Path: csiph.com!eternal-september.org!feeder.eternal-september.org!reader01.eternal-september.org!.POSTED!not-for-mail From: Keith Thompson Newsgroups: comp.theory,comp.ai.philosophy,comp.ai.nat-lang,sci.lang.semantics Subject: Re: Simply defining =?utf-8?Q?G=C3=B6del?= Incompleteness and Tarski Undefinability away V24 (Are we there yet?) Date: Mon, 13 Jul 2020 16:10:24 -0700 Organization: None to speak of Lines: 63 Message-ID: <875zar3tb3.fsf@nosuchdomain.example.com> References:

<2tCdnb0urbddzpfCnZ2dnUU7-b_NnZ2d@giganews.com>

<87k0z85tt0.fsf@nosuchdomain.example.com> <874kqb6e3j.fsf@nosuchdomain.example.com> <87eepf3yr5.fsf@nosuchdomain.example.com> <878sfnnj9s.fsf@bsb.me.uk> Mime-Version: 1.0 Content-Type: text/plain Injection-Info: reader02.eternal-september.org; posting-host="4502a91299fcc6f473e7c6cd344f8c9a"; logging-data="14460"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1+wi1VDXhaTsegSp2XmfOzf" User-Agent: Gnus/5.13 (Gnus v5.13) Emacs/26.3 (gnu/linux) Cancel-Lock: sha1:YmbYyRcZeZwOH0A4+XVg49wkH00= sha1:5u4s0b8rDzSLu2hWr/pYSKiqSUs= Xref: csiph.com comp.theory:21631 comp.ai.philosophy:21961 comp.ai.nat-lang:2370 Ben Bacarisse writes: > Jeff Barnett writes: >> On 7/13/2020 3:12 PM, Keith Thompson wrote: >>> Jeff Barnett writes: > >>>> Recall, we can't prove commutativity in R|Q period. However, our meta >>>> logic can do the proof because it provides additional >>>> mechanisms. Let's look at a Peano-like induction rule where I is the >>>> integers and S is the successor function: If Z subset-of I, 0 in Z, >>>> and whenever x in Z then Sx in Z; then Z = I. How would an >>>> anti-induction rule be expressed? What would it mean? And would it >>>> lead to inconsistency? In other words would it falsify even one >>>> theorem provable in R|Q by the remaining axioms? >>> >>> The answer to most of that is "I don't know", but I can sketch one >>> possible partial answer. >>> >>> https://en.wikipedia.org/wiki/Robinson_arithmetic#Axioms >>> >>> Consider a set consisting of blue natural numbers (integers >= 0, >>> colored blue) and red integers (all integers, negative, 0, and positive, >>> colored red). The successor of any blue number is a blue number, and >>> the successor of any red number is a red number. (red 0 is zero in the >>> model; blue 0 is not. Blue 0 is the successor of blue -1.) CORRECTION: I should have written: ... (blue 0 is zero in the model; red 0 is not. Red 0 is the successor of red -1.) >> If I read this correctly: 1) All non-negative integers are blue. 2) >> All integers less than, equal to, or greater than zero are read => all >> integers are read. 3) Therefore, every blue integer is also red. I >> don't believe that is what you mean so I ask for a clarification >> before continuing to what follows. I understand that I might be >> misreading the above paragraph but I keep getting what I just said in >> my paraphrase. > > The model is ({blue} x N) U ({red} x Z). I.e. the elements are pairs: > > { (blue, 0), (blue, 1), (blue, 2), ... > ... (red, -2), (red, -1), (red, 0), (red, 1), (red, 2), ... } > > The successor is defined as > > S(blue, n) = (blue, n+1) > S(red, n) = (red, n+1) Yes. > Q's zero is (red, 0). No, Q's zero is (blue, 0) (which has no predecessor). I misstated that above. > I've not been following the sub-thread so I won't say if this model > shows what is claimed -- indeed I'm not yet sure what is claimed -- but > since I /was/ pretty sure I know what Keith was suggesting, I thought I'd > say something. -- Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com Working, but not speaking, for Philips Healthcare void Void(void) { Void(); } /* The recursive call of the void */