From: MoeBlee on 7 Jul 2010 14:16 On Jul 7, 12:57 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > "Jesse F. Hughes" <je...(a)phiwumbda.org> writes: > > > Transfer Principle <lwal...(a)lausd.net> writes: > > >> In this "secret" thread which he tried in vain to keep hidden[...] > > > Which he tried to *what*? > > MoeBlee tried to keep this "secret" thread hidden by publicly posting it > on a Usenet newsgroup regularly read by those I presume lwalke thinks he > wanted to hide it from. This nefarious plan was foiled by lwalke's > vigilance! You mean anybody can read this thread? But I thought that if prayed to Moloch and offered my pet raccoon as a sacrifice (which I did), then this thread would be visible only to "standard theorists". Drat! Thwarted by the diabolically clever Transfer Principle again! MoeBlee
From: Tim Little on 8 Jul 2010 05:00 On 2010-07-07, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > No it doesn't. Suppose I say the Riemann hypothesis is true because, > you see, if we look very carefully at the zeta function, and squint > our eyes a bit, we find it's in reality made out of paper clips, > hotdog buns and naval lint. Do you have a suitable function made of fruit buns? I'm somewhat hungry right now, and fruit buns would be a very suitable solution. I would prefer not to have lint mixed in, but could separate out paper clips with a powerful magnet if necessary. I call upon your mathematical expertise to solve this problem, which seems to be worsening with time. > Am I vindicated if someone proves the Riemann hypothesis? Well, obivously. You said it was true and it is true. Insignificant side-claims such as the reason you offer for it being true make no difference. - Tim
From: Tim Little on 8 Jul 2010 05:06 On 2010-07-06, Transfer Principle <lwalke3(a)lausd.net> wrote: > And so rather than continue the discussion in the existing threads, > he started this thread in the hopes that that other type of poster > wouldn't notice this thread at all. [...] > In this "secret" thread which he tried in vain to keep hidden This is the sort of comment that leads me to suspect that Transfer Principle really might be a very good troll after all. Most of his/her posts appear to be filled with clueless stupidity and misreading of a particularly consistent kind, but inferring something like this from MoeBlee's posts surely takes more than mere idiocy can adequately explain. - Tim
From: MoeBlee on 8 Jul 2010 13:08 To whom it may concern: When we say there is no finitistic proof that PA is consistent, we don't contradict that there is a Z-R formal proof of "PA is consistent". It is finitistically "checkable" that a given sequence of formulas in the langauge of Z-R is or is not a proof of the Z-R formula (that we are here nicknaming) "PA is consistent". But that is not what we mean by 'finitistic proof'. Yes, per some formal proof system and formal axioms, it is finitisitc to check whether a given sequence of formulas of a formal language is a proof from said axioms in said system. But what we mean by 'finitistic proof' is that said proof itself uses no more than finitistic methods (often formalized as "proof using no more than PRA"). So, again, to be clear: (1) For a given given formal language, and formal axioms, and formal proof system, it is finitistic to check whether a sequence of symbols is a formula of said language and whether a sequence of formulas of said langauge is a proof from said axioms in said formal proof system of a given formula. (2) However, when we say 'finitistic proof' we are not referring to the above, but rather that said proof system and axioms are themselves finitistic, often in the sense that said proof system and axioms constitute PRA or some subsystem of PRA. MoeBlee
From: MoeBlee on 8 Jul 2010 13:35
P.S. So, Z-R |- Con(PA) That is, there is a formal proof from the axioms of Z-R of the sentence 'Con(PA)' but that formal proof is not a finitistic proof, since Z-R is not PRA nor a subsystem of PRA. IF we had PRA |- Con(PA) then there would be a finitistic proof of 'Con(PA)'. But we do not have PRA |- Con(PA). The incompleteness theorems shows us that we don't have PRA |- Con(PA) and we don't have T |- Con(PA) where T is PA itself (unless, of course PA is inconsistent, but still whether PA is consistent or not, we can, by finitistic means, show that we don''t have T |- Con(PA) where T is PRA or any subtheory of PRA). That "we do NOT have PRA |- PA" we take as saying "there is no finitistic proof of the consistency of PA". But, again, that does not preclude that there are many non-finitistic but still formal proofs of the consistenty of PA. And, again, we (editorial 'we') show that there are such formal proofs but in doing so we don't necessarily assert that they have any epistemological import for anyone who already doubts that PA is consistent. What would have epistemological import would be PRA |- Con(PA), but the incompleteness theorems show us that we don't have PRA |- Con(PA). MoeBlee |