Path: ...!news.misty.com!3.eu.feeder.erje.net!feeder.erje.net!newsfeed.bofh.team!paganini.bofh.team!not-for-mail From: Mikko Newsgroups: comp.theory Subject: Re: Anyone that disagrees with this is not telling the truth --- V5 Date: Tue, 27 Aug 2024 10:28:32 +0300 Organization: To protect and to server Message-ID: References: Mime-Version: 1.0 Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Info: paganini.bofh.team; logging-data="2796729"; posting-host="ArmERdYYIOOJVi41tgCxGQ.user.paganini.bofh.team"; mail-complaints-to="usenet@bofh.team"; posting-account="9dIQLXBM7WM9KzA+yjdR4A"; User-Agent: Unison/2.2 X-Notice: Filtered by postfilter v. 0.9.3 Bytes: 5232 Lines: 92 On 2024-08-20 13:13:57 +0000, olcott said: > On 8/20/2024 3:45 AM, joes wrote: >> Am Mon, 19 Aug 2024 23:33:52 -0500 schrieb 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. >> >>>>>>> _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] >>>>>> 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. >> Changing the code, but not the address, constitutes a change. >> >>>>>>> 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. >> >>>>>>> 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). >>> It does this same same way that HHHn(EEE) >>> predicts the behavior of HHH∞(EEE). >> The bytes of HHH are part of DDD. >> > > *The following criteria only means* > HHHn(DDD) correctly predicts the behavior of HHH∞(EEE) and HHH∞(DDD) > > > If simulating halt decider H correctly simulates its input D > until H correctly determines that its simulated D would never > stop running unless aborted then No, it does not. You have not proven that H determines with a sound method that its simulated D would never stop running unless aborted. Nothe that above D is called H's input so it is important to apply the correct meaning of the word "input". -- Mikko