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 20:11:31 -0700 Organization: None to speak of Lines: 99 Message-ID: <87lfjm3i58.fsf@nosuchdomain.example.com> References: <2tCdnb0urbddzpfCnZ2dnUU7-b_NnZ2d@giganews.com> <87k0z85tt0.fsf@nosuchdomain.example.com> <87d0505kmk.fsf@nosuchdomain.example.com> <5Lmdnehh4P6hLZbCnZ2dnUU7-LdQAAAA@giganews.com> <87365vnik3.fsf@bsb.me.uk> <87wo363ol5.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="4502a91299fcc6f473e7c6cd344f8c9a"; logging-data="11398"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1/UvRAbmYpu84jfTaoVy+jy" User-Agent: Gnus/5.13 (Gnus v5.13) Emacs/26.3 (gnu/linux) Cancel-Lock: sha1:AsP4Tv+ncN1NcQ0UUT1L4ErhQ1E= sha1:Scvk+iDq03PztJWz7GKfw2Hmb8I= Xref: csiph.com comp.theory:21642 comp.ai.philosophy:21974 comp.ai.nat-lang:2380 olcott writes: > On 7/13/2020 7:52 PM, Keith Thompson wrote: >> olcott writes: >>> On 7/13/2020 5:42 PM, Ben Bacarisse wrote: >>>> olcott writes: >>>>> φ = ∀x ∈ ℕ ∀y ∈ ℕ (x + y = y + x) >>>>> >>>>> φ is not true or false in Q because Q lacks a mapping in Q from φ to a >>>>> Boolean value. >>>> >>>> Would you like to learn why that's wrong, or would you rather just keep >>>> repeating it? >>>> >>>> If you'd like to learn, you have to be a student. I'd ask a student to >>>> consider an instance of x + y = y + x in Q, for example this one: >>>> >>>> S0 + SS0 = SS0 + S0. >>>> >>>> and I'd ask them: what can you say about this formula in Q? >>> >>> How do you get from point "A" to point "B" when no path from point "A" >>> to point "B" exists? YOU DON'T !!! >>> >>> How do you show that an expression of language is true when there is >>> no mapping from this expression to Boolean values? YOU DON'T !!! >>> >>> How do you show that an expression of language of a formal system is >>> true in this formal system when there is no mapping in this formal >>> system from this expression to Boolean values? YOU DON'T !!! >> >> Please stop shouting. >> >> The post to which you're responding doesn't ask what you can say about >> x ∈ ℕ ∀y ∈ ℕ (x + y = y + x) > > Ben tried to change the subject and I did not allow it. Ben did not change the subject. Ben asked you a question. The question happens to be about a specific instance (S0 + SS0 = SS0 + S0) of a more general statement (x ∈ ℕ ∀y ∈ ℕ (x + y = y + x)). I believe that addressing Ben's question would bring us one step closer to an understanding of what we've been talking about. You often do something similar yourself, by introducing specific things like dogs and ice cream into a discussion of something more general. >> in Q. It asks what you can say about >> S0 + SS0 = SS0 + S0 >> in Q (or, using more standard notation, 1 + 2 = 2 + 1). > > The whole point of thread since V23 was prove that true and unprovable > is impossible. Changing this sentence: ∀x ∈ ℕ ∀y ∈ ℕ (x + y = y + x) > to this sentence: S0 + SS0 = SS0 + S0 is flat out cheating. No, it's not cheating. > Yes we can prove that each concrete instance of (x + y = y + x) > is true in Q one at a time. We cannot prove that this sentence: > ∀x ∈ ℕ ∀y ∈ ℕ (x + y = y + x) is true in Q. So you do acknowledge that S0 + SS0 = SS0 + S0 is true and provable in Q. Why didn't you just say so? If you had, you could have seen what Ben has to say next that much sooner. >> What you can say about S0 + SS0 = SS0 + S0 in Q, if I'm not mistaken, >> is that it's both true and provable. It doesn't even touch on the >> points of disagreement that have been consuming this newsgroup. I'm >> sure that Ben has more things to say once that's established, things >> that you'll undoubtedly disagree with, but let's just start with >> S0 + SS0 = SS0 + S0, shall we? Without the shouting. > > Capitilization is the way that I put words in bold when people freak > out from seeing things in HTML. And how's that working out for you? > They say that their reader is set to automatically ignored HTML and > refuse to look at it in Google groups. And they're right. You shouldn't post HTML because Usenet is a plain text medium. You shouldn't post in all-caps because it denotes shouting, which does not advance the discussion. > When I say the same trivially simple and obvious point over and over > and over and no one gets it I escalate the attention to this point. And it doesn't help. Nobody is ignoring your points because you're not shouting enough. People disagree with you. We think you're wrong. If I told you the moon is made of green cheese, you'd think I was wrong. If I told you THE MOON IS MADE OF GREEN CHEESE, THE MOON IS MADE OF GREEN CHEESE, THE MOON IS MADE OF GREEN CHEESE!!! you'd think I was wrong and annoying, and you wouldn't waste your time telling me what it's really made of and how we know. Please don't shout. -- 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 */