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: Fri, 10 Jul 2020 12:53:44 -0700 Organization: None to speak of Lines: 64 Message-ID: <87v9iv6t9z.fsf@nosuchdomain.example.com> References: <87k0zc8ps5.fsf@nosuchdomain.example.com> Mime-Version: 1.0 Content-Type: text/plain; charset=utf-8 Content-Transfer-Encoding: 8bit Injection-Info: reader02.eternal-september.org; posting-host="06cd1b9edd2b2564eb5e3b0578cd4d7b"; logging-data="14486"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX19VkYk9YLGuxYLLpyqMqvle" User-Agent: Gnus/5.13 (Gnus v5.13) Emacs/26.3 (gnu/linux) Cancel-Lock: sha1:lihRsWnSHrg772PQrNz9ADCDvFg= sha1:B3wP9b1etHWl3LtIdGRh2P93Ojk= Xref: csiph.com comp.theory:21558 comp.ai.philosophy:21874 comp.ai.nat-lang:2310 olcott writes: > On 7/9/2020 2:14 PM, Keith Thompson wrote: >> olcott writes: >>> On 7/9/2020 8:40 AM, André G. Isaak wrote: >> [...] >>>> I've asked you repeatedly about Robinson's arithmetic, in which x + >>>> y = y + x is not provable. Neither is ¬(x + y = y + x) provable. The >>>> law of the excluded middle demands that one of those be true, so >>>> there exists a true statement in Q which is not provable in Q. >>>> >>>> And one can prove that x + y = y + x is true in Q. You just can't >>>> prove it from within Q. >>> >>> That is the exactly same key mistake that you, Tarski and presumably >>> Gödel made. How do we know that it is true IN Q when it is not >>> provable IN Q (We look outside of Q). THEN IT IS NOT TRUE IN Q, IT IS >>> ONLY TRUE OUTSIDE OF Q. >> >> If it is not true in Q, then there are values x and y in Q such that >> x + y = y + x is false in Q. >> >> In fact there are no such values. (You could refute that if you could >> provide such values.) >> >> I'm assuming that "x + y = y + x is true in Q" and "x + y = y + x is >> false in Q" are the only possibilities (law of the excluded middle). >> Do you accept that assumption? > > This is my current best guess of the correct use of the term > satisfiable if the term satisfiable can even be applied to a single > theory: > > ∃φ (Q ⊢ "x + y = y + x") would seem to be unsatisfiable in Q. > ∃φ ¬(Q ⊢ "x + y = y + x") would also seem to be unsatisfiable in Q. > > This would seem to indicate that Q is incomplete relative to commutativity. > > I am certain that the ideas are correct. I am uncertain if my use of > the term unsatisfiable corresponds to its conventional use. > > I am certain that my use of the term incomplete correctly augments the > conventional use of the term such that my use is more correct than the > conventional use. And this is an example of why trying to have a conversation with you is so frustrating. I asked what I thought was a straightforward yes or no question, "Do you accept that assumption?". Your response did not include the word "yes" or "no", nor did it attempt to demonstrate that neither "yes" nor "no" would be a meaningful answer. Instead you wrote several paragraphs about the meaning of "satisfiable". By all means, write all you like about the meaning of "satisifiable", but please don't do so in a context that makes it look like you're trying to answer my question. Perhaps what you wrote has some relevance to what I asked, but I don't see it. You have not answered my question. "Yes" or "No" would be an answer. -- 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 */