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 3 Jul 2010 17:55 On Jul 3, 2:42 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Then don't attack people on the behalf of Aatu's views I didn't! You're hopeless. MoeBlee
From: MoeBlee on 3 Jul 2010 18:32 It's likely this post will earn me yet more tedium in responding back and forth with nutcases, but I'd like to add this anyway: Consider the axioms of PA aside from the induction schema. My view is that IF there are ANY non-logical simple truths about natural numbers (in some suitable pre-formal ordinary sense of 'natural number', which would be any suitable abstraction of the activity of counting) then those axioms are among them. That is to say, if there is ANYTHING that could be said to be a finitistic mathematical truth, then those axioms are among them. Then, the induction schema seems to me ineluctable too (though, not as "certain" as the finitistic part of PA, or at least not certain in the same way, as there are complications having do with the notion of properties, even confined to those properties that are "captured" by first order formulas, etc.) . Now, I do not hold to this perspective as a matter of dogma, nor do I have any polemical investment in it. Rather, this perspective is just that: a perspective for me. It is a basis for "making sense" of mathematics for me. If someone else rejects it as a basis, then I'm not inclined to try to convince otherwise. In other words, such a perspective is for me a "way of looking at things" that is suitable, that serves me well, at least so far in my life. I don't need to argue that it is a "correct" perspective, let alone a "true" perspective. Rather, it is, for me, just a perspective that serves me well toward making sense of mathematics and of my own mathematical mental experiences. So, in this way, certainty is put in terms that seems as unassailable as could be to me: IF there are any certain (non-logical) mathematical truths, then those of finitistic arithmetic are ones. I can't imagine that they are not certain, but even IF I could imagine them not certain, then I don't know what OTHER (non-logical) mathematical certain truths there can be. And even here, I don't disallow that there might be even more certain non-logical truths than those of finitistic arithmetic. It's just that presently I can't imagine them. I can't imagine what non-logical matter would qualify as certain if simple matching of strings of symbols is not certain. However, if someone doesn't even regard finitistic mathematics (such as results of PRA) - essentially just recognizing whether strings of symbols match or do not match - as "certain", "correct", whatever, then I admit that I can't see what basis for communication I would have with him or her. I don't know how we could even communicate if we couldn't agree that we can look at finite strings of symbols and check for matching. As to the question of infinite sets, I've posted the bulk of my view (in SOME ways drawn from Hilbert's contentual/ideal notion and from other influences along with my own, possibly individual, reasoning too) in various posts over the years. MoeBlee
From: K_h on 3 Jul 2010 20:45 "R. Srinivasan" <sradhakr(a)in.ibm.com> wrote in message news:d97a79f9-9414-4c70-aab5-8d86680b2805(a)g24g2000pra.googlegroups.com... On Jul 2, 7:06 am, "K_h" <KHol...(a)SX729.com> wrote: > "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message > > news:90ce086b-c089-4b90-866d-c7391e95bbd6(a)j8g2000yqd.googlegroups.com... > On Jul 1, 3:15 am, "K_h" <KHol...(a)SX729.com> wrote: > > > > > "R. Srinivasan" <sradh...(a)in.ibm.com> wrote in message > > >news:46d58d89-34b1-40a9-a5a8-1ee250ba57e3(a)e5g2000yqn.googlegroups.com... > > On Jun 29, 8:33 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Jun 29, 2:09 am, "R. Srinivasan" <sradh...(a)in.ibm.com> > > > wrote: > > >> > > ZF-"Inf'+"~Inf" > > >> > > That theory entails that every object is finite. And > >> > > there is no > >> > > definition of any infinite object possible in that > >> > > theory. > > >> > OK. Here I want ~Inf to be stated in the form that you > >> > mentioned, that is, every set is hereditarily finite. > > >> Why do you think the axiom of infinity is false? What is > >> the basis for your belief in ~Inf? To me it is > >> self-evident that all the naturals exist. > > > First of all I happen to work in a logic (NAFL) where I have a *proof* > > of ~Inf. Essentially, if you define truth (as provability) such that > > all vestiges of Platonism are thrown out, infinite sets will not > > > respect to T, meaning "In the theory T+P, ~~P is provable". So you accept double negation -- good! There are a number of flaws in your post, including the flawed assumption that an un-decidable thing in T doesn't exist. Platonism is still strongly intact. > To put it in a nutshell, ~(P&~P) can only be a > theorem of those NAFL theories which either > prove P or prove ~P. If P is undecidable in T, > there is no proof of ~(P&~P) in T. ~(P&~P) is a logical formula and not a theorem from an axiomatic theory of sets. Form a truth table for this and you get: P ------- 1 0 ~P ------- 0 1 P&~P ------- 0 0 ~(P&~P) ------- 1 1 If P is either true or false ~(P&~P) is always true. So it seems like "NAFL" is some type of new logic. Can you post its axioms here? > From the NAFL argument, one can refute the > existence of infinite sets and I have done > this in my paper. Can you provide a link to your paper here? > ... and there cannot be a consistent NAFL theory > of non-Euclidean geometry... Then NAFL theory is instantly falsified because the surface of a sphere is non-Euclidean geometry of a two-dimensional space. If non-Euclidean geometry is inconsistent in NAFL then NAFL cannot handle the surface of a sphere in E^3. > In NAFL, N is a proper class and there is no > such thing as P(N). You cannot quantify over all > the infinite sub-classes of N to get P(N), > because quantification over proper classes is not > permitted. Why is N a proper class in NAFL and what is the basis to even have proper classes in NAFL? The whole reason to bifurcate collections into sets and proper classes is so that mathematics is self-consistent. That is, the problems presented by Cantor's theorem (for example the Cantor and Burali-Forti paradoxes) can only be solved by dividing collections into two types: sets and proper classes. (The problems presented by Russell's paradox are solved by the axiom of foundation, no collection can be a member of itself.) So it's because of Cantor's theorem that such a bifurcation is necessary. If Cantor's theorem is false in NALF then on what basis does NALF bifurcate collections into the two types? As far as I can tell, NAFL has no basis to do so and so N should be a set. As a set, it is easy to show that |N|<|P(N)|. We must show that |N|<|P(N)|. For this it must demonstrated that |N|<=|P(N)| and |N|=/=|P(N)| are both true. To establish |N|<=|P(N)|, note that for all n in N the function f(n)={n} does the job. To establish |N|=/=|P(N)|, a proof by contradiction is needed. Subsets of N, like {2,5} and {1,3,5,7,9,...}, can be written like {_,_,2,_,_,5,_,_,_,...} and {_,1,_,3,_,5,_,7,_,9,_,...} where the underscore denotes the natural numbers that are missing from the subset. Since one subset of N is N itself, N={0,1,2,3,4,5,6,7,8,9,...} is the only subset without underscore characters and the empty set will have nothing but underscore characters. Assume there is a bijection, f, between N and P(N) as follows: f(0) = {_,_,2,_,_,5,_,_,_,...} f(1) = {_,1,_,3,_,5,_,7,_,9,_,...} f(2) = {0,1,2,3,4,5,6,7,8,9,...} f(3) = {0,1,_,_,4,_,6,_,_,9,...} f(4) = {0,_,2,_,_,5,6,_,8,_,...} .... Now a new subset, S, of N is constructed diagonally as follows. For each f(n) look at the nth place in its set and if it is an underscore then the nth place in S contains n otherwise it contains an underscore. For example, the zeroth place of f(0) is an underscore so the zeroth place of S contains a zero. The first place in f(1) has a 1 so the first place in S has an underscore. The second place in f(2) has a 2 so the second place in S has an underscore. The third place in f(3) has an underscore so the third place in S has a 3. And so on. Eventually S is the set: S = {0,_,_,3,4,...} It is clear that the subset S of N is not in the list because it is different from every set in the list according to the axiom of extensionality. Recall that the axiom of extensionality says that two sets are equal iff they contain the same elements. The set S={0,3,4,...} is clearly a subset of N and it differs from each subset in the list. This contradicts the assumption that f is a bijection between N and P(N). So no bijection between N and P(N) is possible and that means |N|=/=|P(N)|. > Again, you are arguing backwards. In NAFL, > truth as provability is a fundamental logical > requirement (as I have illustrated above while > discussing the law of non-contradiction). > Therefore the incompleteness theorems do not > hold for NAFL theories (e.g. F, the theory of > finite sets). You haven't demonstrated what you say you have. All formal systems are limited in what they can mechanically prove but those limitations are not limitations on existence. _
From: MoeBlee on 5 Jul 2010 14:01 On Jul 3, 3:07 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > MoeBlee wrote: > > On Jul 3, 2:39 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > >> MoeBlee wrote: > > I don't know what you find bundled with the word 'confirm', so I > > prefer to stand by what I posted, which is clear enough, especially as > > it is an extremely common notion discussed widely. > > Oh very simple. I can _confirm_ by sheer formal proof that the theory > T = {(x=x) /\ ~(x=x)} is inconsistent. There you give an EXAMPLE of what you mean by 'confirm' but not a definition. But we might not need to get bogged down in that if possibly your notion of 'confirm' is close enough to the notion of 'finitistic proof'. Very roughly stated: a finitistic proof is one that uses merely primitive arithmetic (which can also be viewed as purely mechanistic pattern matching of finite strings of symbols). I would think this would at least be subsumed by what you might mean by "syntactic proof". And, in that regard, I have said all along that there is no finitistic proof of the consistency of PA. I base that on the second incompleteness theorem. (I haven't personally verified every detail in a proof of the second incompleteness theorem, so my remarks are to the extent we can be confident that the second incompleteness theorem does indeed withstand complete scrutiny, as it is reported in the literature that it does.) Meanwhile, I have no further comments about your own notion of 'confirm'. MoeBlee
From: MoeBlee on 5 Jul 2010 14:01
On Jul 3, 3:12 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > Sorry, MoeBlee, you were wrong (if not hopeless) about attacking people, > when representing a portion of somebody else's views! (Note the your > word "these subjects"!) Life Too Short. MoeBlee |