Path: ...!3.eu.feeder.erje.net!feeder.erje.net!weretis.net!feeder8.news.weretis.net!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail From: olcott Newsgroups: comp.theory Subject: Re: Anyone that disagrees with this is not telling the truth --- V5 Date: Mon, 19 Aug 2024 22:50:16 -0500 Organization: A noiseless patient Spider Lines: 105 Message-ID: References: MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Tue, 20 Aug 2024 05:50:16 +0200 (CEST) Injection-Info: dont-email.me; posting-host="5377c0fdd88ce017cc92254daa4bcf0b"; logging-data="3381709"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX19RCkiYmviH06xkZlZ5vzhv" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:J3NhyMTcBQdeM8l4kvTJWdWgx40= In-Reply-To: Content-Language: en-US Bytes: 4352 On 8/19/2024 10:32 PM, Richard Damon wrote: > On 8/19/24 10:47 PM, olcott wrote: >> *Everything that is not expressly stated below is* >> *specified as unspecified* > > Looks like you still have this same condition. > > I thought you said you removed it. > >> >> void DDD() >> { >>    HHH(DDD); >>    return; >> } >> >> _DDD() >> [00002172] 55         push ebp      ; housekeeping >> [00002173] 8bec       mov ebp,esp   ; housekeeping >> [00002175] 6872210000 push 00002172 ; push DDD >> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD) >> [0000217f] 83c404     add esp,+04 >> [00002182] 5d         pop ebp >> [00002183] c3         ret >> Size in bytes:(0018) [00002183] >> >> *It is a basic fact that DDD emulated by HHH according to* >> *the semantics of the x86 language cannot possibly stop* >> *running unless aborted* (out of memory error excluded) > > But it can't emulate DDD correctly past 4 instructions, since the 5th > instruciton to emulate doesn't exist. > > And, you can't include the memory that holds HHH, as you mention HHHn > below, so that changes, but DDD, so the input doesn't and thus is CAN'T > be part of the input. > > >> >> X = DDD emulated by HHH∞ according to the semantics of the x86 language >> Y = HHH∞ never aborts its emulation of DDD >> Z = DDD never stops running >> >> The above claim boils down to this: (X ∧ Y) ↔ Z > > And neither X or Y are possible. > >> >> x86utm takes the compiled Halt7.obj file of this c program >> https://github.com/plolcott/x86utm/blob/master/Halt7.c >> Thus making all of the code of HHH directly available to >> DDD and itself. HHH emulates itself emulating DDD. > > Which is irrelevent and a LIE as if HHHn is part of the input, that > input needs to be DDDn > > And, in fact, > > Since, you have just explicitly introduced that all of HHHn is available > to HHHn when it emulates its input, that DDD must actually be DDDn as it > changes. > > Thus, your ACTUAL claim needs to be more like: > > X = DDD∞ emulated by HHH∞ according to the semantics of the x86 language > Y = HHH∞ never aborts its emulation of DDD∞ > Z = DDD∞ never stops running > > The above claim boils down to this: (X ∧ Y) ↔ Z > Yes that is correct. > Your problem is that for any other DDDn / HHHn, you don't have Y so you > don't have Z. > >> >> void EEE() >> { >>    HERE: goto HERE; >> } >> >> HHHn correctly predicts the behavior of DDD the same >> way that HHHn correctly predicts the behavior of EEE. >> > > Nope, HHHn can form a valid inductive proof of the input. > > It can't for DDDn, since when we move to HHHn+1 we no longer have DDDn > but DDDn+1, which is a different input. > You already agreed that (X ∧ Y) ↔ Z is correct. Did you do an infinite trace in your mind? If you can do it and I can do it then HHH can do this same sort of thing. Computations are not inherently dumber than human minds. -- Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius hits a target no one else can see." Arthur Schopenhauer