How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
in [Math]
|
Prev: ? theoretically solved
Next: How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
From: Frederick Williams on 1 Jul 2010 12:23 Charlie-Boo wrote: > > Of course infinity = PA = arithmetic. Of course. PA doesn't mention infinity. Just because these chaps 0, 0', 0'', 0''', ... exist, doesn't mean that this chap {0, 0', 0'', 0''', ...} exists. Which--speaking loosely--is why ZF needs an axiom of infinity and PA doesn't. -- I can't go on, I'll go on.
From: MoeBlee on 1 Jul 2010 12:25 On Jul 1, 9:01 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > MoeBlee wrote: > > On Jun 29, 10:47 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > >> Aatu Koskensilta wrote: > >>> Frederick Williams <frederick.willia...(a)tesco.net> writes: > >>>> Yes, you can: take Gentzen's proof (or Ackermann's etc) and formalize > >>>> it in ZFC. > >>> This is a pretty silly way of proving the consistency of PA in set > >>> theory. > >>> That PA is consistent is a triviality. > >> In what formal system is this triviality in? > > > It's a theory of Z-R, for example. Whether it's "trivial" to prove in > > Z-R depends on what strikes one as trivial. > > >> (Iow, you didn't mean > >> it's a fact that PA is syntactically consistent, did you?) > > > Consistent IS syntactically consistent. > > But there's also such thing as relative consistency proof! Yes, of course. I don't know why you're excited about that fact though. > For example, > from T = {Ax[xex] /\ ~Ax[xex]}, it's a triviality to prove the consistency > of PA, Sure, as long as there is some sentence in the language of T that we read as "PA is not consistent". Of course, such a proof does not in itself give evidence that there is a PA proof of a formula P&~P. Rather, such a proof gives evidence merely that in T there is a certain derivation of a formula that we are reading as "PA is not consistent". > but should I proclaim that PA is consistent, as in, "that PA is > consistent is a triviality", as Aatu put it? Right, we agree you should not take such a proof as evidentiary in that way. But, just to be clear (since I'm not sure exactly what you're saying) Aatu is not claiming that you should. > The question I had for him was a clarification request to see if he meant > PA is really consistent, Yes, he means that PA is consistent, really consistent. > or if he meant that was just a relative consistency > proof he had referred to. The above you referred to is not a relative consistency. A relative consistency is of the form: T |- G consistent -> G* consistent The proof you mentioned is of the form: T |- G consistent. Anyway, Aatu is not saying just that there exists a relative consistency proof nor just that, say, ZF or some other formal system proves Con(PA), but rather he's saying that PA IS consistent. He's saying that aside from whatever FORMAL proofs, PA is consistent - PERIOD. His basis is for that is not a FORMAL proof, but rather his conviction that the axioms of PA are true (and not even in confined to a FORMAL model theoretic sense of truth, but rather that the axioms are simply true about the natural numbers, as we (editorial 'we') understand the natural numbers even aside from any formalization. Haven't you read Franzen's incompleteness book? > (You should read people's conversation more carefully, before jumping to > conclusion whether or not people understand this or that.) I didn't post anything that shows lack of context of the conversation. > > Here's one among equivalent definitions: > > > DEFINITION OF CONSISTENT: > > > A set of formulas S is in a language is consistent iff there is no > > formula P and the negation of P in S. Typo: delete the first 'is'. > > PERIOD. > > I was about to ignore your incorrectness here, but the tallness of your > ending "PERIOD" seemed defying any, say, "forgiveness". So here it is. Nope, I'm not gonna go down the suckhole with you again. MoeBlee
From: MoeBlee on 1 Jul 2010 12:32 On Jul 1, 11:00 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > how about a high > level summary of the proof and how ZFC axioms are needed to formalize > that? That's been posted already! (No, I'm not gonna get post numbers for you; It's enough that at least a few posters have been providing such summaries all along.) As to the axioms used, the axioms of Z-R suffice. Whether there is a proper subset of those axioms that suffice, I don't opine. MoeBlee
From: R. Srinivasan on 1 Jul 2010 12:49 On Jul 1, 5:01 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > "R. Srinivasan" wrote: > > On a more intuitive level, how can we fault the existence of the > > infinite set N? If you consider the statement "All natural numbers are > > not upper bounds for N" and ask "how many natural numbers are exceeded > > by some element of N?", the answer has to be "infinitely many". > > The answer is finitely many. Let's suppose N starts with 0 (if you take > it as starting with 1 my comment need only a slight adjustment). Let's > take the "some" element to be n. How many natural numbers are exceeded > by n? Answer n. E.g. take n = 4: > > 0, 1, 2, 3, *4*, 5, 6, ... > > and count the numbers exceeded by 4: 0 < 4, 1 < 4, 2 < 4, 3 < 4, and > nothing else is exceeded by 4, so that makes 4. > > Btw, your "All natural numbers are not upper bounds for N" is more > idiomatically expressed as "No natural numbers are upper bounds for N", > which I agree with. > I didn't ask how many natural numbers are exceeded by some specific, fixed n. I am asking you to count the number of natural numbers that are not upper bounds for N. Since all natural numbers fit this description, the answer has to be "infinitely many". The assertion that infinitely many natural numbers have been exceeded within N is paradoxical, as I pointed out. The root cause of the paradox is the existence of infinitely many finite natural numbers. RS
From: Frederick Williams on 1 Jul 2010 12:59
"R. Srinivasan" wrote: > > > I didn't ask how many natural numbers are exceeded by some specific, > fixed n. You asked "how many natural numbers are exceeded by some element of N?" But never mind... > I am asking you to count the number of natural numbers that > are not upper bounds for N. Since all natural numbers fit this > description, the answer has to be "infinitely many". Agreed. > The assertion > that infinitely many natural numbers have been exceeded within N is There is nothing in N that exceeds infinitely many natural numbers. > paradoxical, as I pointed out. The root cause of the paradox is the > existence of infinitely many finite natural numbers. > > RS -- I can't go on, I'll go on. |