that we do this in Z with the language of Z = the language of Y). > > The language of Y is the language of set theory (i.e., the 2-place > > relation symbol 'e' and the logical symbols). And in that language > > (here I'm only giving an informal English rendering), we may define > > 'PA' as "the set of sentences closed in the language with '0', '+', > > 'S', and '*', under the [such and such] a set of axioms'. > > OK. I think I can restrict the language of Y such that no explicit > references to any infinite sets are allowed. So I want the language of > Y to only contain references to finite sets and the axiom that > "Infinite sets do not exist" is stated in the form you mentioned: So you've got another theory in another language. What EXACTLY is your language? The language of Y has only ONE non-logical symbol, viz. 'e', any further restriction is the language of equality theory. > Ax En x in P_n(0) > > without any explicit reference to infinite sets What do you mean "explicit reference to infinite sets" in a manner that "restricts" a language? Y is just a theory in a language with one 2-place relation symbol. Of course the predicate 'is infinite' has a DEFINITION in Y. There's no way to BLOCK that. Sure, we could dispense ever using the defined predicate, but we could still use the equivalent formulation in the primitive language. > (By the way, I think > that the axiom D=0 which I wanted to add is very interesting in its > own right, but for the time being, let us use your definition). The way you wrote it is UNSUPPORTED. Clearly, you're not reading what I write. And I didn't provide a DEFINITION; rather, I just told you how to restate the AXIOM. > If I do that you will not be able to define any object called 'PA' in > the language of Y. The propositions in this language can only contain > explicit references to finite sets. Please give your TECHNICAL definition of 'explicit references to finite sets'. Anyway, it's aside the point. I ALREADY have to told you that Y cannot define 'PA' in a way that corresponds to such a definition in Z. (Of course, we can use 'PA' as name for something else, that is not at issue, but we cannot define ANY symbol, whether 'PA' or whatever, in Y the way we define 'PA' in Z because Z proves there EXISTS object such as described in the definens of the definition while Y proves there does NOT exist such an object.) Why aren't you getting this yet? > But I do not even need to do this. Let me play along with you. Suppose > that Y proves that there is no infinite object called PA. No, that is NOT what I said. What I said is that Y proves "there is no infinite object". (By the way, I'm taking that on imperfect memory from a few years ago when, as I recall, Aatu showed a proof (which I should look up in my notes) that ZF-I proves "I iff there does not exist an infinite set".) > It still > does not undermine my argument in any way, as I will argue below. I'll look for some NEW argument you might have. But your original argument is kaput. > > Probably what you mean is that one sentence (in the language of Y) > > "infinite sets do not exist" is LOGICALLY equivalent to another > > sentence (in the language of PA) "~Con(PA)". > > There is a sentence in the language of Y that encodes ~Con(PA). So, you've backed off this business of LOGICAL equivalence? Instead of using undefined terminology "encodes" why not just refer to this EXACT statement: There is a sentence S in the language of Y such that when S is interpreted to a sentence S* in the language of PA, we have that that S* is true in the standard model of PA iff PA is inconsistent. (And, from now on, in this context, I'll say 'true' as abbreviation for 'true in the standard model of PA'.) If that is not suitable then please state EXACTLY what assertion you mean by "encodes". > What I > am saying is that the sentence "infinite sets do not exist" is > equivalent to this sentence. Equivalent in WHAT THEORY? Equivalent in Y? Equivalent in Z? Then, once you answer that, please prove your assertion. And then once you do that, please say what point you draw from this. And if you say your point is that Y proves that PA is inconsistent, then please provide demonstration. > Yes, there was a typo in what I said > above (although I do believe truth is provability, but for the time > being, let me stay within classical thinking). I don't know whether you're referring to the bit about LOGICALLY equivalent (or what that has to do with "truth is provability" or even what you MEAN by that). > > Y proves "there does not exist an infinite set", so Y proves "there > > does not exist a set of sentences that is infinite" (since Y can > > express 'set of sentences'), and (and it can be proven IN Y) every > > theory is an infinite set of sentences, therefore Y proves there does > > not exist a theory. > > > Just STOP for a minute right there. Whatever else about your "coding", > > the above argument stands. You cannot refute it. > > OK. So what? It does not undermine the point I want to make. It undermines (in the exact sense I originally posted) your argument as you orginally posted it. If you have some point you want to make, then what exactly is the point you want to make. You sure as hell don't show in your remarks below that Y proves that PA is inconsistent. > As I > said, let me play along with you. Here is what I want to drill into > your head: Don't drill anything into my head. Just posting with you is torture enough. Let's put aside your machine tools if we may. > 1. The theory PA can be interpreted in Y. I ALREADY know this. No need to "drill" it in my head. > 2. Y formalizes sentences like Con(PA) (via suitable encoding). See my above remarks. > 3. Y denies the existence of infinite sets. You don't need to "drill" that in my head, since I, not YOU, can dig up (as I recall) a PROOF of that. > Basically these three points are to be understood as follows. Y allows > all the arithmetic operations of PA, Y understands what it means for > PA to be consistent (or for a model of PA to exist) What does "Y understands" MEAN? Y is not a human being that has understandings. Y is a formal theory, which is a set of finite strings of finite strings of symbols. And AGAIN: Y ITSELF does not refer to PA. What we have is that there is a sentence S of the language of Y that when interpreted into S*, a sentence in the language of PA, we have that S* is true iff PA is consistent. If THAT is what you mean by Y "understands" what it "means" for PA to be consistent, then okay, but so what? IF it is not what you mean by Y "understands" what it "means" for PA to be consistent, then what EXACTLY do you mean? > and Y denies the > existence of infinite sets. These three facts together mean that Y > proves ~Con(PA) in a meaningful way, for Y denies that models of PA > can exist. NO, WRONG. You've shown no such thing, as well as AGAIN we have undefined terminology from you, viz. "meaningful way". What would at least MAKE SENSE is for you to say: Y proves S*, which when interepreted back to S in the language of PA is true in the standard model for PA iff PA is inconsistent. But (1) you have not proved that, and (2) you have not shown that it entails that Y proves that PA is inconsistent. And, if PA is consistent, then Y is consistent. So Y does NOT prove S* nor the negation of S*. Moreover, S* is only a sentence such that its interpretation S in the language of PA is true in the standard model of PA iff PA is consistent. Y ITSELF does not prove, IN Y, ANYTHING about PA. Rather, when we STEP back into a META-THEORY about Y we find that there is a sentence S* in the language of Y such that the interpretation of S* into the language of S is true in the standard model of PA iff PA is consistent. > Now let us get to your point. Y does not recognize any explicit object > that stands for the theory 'PA' and even denies its existence. It denies that there exists an object so desribed as "a theory with axioms and such and such...". > Perfectly in tune with the fact that Y proves ~Con(PA). What PARTICULAR FORMULA in the language of Y are you taking as ~Con(PA)? Please state the formula and then show the proof in Y (or even prove non-constructively, in ZFC or whatever stated meta-theory, that such a proof exists). > For if PA is > inconsistent, infinite sets cannot exist and indeed there cannot exist > any infinite objects like the theory 'PA' from Y's perspective. How do you derive: If PA inconsistent, then there do not exist infinite sets? In what THEORY are you making such a derivation? > So what? In what way does this undermine my argument above based on > the 3 points noted above? Your argument above is a DIFFERENT argument from the one you gave originally. And your argument above is shot through with holes by (1) lack of defining certain crucial terms in your argument and (2) not proving certain of your crucial claims, and (3) fundamental confusion about what is stated and proven in a meta-theory and what is stated and proven in an object theory. > The point is that Y understands what Con(PA) > is via encoding. "understands", "via encoding". Waffle words as YOU use them. The EXACT situation is as I gave it: There is a sentence S in the language of Y such that when S is interpreted to a sentence S* in the language of PA, we have that that S* is true in the standard model of PA iff PA is inconsistent. Show me an EXACT result about what Y proves in terms of the consistency of PA, please don't waste my time with waffle about "Y understands" and "Y encodes". > It does not need to recognize any explicit object > called 'PA' to prove ~Con(PA) (or its encoding in the language of Y). See above. > Kindly keep your focus on the 3 points I mentioned above and avoid > nitpicking. To YOU getting this stuff technically correct (so that you can't argue via undefined waffel words, claimed but unproven assertions, and confusion of meta and object) is nitpicking. > > Now, don't be a pouting little baby and cry "rubbish" et. al. Instead, > > look at EXACTLY what I've written, then read a good textbook in > > mathematical logic. > > This is your problem. You want to go by the book, but here you need to > think out of the box. NO, YOU claimed to have all these proofs in terms of Y. But Y is a certain formal theory. If you claim to show things about it, then don't evade by saying "oh, have to think outside the box". I'm not opposed to approaches that aren't mentioned in textbooks. But YOU raised a matter that IS a textbook subject, and you do so in a way, as usual for you, that indicates you don't understand the very material you're pontificating about. > I certainly do not want to get dragged into the > same mess that you have fallen into. I'm in no mess in this current regard (I'm in other messes in my life, but you don't need me to expatiate on those). And reading a textbook on mathematical logic will not draw you into a "mess". MoeBlee
From: MoeBlee on 28 Jun 2010 18:40 By the way, here's Aatu's proof: Theorem (of Z-"ax regularity"-"ax infinity"+"~ ax infinity"): Ax x is finite. PROOF ['P' stand for 'power set'] Let 'F' be a 1-place operation symbol defined as follows: F(y) = the n such that ((y is finite & n is a natural number & y is equnumerous with n) or (y is infinite & n=0)). Toward a contradiction, suppose x is infinite. So {F(y) | y in Px} is the set of natural numbers, which is successor inductive. MoeBlee
From: MoeBlee on 28 Jun 2010 18:46 One thing I don't know how to do is show the mutual-interpretability of PA and Y=ZF-"ax inf"+"~ax inf" One direction seems not too difficult: interpreting PA in Y. But how do we interpret Y in PA? Specifically, how do we define 'e' in PA and then prove, in PA, all the axioms of Y as interpreted in the language of PA? Aatu, Daryl, or anyone help me out? MoeBlee
From: Transfer Principle on 28 Jun 2010 18:48 On Jun 28, 9:14 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 27, 10:49 pm, "R. Srinivasan" <sradh...(a)in.ibm.com> wrote: > > Here is one possible approach. Consider the theory F that I had > > defined in another post, where > > F = ZF - Inf + D=0, > > Where 0 is the null set and D is defined as > > D = {x: An(x not in P_n(0))} I wonder why Srinivasan can't simply use the nth level of the cumulative hierarchy, V_n, so that we don't have to worry about whether P_n(0) is welldefined or not. > Prove (and please state the formal language, logic, and axioms in > which you conduct this proof) that there exists a D such that for all > x, we have x in D iff An x not in P_n(0). > Unless you can do that, then your 'D' is not well defined. OK, I see the problem here. In standard ZF, the object that Srinivasan calls "D" would be too large to be a set. Thus, it's not evident that the set D exists. Rather than attempt to prove it, why can't we just let this be the axiom itself -- so instead of writing D=0, the axiom becomes: ED (Ax (xeD <-> An (~xeV_n))) exactly the statement that MoeBlee asks Srinivasan to prove (except inserting V_n). We can worry about whether such a set can equal 0 down the road. > > Here P_n(0) is power set operation iterated n times on 0, and P_0(0) = > > P(0), P_1(0)=P(P(0)), etc. Clearly only hereditarily finite sets can > > exist in models of F. > Prove it. OK, I see the problem here. Even if F has a model in which every set is HF, there might be non-isomorphic models of F with sets which aren't HF at all. Still, I believe that Srinivasan is on his way to giving a theory that will satisfy most finitists.
From: Transfer Principle on 28 Jun 2010 19:06
On Jun 28, 6:38 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jun 27, 2:29 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > What's CBL? Is it "Charlie-Boo logic?" If so, then I'd like to > > learn more about this challenger to FOL. > It's not really a challenger to FOL. In fact, it uses FOL for what it > does well: represent functions and relations over a single universal > set - or subsets of it. > CBL replaces FOL in certain contexts where we don't have a single > universal set, because the domains of the components of functions and > relations are not only different, they have diferent cardinalities! > This is why the wffs for ZFC are so long, complex and sometimes > debated as to validity. The primitives of something so primitive as > set theory should be able to please Occam. And they can - in CBL. > The ZFC axioms can be stated in a fraction of the size as using FOL. > And how is the Theory of Computation formalized? How do we express > its fundamental theorems and proofs? In CBL it is just a wff with a > particular value of Q. Interesting. Thanks for the explanation. > Where does that occur? That is the definition of metamathematics. We > are trying to draw relationships between sets of differing cardinality > - e.g. to equate the aleph-1 set of wffs Are there really aleph-1 wffs? In the MoeBlee-Srivinasan subthread, we have MoeBlee explaining that PA is an infinite theory: MoeBlee: In particular, we may specify a certain set of symbols and arity function so that that system is a language for a first order theory such as PA, then specify PA to be the theory that is the closure of the INFINITE set of axioms (the induction schema is an infinite set of axioms) that we specify. but MoeBlee doesn't state whether this "infinite" is countable, aleph_1, or larger. (Similarly ZFC is also infinitely axiomatized, so we know that it's infinite.) I could've sworn that someone posted that there are only countably many wff's (in the language of either PA or ZFC) since each wff is a finite formula, each taken from a countable set of symbols (which includes the countably many variables, as also mentioned by MoeBlee). > with the aleph-2 set of functions over wffs. If by "functions over wffs," Charlie-Boo means functions from the set of wffs to itself, then there are aleph_0^aleph_0 = continuum many such functions. Of course, if the metatheory is ZFC+"c=aleph_2," there could be aleph_2 many such functions. > In metamathematics, we want to know if a certain set can be > represented within another set by substituting a constant for one > component. We may also want to know and use this constant. So we > relate a wff to a set. The universal sets having different > cardinalities, we need a relation over sets of different > cardinalities. OK, I somewhat see what this is leading to. Charlie-Boo does mention "universal sets" such as V (a proper class in NBG, but apparently a set in CBL), so that one can replace the wff such as "x=x" with "xeV." This isn't much shorter, but I assume that wffs such as "x is an ordinal" (a bit lengthy when expanded to full primitives) would become something like "xeOn" in CBL. Am I on the right track here? |