Path: ...!news.mixmin.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: Wed, 21 Aug 2024 09:59:46 +0200 Organization: A noiseless patient Spider Lines: 140 Message-ID: References: MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Wed, 21 Aug 2024 09:59:47 +0200 (CEST) Injection-Info: dont-email.me; posting-host="c35bd23cdeef2c979c84d642c8787936"; logging-data="3983691"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX18dDDIkJG+WYEJ1j6hvzzJJ" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:VELOra3K7kuffkF9gTadnV7RSu4= In-Reply-To: Content-Language: en-GB Bytes: 6337 Op 20.aug.2024 om 15:18 schreef olcott: > On 8/20/2024 5:29 AM, Fred. Zwarts wrote: >> 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. > > Counter factual. HHH∞ is hypothetical thus takes no memory. > HHH and DDD remains at the same physical machine address locations. > >> 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. > > HHH∞ is hypothetical thus takes no memory. HHHn(DDD) predicts > the behavior of a hypothetical HHH∞(DDD) as described below Which makes HHHn incorrect, because it should simulate its input, not a hypothetical non-input HHH∞.