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 1 Jul 2010 10:13 Nam Nguyen <namducnguyen(a)shaw.ca> writes: > but should I proclaim that PA is consistent, as in, "that PA is > consistent is a triviality", as Aatu put it? You will of course decide for yourself what you wish to proclaim. > The question I had for him was a clarification request to see if he > meant PA is really consistent, or if he meant that was just a relative > consistency proof he had referred to. The trivial consistency proof for PA is no more relative than any other proof in mathematics. Since you apparently take the view that pretty much everything in mathematics is relative you will naturally regard the proof as relative, just as you will regard Dirilecht's theorem, the deduction theorem, the Beurling-Lax theorem, and so on. Concentrating on trivial consistency proofs is totally arbitrary. -- 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 1 Jul 2010 10:52 Nam Nguyen <namducnguyen(a)shaw.ca> writes: > I'd prefer a straightforward kind of answer that this is a relative > proof of (PA's) consistency in another formal system, ZFC, ZF-R,..., > rather than a less straightforward kind answer like "no more > relative". We have in mathematics results such as If the generalized Riemann-Hypothesis holds, the deterministic Miller-Rabin primality test runs in polynomial time. If ZFC + "there are infinitely many Woodin cardinals" is consistent, so is ZF + AD + DC. and so on, which are naturally regarded as relative, in that they state that something holds relative to the assumption that some as yet unproven statement holds. The theorem that PA is consistent, that 4 + 4 = 8, Dirilecht's theorem, and so on, are not relative in this ordinary sense. We will of course regard them as relative if we consider the principles usually accepted in ordinary mathematics as conjectural or doubtful. Even so concentrating on consistency results is totally arbitrary. > Again this is just an over-exaggeration or misunderstanding of my view. Quite possibly. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Alan Smaill on 1 Jul 2010 12:09 Charlie-Boo <shymathguy(a)gmail.com> writes: > On Jul 1, 9:31�am, Alan Smaill <sma...(a)SPAMinf.ed.ac.uk> wrote: >> Charlie-Boo <shymath...(a)gmail.com> writes: >> > On Jun 29, 5:28�pm, MoeBlee <jazzm...(a)hotmail.com> wrote: .... >> >> As to ZFC, you don't need such fancy stuff as epsilon_0. Just do the >> >> routine proof that with the system of omega with 0, successor, >> >> addition, and multiplication we get a model of all the PA axioms. >> >> > The problem isn't to conclude that a model exists, using ZFC. �The >> > problem is to prove that PA is consistent, using ZFC. >> >> If you admit that ZF can express "structure S is a model for theory T", >> then have a go at working out how the provable existence of a model >> shows consistency. >> >> Hint: �a semantic definition of consistency of T is that there is >> no formula P such that P and ~P are logical consequences of T. >> >> Hint: recall the definition of satisfaction of a negated formula >> in a structure. >> >> Hint: try using FOL. >> >> Is that enough hints? > > Instead of nibbling at the edges and playing games, how about a high > level summary of the proof and how ZFC axioms are needed to formalize > that? And then give a reference to where someone has carried it out? It's more educational to work out the ideas for oneself, I find. Your own claim that I responded to was this: The problem isn't to conclude that a model exists, using ZFC. The problem is to prove that PA is consistent, using ZFC. Are you really not interested in finding out that doing the first leads easily to doing the second? > C-B > -- Alan Smaill
From: Jesse F. Hughes on 1 Jul 2010 14:31 Charlie-Boo <shymathguy(a)gmail.com> writes: > Yes, and as a Conservative you know the creed that to name a book is > to defeat your enemy - especially if it is very hard to obtain. (And > to make sure, refuse to quote from it!) You live in Cambridge, MA. You can't throw a rock without hitting a nearby university library. How hard is it to get the book that Moe mentioned? Heck, ask one of your MIT grad student party buddies to get it for you. -- "The math doesn't care about their mortgages. It doesn't care about their political needs. [...] Today's mathematicians have to hate mathematics because mathematics doesn't look out for them. It doesn't pay attention to their needs." --- JSH analyzes mathematicians
From: Aatu Koskensilta on 2 Jul 2010 09:33
herbzet <herbzet(a)gmail.com> writes: > herbzet wrote: > >> For classical logic the following sets are extensionally equivalent: > > Well, that's not exactly right, but you know what I'm getting at. Such imprecision! And on sci.logic! You should be deeply ashamed of yourself. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |