How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: MoeBlee on 29 Jun 2010 17:07 On Jun 29, 11:30 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jun 29, 12:13 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > On Jun 29, 10:25 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > On Jun 29, 10:55 am, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > On Jun 28, 9:07 pm, Charlie-Boo <shymath...(a)gmail.com> wrote: > > > > > > On Jun 28, 12:24 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > > And it's easy enough to see that if a theory has a model then that > > > > > > theory is consistent. > > > > > > Is that an axiomatic proof in ZFC? > > > > > Plain Z-regularity proves that if a theory has a model then the theory > > > > is consistent. It's quite simple; you would come up with it yourself > > > > on just a moment's reflection. > > > > Sorry, but I don't know what proof you have in mind, so I can't > > > determine how the Axiom of Regularity would play a role. > > > It DOESN'T play a role, which is why I took it out. > > I asked what axiom is essential and would be needed to carry out the > proof in PA. I thought you were answering that. So what is the > answer? > > > In other words, we > > can prove in Z set theory even without the axiom of regularity > > (possibly without certain other axioms? but I've never done such > > detailed bookkeeping, as my only claim is that Z-R is SUFFICIENT). THAT is my answer. Damn, please READ my post if you're going to ask question about it. > > As to the proof. Would you just TRY to do it in your mind one time? > > I did. You can't just say the Axiom of Infinity provides a model as > you have to also prove that implies consistency. You claim to know > how, so be a mathematician and substantiate your claim. Of course, that is not the proof. Damn, I'm really using up my time with you. Look, in Z set theory we have: w (omega), 0, the successor function on w, the addition function for w, and the multiplication function for w. Now, take each axiom of PA and verify it is a theorem of Z as translated per the above and for all members of w. For example: 'Sn' in PA translates to 'nu{n}' in Z. PA axiom: Anm(Sn = Sm -> n=m) Then just verify that this is a theorem of Z: Anm((n in w & m in w & nu{n} = mu{m}) -> n=m). Then go on down the line verifying each axiom in PA as translated into Z. The induction schema for PA will be a bit of paperwork, but nothing conceptually too difficult. Then the model for PA will be: <w 0 S' +' *'> where S', +' and *' are the set theoretic operations on w corresponding to the operation sybmols for PA. (As, for example, I mentioned nu{n} corresponds to Sn). If you want to know more, then please just read up about the construction of the system of naturals in Z and about models in mathematical logic. I need to not type all day telling you things you can read for yourself. > > If > > you still can't see it, then, if I'm feeling generous, I'll outline it > > for you. As to showing an exact sequence of primitive formulas of the > > language of Z, no, that's just a chore. > > I don't know what you're referring to. I did ask for the statement of > the theorem in ZFC, but nobody has come up with that either. > > So in summary, > > 1. ZFC can prove PA consistent - it's easy and lots of people have > done it. > 2. Nobody can give a reference to its being done. It's done easily as an exercise, just as I gave you a start abovel. > 3. Nobody can describe the proof that has been done in ZFC. NO, I have DESCRIBED a proof. > 4. Nobody can give even the ZFC expression for the theorem itself. NO. If you paid me enough money, I'd do the labor of translating the informal proof to a perfectly formal one. > > > What is the proof and how is Regularity essential? > > > You're mixed up. Regularity is NOT needed. That's why I put Z- > > regularity, which means "Z without the axiom of regularity". Why don't you have the courtesy even to recognize that I've answered and corrected you? Even if not a matter of courtesy, but at least of communication so that I would know that you do recognize that a needed correction has been made to your misunderstanding. MoeBlee
From: MoeBlee on 29 Jun 2010 17:12 On Jun 29, 11:49 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > Didn't you just ask for a reference yourself a minute ago? So what? Aatu mentioned some material I'm not familiar with. I asked where I could read more about it. Meanwhile, you've been told that if you read a textbook in logic and one in set theory (even just the relevant portions, actually) then you'd see how to show that Z proves the consistency of PA as even just an added easy exercise, whether such books even make a point of the matter. And today I also gave you a basic outline and started you off with one of the "subsections" to complete. You are just a time/energy suck. Men of better discipline than I are lucky not to waste their time/energy on your juvenile silliness. MoeBlee
From: MoeBlee on 29 Jun 2010 17:14 On Jun 29, 4:05 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > What do you mean "obviously"? There are very few set theorists to > believe it the case. Sorry, delete that. I cut your sentence off and unintentionally misconstrued you. MoeBlee
From: Transfer Principle on 29 Jun 2010 17:23 On Jun 29, 9:30 am, Charlie-Boo <shymath...(a)gmail.com> wrote: > On Jun 29, 12:13 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > > If you still can't see it, then, if I'm feeling generous, I'll outline it > > for you. As to showing an exact sequence of primitive formulas of the > > language of Z, no, that's just a chore. > I don't know what you're referring to. I did ask for the statement of > the theorem in ZFC, but nobody has come up with that either. > So in summary, > 1. ZFC can prove PA consistent - it's easy and lots of people have > done it. > 2. Nobody can give a reference to its being done. > 3. Nobody can describe the proof that has been done in ZFC. > 4. Nobody can give even the ZFC expression for the theorem itself. > In other words, business as usual. I'm not skeptical (in the way that Charlie-Boo is skeptical) about the provability of Con(PA) in either ZFC or PRA+epsilon_0. But I am suspicious of the fact that the induction needs only to be taken up to epsilon_0, which the the smallest ordinal not reachable from omega via finitely many additions, multiplications, and exponentiations, but can be reached via finitely many _tetrations_ since epsilon_0 = omega^^omega. This is why I so often mention Ed Nelson and his proof attempt of ~Con(PA) involving tetration.
From: MoeBlee on 29 Jun 2010 17:28
On Jun 29, 4:23 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > I'm not skeptical (in the way that Charlie-Boo is skeptical) about > the provability of Con(PA) in either ZFC or PRA+epsilon_0. But I > am suspicious of the fact that the induction needs only to be > taken up to epsilon_0, As to ZFC, you don't need such fancy stuff as epsilon_0. Just do the routine proof that with the system of omega with 0, successor, addition, and multiplication we get a model of all the PA axioms. MoeBlee |