Path: ...!weretis.net!feeder8.news.weretis.net!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail From: "Fred. Zwarts" Newsgroups: comp.theory Subject: Re: Anyone that disagrees with this is not telling the truth --- V5 Date: Tue, 20 Aug 2024 12:29:31 +0200 Organization: A noiseless patient Spider Lines: 127 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 12:29:31 +0200 (CEST) Injection-Info: dont-email.me; posting-host="72bbefecc85a70f3af7b7eb1065881ef"; logging-data="3517383"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1//uC1jAHzeWB+1N0GMRlhx" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:T8hFCjnJbn522KBNtthlIN/4J3o= In-Reply-To: Content-Language: en-GB Bytes: 5552 Op 20.aug.2024 om 06:33 schreef olcott: > On 8/19/2024 11:02 PM, Richard Damon wrote: >> On 8/19/24 11:50 PM, olcott wrote: >>> 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. >> >> So, you only prove that the DDD∞ that calls the HHH∞ is non-halting. >> >> >> Not any of the other DDDn >> >>> >>>> 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? >> >> But only for DDD∞, not any of the other ones. >> >>> >>> 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. >>> >> >> But HHHn isn't given DDD∞ as its input, so that doesn't matter. >> > > All of the DDD have identical bytes it is only the HHH that varies. > HHHn(DDD) predicts the behavior of HHH∞(DDD). Not all HHH can be at the same memory at the same time. When HHHn is in the memory, then DDD calls HHHn, not HHH∞. When HHHn is doing the simulation, HHHn is in that memory, therefore, it should simulate HHHn, not HHH∞. They cannot be at the same memory location at the same time, unless you are cheating with the Root variable to switch between HHHn and HHH∞, which causes HHHn to process the non-input HHH∞ instead of the input HHHn.