How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Charlie-Boo on 25 Jun 2010 00:21 On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy to > > prove that PA is consistent (its axioms and rules preserve truth) yet > > (by Godel-2) PA can't do such a simple proof as that. > > So what? ZFC can prove it. > ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove > that PA is consistent. This is why you can't say that "ZFC/PA > doesn't prove PA is consistent." "ZFC/PA" is just a meaningless > locution in any case. > ZFC is one thing. PA is another. PA is a subset of ZFC, so I emphasize that by calling it ZFC/PA (it makes more sense to distinguish the two anyway.) This is besides the point. Who has proved PA consistent using ZFC? If it were possible then I assume someone would have done it. It certainly would be a very educational exercise. In any case, it shows the weakness of PA. I added ZFC as that is so popular. C-B
From: Chris Menzel on 25 Jun 2010 14:16 On Thu, 24 Jun 2010 21:21:05 -0700 (PDT), Charlie-Boo <shymathguy(a)gmail.com> said: > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: >> On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: >> >> > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy >> > to prove that PA is consistent (its axioms and rules preserve >> > truth) yet (by Godel-2) PA can't do such a simple proof as that. >> >> So what? ZFC can prove it. >> ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove >> that PA is consistent. This is why you can't say that "ZFC/PA >> doesn't prove PA is consistent." "ZFC/PA" is just a meaningless >> locution in any case. >> ZFC is one thing. PA is another. > > PA is a subset of ZFC, No it isn't. (Curious that you continue to assert otherwise.) > so I emphasize that by calling it ZFC/PA (it makes more sense to > distinguish the two anyway.) This is besides the point. Who has > proved PA consistent using ZFC? It is an elementary exercise. A proof can be found in almost any reasonably thorough text in set theory.
From: Frederick Williams on 25 Jun 2010 16:58 Charlie-Boo wrote: > > PA is a subset of ZFC, No it isn't, but even if it were... > so I emphasize that by calling it .... there would still be two things: something _and_ a subset of it which would deserve two names. > ZFC/PA (it > makes more sense to distinguish the two anyway. Well, yes, since they're different. > ) This is besides the > point. Who has proved PA consistent using ZFC? To "construct" a model of PA in ZFC is rather easy. Also, see Gentzen and Ackermann. Gentzen's proof used far less than full ZFC. > If it were possible > then I assume someone would have done it. You assume correctly. > It certainly would be a > very educational exercise. > > In any case, it shows the weakness of PA. PRA + induction up to epsilon_0 (which is what Gentzen used) is incomparable with PA. > I added ZFC as that is so > popular. So is watching telly (among a certain class of people at least). You may wish to know that ZFC with the axiom of infinity replaced by its negation is a model of PA and vice versa. -- I can't go on, I'll go on.
From: George Greene on 25 Jun 2010 17:14 On Jun 25, 12:21 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > PA is a subset of ZFC, NO, IT ISN'T. I am not going to belabor this further since I ALREADY EXPLAINED why. Typical that you did not reply at all to the explanation.
From: George Greene on 25 Jun 2010 17:29
> >> On Jun 14, 11:42 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > >> > ZFC/PA is supposed to do all ordinary mathematics*. But it is easy > >> > to prove that PA is consistent (its axioms and rules preserve > >> > truth) yet (by Godel-2) PA can't do such a simple proof as that. > > On Jun 24, 6:04 pm, George Greene <gree...(a)email.unc.edu> wrote: > >> So what? ZFC can prove it. > >> ZFC can't prove that ZFC is consistent, but it CAN AND DOES prove > >> that PA is consistent. This is why you can't say that "ZFC/PA > >> doesn't prove PA is consistent." "ZFC/PA" is just a meaningless > >> locution in any case. > >> ZFC is one thing. PA is another. > > > PA is a subset of ZFC, On Jun 25, 2:16 pm, Chris Menzel <cmen...(a)remove-this.tamu.edu> wrote: > No it isn't. Thank you kindly for taking the simple side, for once. I was living in fear of again being attacked by irrelevant complexity from superior education, on the grounds that one can prove "a certain sort of equivalence" between PA and ZF withOUT the axiom of infinity. As I said when I first rebutted C-B, > A stronger theory can often prove the consistency of a weaker one. > ZFC *can* (and does) prove that PA is consistent: the set required > to exist by ZFC's Axiom of Infinity is (the domain of) a model for PA. In addition to being able to use the von Neumann encoding to interpret PA in ZF, one can also interpret natnums as finite sets in order to encode a theory of finite sets in PA (infinite sets would get encoded as non-standard hyper-finite numbers, in that case). But, of course, just because something CAN be defined under something else Does Not Make the something- defined into a "subset" of the larger something-else under which it WAS defined. The newer smaller thing would've needed to be present in the larger one FROM THE BEGINNING in order for it to be properly deemed a "subset" -- BUT IT WASN'T -- all the WORK OF DEFINITION AND ENCODING had to be ADDED to the original environment to produce the smaller thing within it. In any case, if you take this mutual encodability as connoting "some sort of" equivalence, then PA "is like" ZF without the axiom of infinity, so to the extent that the set of axioms (and theorems) without that one axiom is a subset of the one with, There Could Sort Of Be A Point to what C-B was trying to say. But it is not one that he would understand and certainly not one that would justify anyone's talking about "ZFC/PA". |