How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
in [Logic]
|
Prev: equivalence
Next: How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
From: George Greene on 24 Jun 2010 18:02 On Jun 24, 1:45 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > ZFC includes PA. I did NOT ASK for your ignorant OPINION, DUMBASS. I HAVE TWO RELEVANT DEGREES. You don't. I TOLD you that that ZFC does not "include" PA, SO IT DOESN'T. More to the point, you were complaining that it couldn't prove PA consistent. Well, IT CAN, but that is NOT because it INCLUDES PA; rather, it is because it ENCODES PA, or can. It ALLOWS things THAT SATISFY THE AXIOMS of PA to be constructed by interpretation. PA has a ZERO. ZFC *DOES*NOT*, so it does NOT INCLUDE PA. ZFC does, however, have AN EMPTY SET, so you may, if you like, ENCODE the zero from PA as the empty set from ZFC. Or as any other set you like, for that matter. There is, however, a "natural" or "von Neumann" encoding in which the successor of a set s (which we might write S(s), and which would be defined for ANY set s, not JUST a PA natural number n) is s U {S} , and the natural numbers are the successors of the empty set, with zero encoded as the empty set. You can then define addition via the cardinality of disjoint union and multiplication via the cardinality of cross-product, with cardinality being defined bijectively. But all that IS A WHOLE LOT OF WORK, and was definitely NOT just INCLUDED in ZFC to begin with!
From: George Greene on 24 Jun 2010 18:04 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.
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. |