How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Aatu Koskensilta on 5 Jul 2010 10:05 "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > Aatu said PA is consistent, _period_, without any formal proof? There seems to be some confusion over my (perfectly standard as always) take on these matters. Allow me to clarify. I know that PA is consistent because I know how to produce a mathematical argument for that fact which I fully understand -- in the sense that I grasp the key idea and can explain all the technical details that go into its execution -- and find compelling. The proof I have in mind has the following structure: 1. We give an inductive definition of a property T(x) of arithmetical sentences. 2. We show that T("P") iff not T("~P"), that T("P & Q") iff T("P") and T("Q"), and so on. 3. We conclude from this that T does not hold of any contradiction. 4. We show that T holds of all the axioms of PA. 5. We show that the rules of inference of first-order logic preserve T. 6. We conclude that no contradiction is formally derivable in PA. We can of course formalize this proof in any number of theories -- ACA, ZFC, ... -- but this is just an incidental technical observation of no immediate interest as far as consistency of PA is concerned. We may also be interested in knowing exactly which set theoretic axioms are needed e.g. to justify the sort of inductive definitions involved, and so on. Again, this has no immediate connection to the question of consistency of PA. As with any proof in ordinary mathematics we take in the above proof many mathematical principles and modes of reasoning for granted. For example, we apply induction to a property involving a predicate defined by means of an infinitary inductive definition. Everyone must decide for themselves whether they find such invocations evident, compelling, legitimate mathematics. It's a brute fact of life that in ordinary mathematics we do in fact take them for granted. In Usenet debates, and elsewhere too, people like to make much of and pontificate tediously on who has or does not have the "burden of proof". In reality, such "logical" rules of formal debate have about as much to do with real arguments, persuasion, conversion, as the rules of cricket. Whoever wants to convince someone else of something must of course present arguments, questions, reflections, examples, illustrations that have some real force to the receiving person, regardless of whether they make a "positive claim" according to some rulebook of debating. I don't have any real interest in convincing anyone who does not take ordinary mathematics for granted of much anything. I'm content with simply noting that the consistency of PA is a triviality given what we in fact usually take for granted in mathematics. If someone tells me they find the usual modes of reasoning horribly dubious or unclear, even after careful explanation and meticulous elucidation, I simply shrug and move on. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 5 Jul 2010 18:08 MoeBlee <jazzmobe(a)hotmail.com> writes: > 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. The first incompleteness theorem already suffices. This elementary point is often overlooked, and has recently been stressed by Richard Zach. I said a bit about this in an earlier post, not that it matters much in the grand scheme of things. > (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.) Come now, you're surely not hedging your bets and saying you're not absolutely certain the second incompleteness theorem, which has withstood the most exacting and extreme scrutiny humanly possible, far, far beyond what any run of the mill result in mathematics -- the classification of finite groups, the Heine-Borel theorem, most of stuff in analysis, Wiles's proof of Fermat's last theorem, the fundamental theorem of arithmetic ... -- usually has to endure to prove its good name, holds. (I apologise for the clumsiness in the construction of the previous sentence. My only excuse is that I'm again suffering of severe lack of sleep.) -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 5 Jul 2010 18:14 MoeBlee <jazzmobe(a)hotmail.com> writes: > On Jul 5, 7:05�am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: >> "Jesse F. Hughes" <je...(a)phiwumbda.org> writes: >> >> > Aatu said PA is consistent, _period_, without any formal proof? >> >> There seems to be some confusion over my (perfectly standard as always) >> take on these matters. > > Is there anything in my paraphrase several posts ago (in the full > context I gave it) that is inaccurate? Not that I can see. It seems that it has given an incorrect impression of what I think about these matters to Nam, though. I took your paraphrase to be simply an explanation of what I mean when I say I don't regard the usual consistency proofs for PA as establishing any "relative" sort of consistency -- that is, I took you to be explaining I were asserting PA does not prove P and not-P for any sentence P. and not merely ZF proves "PA does not prove P and not-P for any sentence P". or It is provable in ZF that PA does not prove P and not-P for any sentence P. which is indeed what I intended. > However that quote is to be understood, for the record, I did not say > that you hold there is no formal proof that PA is consistent, but > rather that you hold PA is consistent on (for lack of better term I > can think of right now) even more basic grounds than formal proof. Right. I hold PA is consistent because there's a mathematical proof of this fact, that is, a piece of compelling mathematical reasoning establishing the claim, invoking only principles we usually take for granted when doing mathematics. I can't really see how anyone who didn't take such things for granted could conclude from the "relative consistency" theorems anything about the consistency of PA. > And that is what I have highlighted as to your view. It seems your posts have nevertheless led Nam and lwalke to incorrect ideas about my views. This is not any fault of yours. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 7 Jul 2010 13:27 Nam Nguyen <namducnguyen(a)shaw.ca> writes: > Aatu Koskensilta wrote: > >> 1. We give an inductive definition of a property T(x) of >> arithmetical sentences. > > So, would T(x) be a formula in L(PA)? No. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 7 Jul 2010 13:29
Nam Nguyen <namducnguyen(a)shaw.ca> writes: > MoeBlee wrote: > >> And, in that regard, I have said all along that there is no >> finitistic proof of the consistency of PA. > > So you've agreed that there's no formal proof for PA's consistency and that > if you go by formal proof only then you don't have knowledge of PA's > consistency. There are many formal proofs of the consistency of PA. None of them are finitistic. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |