How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF)
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From: Aatu Koskensilta on 30 Jun 2010 14:27 "R. Srinivasan" <sradhakr(a)in.ibm.com> writes: > If I had wasted my time trying to dig into the rubbish that you have > laid out above, I would not have had much time or energy left to deal > with the kind of stuff that *i* consider worth doing. You're of course free to spend your time and energy however you choose. But why do you think others should take any notice of your interests and inclinations? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 30 Jun 2010 20:32 MoeBlee <jazzmobe(a)hotmail.com> writes: > Okay, I'm hanging in here. Through coding, we can "talk about" PA > formulas, thus, "talk about" the sets those formulas define, right? Sort of. > (I'm lost about the restricted quantifier complexity part, but > nevermind for now, I'm just trying to get the big picture at this > point). The restriction on quantifier complexity is necessary because truth of sentences of unrestricted quantifier complexity is not arithmetical. In contrast, for any n there is a Sigma-n formula True-n(x) such that o Every instance of the T-schema restricted to Sigma-n sentences provably holds in PA for True-n, i.e. for every Sigma-n sentence P it's provable in PA that P iff True-n("P") o The inductive Tarskian clauses hold for each n, i.e. it's provable in PA that P & Q is True-n iff P is True-n and Q is True-n; (x)P(x) is True-n iff P(a) is True-n for all a; and so on. To translate a statement containing quantifiers over Sigma-n sets ("for all Sigma-n sets X", "there is a Sigma-n set X") into the language of arithmetic we simply replace "for all Sigma-n sets X" with "for all Sigma-n formulas with one free variable x"; "there is a Sigma-n set X" with "there is a Sigma-n formula with one free variable x"; and "x in X" with "substituting the numeral for x in X yields a True-n sentence". (All this holds when we replace "Sigma-n" with "Pi-n.") Now, here's an exercise for you: show that adding to PA a predicate T and all instances of the T-schema for T, and extending the induction schema to cover formulas containing T, yields a conservative extension of PA. (Hint: use compactness, and the observation that for any finite number of instances of the T-schema we can find a restricted truth predicate that satisfies them -- one of the restricted truth predicates from above.) On the other hand, if we also add the usual Tarskian clauses (and extend induction to cover formulas containing the truth predicate) we get a theory equivalent to the subsystem ACA of second-order arithmetic -- essentially by using the truth predicate to translate statements that quantify over sets into quantification over arithmetical formulas, as we did above with the restricted truth predicates. I believe there's a bit about this in Torkel's _Inexhaustibility_. > I know what a Sigma-n set of sentences is. But what's a Delta-n+1 > model? It is a model the parts of which -- domain, relations, functions, ... -- are Delta-n+1 sets, that is, both Pi-n+1 and Sigma-n+1. > A followup question: Is there ANYTHING in mathematical logic you don't > know about? Lots. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 1 Jul 2010 00:17 Charlie-Boo <shymathguy(a)gmail.com> writes: > If so, what is the formal expression in ZFC that PA is consistent? On any scheme for expressing such statements in the unextended language of set theory it's a virtually incomprehensible string of symbols. If you really want to have a look, go through a set theory text and write a computer program to output it for you. Why would you want to see it in the first place? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 1 Jul 2010 02:04 Charlie-Boo <shymathguy(a)gmail.com> writes: > To see it means to have actually created it, and its actual creation > would answer the very interesting question of whether ZFC can prove > that PA is consistent even though PA can't. We already know the answer. The axioms used in the proof are a restricted form of comprehension, infinity, union and pairing. Spelling out the formalization of "PA is consistent" in the language of set theory would be a tedious, trivial and pointless undertaking, of no apparent mathematical interest whatsoever. > It would substantiate your assertions. Is that of value? Well, no. I can't think of any pressing reason I should want to convince you of anything. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 1 Jul 2010 02:45
Ki Song <kiwisquash(a)gmail.com> writes: > Perhaps an analogy is in order! > > I feel like what Charlie-Boo is asking people to do is analogous to > asking someone to perform the addition: > > Sqrt{2}+Sqrt{3}, with a decimal precision of 10^(10^(10^100)) place. The analogy is apt, but it's not really quite that bad. Based on nothing at all I'd estimate that a formalization of "PA is consistent", fully spelled-out in the language of set theory, fits in less than ten pages. But perhaps you were thinking of Bourbaki, whose term for the number 1 already consists of 4,523,659,424,929 symbols and 1,179,618,517,981 disambiguatory links? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |