How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Frederick Williams on 30 Jun 2010 11:22 MoeBlee wrote: > Great, thanks, I think that NDJFL is free online. I'll look. I don't think so, but you'll find it at Kaye's site. -- I can't go on, I'll go on.
From: MoeBlee on 30 Jun 2010 11:28 On Jun 30, 9:42 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > Hmmm. I am not very conversant with classical model theory. So > > according to you, there is a "standard model for the LANGUAGE of PA" > > even if the theory PA is inconsistent. > > No just according to me. Oopsie doopsie. I meant: NoT just according to me. As I said, if Z is inconsistent, then Z proves every formula in the language. So, if we find something to be a theorem of Z (such as the existence and uniqueness theorems that provide for a definition of the constant nicknamed "the standard model for the language of PA"), then if Z is inconsistent, PERFORCE Z proves those theorems. MoeBlee
From: Frederick Williams on 30 Jun 2010 11:43 "R. Srinivasan" wrote: > Hmmm. I am not very conversant with classical model theory. So > according to you, there is a "standard model for the LANGUAGE of PA" > even if the theory PA is inconsistent. May I infer that you have used > infinite sets to define this model? No sets, and a fortiori no infinite sets, are required (not for first-order PA anyway): the individuals are 0, 1, 2, 3, ...; the constant 0 is 0; successor is x |-> x + 1; sum is (x, y) |-> x + y; and product is (x, y) |-> x * y. Mathematicians knew all about those before model theory was invented and before Peano was an eye in his father's twinkle. > How can you do that if the theory > PA is inconsistent (which would make ZFC inconsistent as well)? The above is a model of PA whether or not it and ZFC are consistent. -- I can't go on, I'll go on.
From: MoeBlee on 30 Jun 2010 11:52 On Jun 30, 10:43 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > "R. Srinivasan" wrote: > > Hmmm. I am not very conversant with classical model theory. So > > according to you, there is a "standard model for the LANGUAGE of PA" > > even if the theory PA is inconsistent. May I infer that you have used > > infinite sets to define this model? > > No sets, and a fortiori no infinite sets, are required (not for > first-order PA anyway): Perhaps in an informal sense of 'model'. My remarks are in the formal sense of a model, in which there is a set that is the universe for the model. > The above is a model of PA whether or not it and ZFC are consistent. Just to be clear, I was speaking of a model ('structure', if you prefer) FOR the LANGUAGE of PA, whether or not it is a model OF the THEORY PA. That there are models ('structures" if you prefer) for the language of a theory does not entail whether or not the theory is consistent. MoeBlee
From: Frederick Williams on 30 Jun 2010 11:57
MoeBlee wrote: > > On Jun 30, 10:43 am, Frederick Williams > <frederick.willia...(a)tesco.net> wrote: > > "R. Srinivasan" wrote: > > > Hmmm. I am not very conversant with classical model theory. So > > > according to you, there is a "standard model for the LANGUAGE of PA" > > > even if the theory PA is inconsistent. May I infer that you have used > > > infinite sets to define this model? > > > > No sets, and a fortiori no infinite sets, are required (not for > > first-order PA anyway): > > Perhaps in an informal sense of 'model'. My remarks are in the formal > sense of a model, in which there is a set that is the universe for the > model. Indeed so, I was thinking of what is _in_ the model modelling the language elements of PA: variables, constant and functions. Sorry if I misunderstood. And sorry to R. Srinivasan if I have added to his confusion. -- I can't go on, I'll go on. |