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 1 Jul 2010 19:23 On Jul 1, 3:50 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > Charlie-Boo wrote: > > No, PA and ZFC both have Peano's Axioms. > > What formulations of first order PA and ZFC have _any_ of their > non-logical axioms in common? None. Why? Well, for starters, all of > the proper axioms of ZFC have the predicate symbol $\in$ in them, none > of PA's axioms do because $\in$ is not even in the language of PA > > > The only difference is the > > universal set, which is N in PA > > There are no sets, universal or otherwise, in first order PA. What does (existsX)P(X) mean? C-B > > and sets in ZFC. That's why it is > > expressed differently - different alphabets as well. > > -- > I can't go on, I'll go on.
From: Charlie-Boo on 1 Jul 2010 19:26 On Jul 1, 4:09 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jul 1, 2:30 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > The point is that ZFC would have to have an axiom other than Infinity > > (i.e. PA) that is necessary to prove PA consistent in order for it to > > be impossible in PA and possible in ZFC. But again you don't show > > that. > > Talking to myself: PLEASE MoeBlee don't waste a second more of your > time trying to get through to the ineducable Charlie-Boo. Listen, > MoeBlee, no matter how much you explain, no matter how detailed or > general, Just the title of a reference with a proof of the consistency of PA carried out in ZFC would do. Or was this one of those cases where nobody really does that because it's too messy, or because it's so easy, one of those "Goldilocks" problems? C-B > Charlie-Boo will still persist with yet more of his > confusions over even the most simple matters such as the difference > between Z set theory and first order PA. > > MoeBee
From: MoeBlee on 1 Jul 2010 19:55 On Jul 1, 6:26 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > Or was this one of those cases where nobody really does that because > it's too messy, or because it's so easy, one of those "Goldilocks" > problems? Gotta go. Life too short for fruitlessly trying to communicate with you. I figured this stuff out pretty much by reading the books and going a step further to formalize every detail I needed so that I could see that complete formalization is possible and to see what it would be, so you should be able to do it to. I'd recommend starting at square one, with what I think is the best book on learning to work in the first order calculus: "Logic: Techniques Of Formal Reasoning" by Kalish, Monatague, and Mar. My very best practical answer to address the root of your confusions is for you to complete that book so that you know it throroughly. THEN we might be able to talk about even more technical matters. MoeBlee
From: herbzet on 1 Jul 2010 20:52 MoeBlee wrote: > > CORRECTION below: > > On Jul 1, 5:10 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > However, even though the definition I gave is okay, and provides these > > theorems, > > > > a theory T is inconsistent <-> T proves a contradiction > > > > a set of axioms X for a theory T is inconsistent <-> X proves a > > contradiction > > Oops, delete that. Not sure if "that" is the previous sentence, or the previous 2 sentences. > > a theory T that is axiomatized by X is inconsistent <-> X proves a > > contradiction, > > > > I realize that in my own notes I actually do use a different > > definition, which, again, is equivalent for THEORIES but different for > > arbitrary sets of formulas: Missing the nuance here, unless you are distinguishing between a set Gamma of formulas *implying* both P and ~P, and a set Gamma of formulas *containing* both P and ~P. But let that go ... > > a set of formulas G is consistent <-> there is no formula P such that > > P and ~P are provable from G OK, so far so good. > > a set of formulas G is inconsistent <-> G is not consistent Can't object to that. > and I think that is an easier and nicer definition to work with as it > gives all at once: > > a set of formulas G is inconsistent <-> G proves a contradiction > > a set of formulas G is inconsistent <-> G proves every formula (in the > language) No doubt missing some of the issues here, but: For classical logic the following sets are extensionally equivalent: 1) a set of formulas G that proves both P and ~P for some formula P. 2) a set of formulas G that proves a contradiction, i.e., some formula (P & ~P). 3) a set of formulas G that proves every formula (in the language). For non-classical logics, these may diverge. What (3) describes is called a trivial theory. We don't necessarily care so much about inconsistency -- a paraconsistent logic will tolerate some inconsistency of type (1) or (2) in a theory T without T's necessarily becoming trivial. We don't worry so much about inconsistency per se -- we worry about triviality instead. Tra-la-la! -- hz
From: herbzet on 1 Jul 2010 20:55
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. -- hz |