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 17:35 On Jul 3, 2:04 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > MoeBlee wrote: > > > I didn't state that as MY view. I was telling you AATU's view. For > > Christ sakes! > > I was asking Aatu, _not you_ to clarify about relative consistency > and you "represented" him, even though he had never asked you to do > so! I don't presume to represent him in general. Only as to those specific remarks, I chose to state what his view is. At worst, I'm guilty of being presumptuous to state what his view is. But I have sufficiently discussed the subject of consistency of PA with Aatu and read enough of his posts to have a clear understanding of his view to the extent I stated it. > So you did represented Aatu in his absence! Therefore in this context > I can't make a distinction between you or him. Then there's no basis for rational discussion with you. I said that I was stating what Aatu's view is but that it is not necessarily my view (the ways in which I depart from Aatu are somewhat philosophically complicated and our differences are found in certain posts we made to each other; as well as some further considerations I would have to explain). By saying "This is what Aatu is saying, ..." doesn't entail that my own views are the same as his. I can't imagine a reasonable person not allowing that. > If you don't want it > that way, then don't represent anybody! Just argue on your own. I'll do as I please. I chose to state that Aatu's view is such and such in order to make that matter stark for you, as pleased me to do. If you claim that obligates me to be regarded as having the same view as Aatu then that's just more of your mulishness. > As far as I'm concerned, you just hide behind Aatu's name to "blast" > me (and you did here a few times) without using your own credibility! You really are a nut. MoeBlee
From: MoeBlee on 3 Jul 2010 17:54 On Jul 3, 2:39 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > MoeBlee wrote: > >> So there are many (formal) proofs of the consistency of PA, after all! > > > OF COURSE! I've said that all along. And I've said that one is quite > > reasonable to view that such proofs have no epistemological value. > > This is all discussed beautifully in layman's terms in Franzen's book. > > OF COURSE if one doubts the consistency of PA, then a proof, from an > > even STRONGER theory, such as Z, provides no basis for alleviating > > said doubts. > > > When I say there is a proof in this contexgt, I mean 'proof' in the > > technical sense of a formal proof, a derivation using recursive rules > > of inference with a recursive set of axioms, not necessarily in the > > sense of "indisputably convincing basis for belief" or related such > > senses. Such a thing may well not provide adequate basis for BELIEF if > > one does not already have adequate basis to believe said axioms are > > true. > > Let me try to summarize what you said above. There's no syntactical > proof that can possibly confirm the fact that PA is consistent, > if PA is in fact consistent. If this is what you meant I'm OK with > that. By 'proof' in this context I mean formal proof. 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. > And if that's the case, would you be able to make a stand and say we > in fact just don't know PA if is consistent? I don't view the matter in that simple way of putting it even. > That would help to save a lot of ridiculous arguments. You'd do best not to argue as if I hold certain views I've never stated holding. A good many of my views are posted over the years. Moreover, on those matters in which I have not filled in my views, then, to be reasonable, you can only conclude that I have not yet had time, inclination, or achieved a conclusion to post on the matter. MoeBlee
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. _ |