Path: csiph.com!xmission!news.snarked.org!border2.nntp.dca1.giganews.com!nntp.giganews.com!buffer2.nntp.dca1.giganews.com!buffer1.nntp.dca1.giganews.com!news.giganews.com.POSTED!not-for-mail NNTP-Posting-Date: Sun, 26 Jul 2020 13:00:44 -0500 Subject: =?UTF-8?Q?Re=3a_Simply_defining_G=c3=b6del_Incompleteness_and_Tarsk?= =?UTF-8?Q?i_Undefinability_away_V33_=28Mendelson_Satisfiability=29?= Newsgroups: comp.theory References: <871rl8dyg1.fsf@bsb.me.uk> <87lfjfovhm.fsf@bsb.me.uk> <87zh7tok63.fsf@bsb.me.uk> <6MednYs8F9v7qYvCnZ2dnUU7-R3NnZ2d@giganews.com> <87lfjcmg9p.fsf@bsb.me.uk> <87tuxzkswv.fsf@bsb.me.uk> <87zh7ojzp8.fsf@bsb.me.uk> <877dusjsda.fsf@bsb.me.uk> From: olcott Date: Sun, 26 Jul 2020 13:00:44 -0500 User-Agent: Mozilla/5.0 (Windows NT 10.0; WOW64; rv:68.0) Gecko/20100101 Thunderbird/68.10.0 MIME-Version: 1.0 In-Reply-To: <877dusjsda.fsf@bsb.me.uk> Content-Type: text/plain; charset=utf-8; format=flowed Content-Language: en-US Content-Transfer-Encoding: 8bit Message-ID: Lines: 43 X-Usenet-Provider: http://www.giganews.com X-Trace: sv3-YfsfhZy3XJED1gXvCbZESZ1S3eK4boZrhqSNhy7VPSa0mM3RNyXN+zozYau5r088pf+VoYdIRC5cJej!RlJjfXq7YjaZE9A+TYnI12e4/n4NkPX//kS/wPLvJoOnUdCpAFBWW6+5rPBiXUHvO9fRDPUqYjA= X-Complaints-To: abuse@giganews.com X-DMCA-Notifications: http://www.giganews.com/info/dmca.html X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly X-Postfilter: 1.3.40 X-Original-Bytes: 3349 Xref: csiph.com comp.theory:21947 On 7/24/2020 8:28 PM, Ben Bacarisse wrote: > olcott writes: > > Do the exercises first. That way I can be sure you really have a sound > intuitive understanding of what the formal stuff is intended to pin > down. It's much easier to understand the next two pages if you are 100% > sure about the informal notion of satisfiable. > - // I am not sure what this: "⇒" means in this context 2.10(a)(ii) - ii. A^2_1(x1, x2) ⇒ A^2_1(x2, x1) Taking 2.10 (a)(ii) literally (exactly what it says) Since the ⇒ (from page 12) requires a Boolean we must somehow convert A^2_1(x1, x2) and A^2_1(x2, x1) into Boolean. The sets that satisfy them are sets and thus not Boolean. If we consider the set that satisfies A^2_1(x1, x2) as its "true" set and the set that does not satisfy A^2_1(x2, x1) as its false set then then the only ordered pairs that are excluded are those ordered pairs that satisfy ^2_1(x1, x2) and do not satisfy A^2_1(x2, x1): (x1 > x2) thus leaving the ordered pair of positive integers (x1 = x2) as the solution set. 2.10(a)(iii) - iii. (∀x1)(∀x2)(∀x3) (A^2_1(x1,x2) ∧ A^2_1(x2,x3) ⇒ A^2_1(x1,x3)) The LHS of ⇒ is positive integers x,y,z such that x >= y >= z The RHS of ⇒ is positive integers x,z such that x >= z The expression is only false when (x >= y >= z) and x < z Thus it is true whenever x >= z, thus the solution set is x >= y I will wait for these to be verified before I move on to 2.10(b)(ii) 2.10(c)(ii) 2.10(b)(iii) 2.10(c)(iii) -- Copyright 2020 Pete Olcott