XPost: comp.theory, sci.logic   
   From: polcott333@gmail.com   
      
   On 7/13/2024 3:15 AM, Fred. Zwarts wrote:   
   > Op 13.jul.2024 om 01:19 schreef olcott:   
   >> On 7/12/2024 5:56 PM, Richard Damon wrote:   
   >>> On 7/12/24 10:56 AM, olcott wrote:   
   >>>> We stipulate that the only measure of a correct emulation is the   
   >>>> semantics of the x86 programming language.   
   >>>   
   >>> Which means the only "correct emulation" that tells the behavior of   
   >>> the program at the input is a non-aborted one.   
   >>>   
   >>>>   
   >>>> _DDD()   
   >>>> [00002163] 55         push ebp      ; housekeeping   
   >>>> [00002164] 8bec       mov ebp,esp   ; housekeeping   
   >>>> [00002166] 6863210000 push 00002163 ; push DDD   
   >>>> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)   
   >>>> [00002170] 83c404     add esp,+04   
   >>>> [00002173] 5d         pop ebp   
   >>>> [00002174] c3         ret   
   >>>> Size in bytes:(0018) [00002174]   
   >>>>   
   >>>> When N steps of DDD are emulated by HHH according to the   
   >>>> semantics of the x86 language then N steps are emulated correctly.   
   >>>   
   >>> And thus HHH that do that know only the first N steps of the behavior   
   >>> of DDD, which continues per the definition of the x86 instruction set   
   >>> until the COMPLETE emulation (or direct execution) reaches a terminal   
   >>> instruction.   
   >>>   
   >>>>   
   >>>> When we examine the infinite set of every HHH/DDD pair such that:   
   >>>> HHH₁ one step of DDD is correctly emulated by HHH.   
   >>>> HHH₂ two steps of DDD are correctly emulated by HHH.   
   >>>> HHH₃ three steps of DDD are correctly emulated by HHH.   
   >>>> ...   
   >>>> HHH∞ The emulation of DDD by HHH never stops running.   
   >>>   
   >>> And thus, the subset that only did a finite number of steps and   
   >>> aborted its emulation on a non-terminal instrucition only have   
   >>> partial knowledge of the behavior of their DDD, and by returning to   
   >>> their caller, they establish that behavior for ALL copies of that   
   >>> HHH, even the one that DDD calls, which shows that DDD will be   
   >>> halting, even though HHH stopped its observation of the input before   
   >>> it gets to that point.   
   >>>   
   >>>>   
   >>>> The above specifies the infinite set of every HHH/DDD pair   
   >>>> where 1 to infinity steps of DDD are correctly emulated by HHH.   
   >>>>   
   >>>> No DDD instance of each HHH/DDD pair ever reaches past its   
   >>>> own machine address of 0000216b and halts.   
   >>>   
   >>> Wrong. EVERY DDD of an HHH that simulated its input for only a finite   
   >>> number of steps WILL halt becuase it will reach its final return.   
   >>>   
   >>> The HHH that simulated it for only a finite number of steps, only   
   >>> learned that finite number of steps of the behaivor, and in EVERY   
   >>> case, when we look at the behavior past that point, which DOES occur   
   >>> per the definition of the x86 instruction set, as we have not reached   
   >>> a "termial" instruction that stops behavior, will see the HHH(DDD)   
   >>> that DDD called continuing to simulate its input to the point that   
   >>> this one was defined to stop, and then returns 0 to DDDD and then DDD   
   >>> returning and ending the behavior.   
   >>>   
   >>> You continue to stupidly confuse the PARTIAL observation that HHH   
   >>> does of the behavior of DDD by its PARTIAL emulation with the ACTUAL   
   >>> FULL behavior of DDD as defined by the full definition of the x86   
   >>> insttuction set.   
   >>>   
   >>>   
   >>>>   
   >>>> Thus each HHH element of the above infinite set of HHH/DDD   
   >>>> pairs is necessarily correct to reject its DDD as non-halting.   
   >>>>   
   >>>   
   >>> Nope.   
   >>>   
   >>> NONE Of them CORRECTLY rejected itS DDD as non-halting and you are   
   >>> shown to be ignorant of what you are talking about.   
   >>>   
   >>> The HHH that did a partial emulation got the wrong answer, because   
   >>> THEIR DDD will halt. and the HHH that doen't abort never get around   
   >>> to rejecting its DDD as non-halting.   
   >>   
   >> *Here is the gist of my proof it is irrefutable*   
   >> When no DDD of every HHH/DDD that can possibly exist   
   >> halts then each HHH that rejects its DDD as non-halting   
   >> is necessarily correct.   
   >>   
   >> *No double-talk and weasel words can overcome that*   
   >>   
   >   
   > This is double talk, because no HHH can possibly exist that simulates   
   > itself correctly.   
      
   Your definition of correct contradicts the semantics of   
   the x86 language making it wrong.   
      
   --   
   Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius   
   hits a target no one else can see." Arthur Schopenhauer   
      
   --- SoupGate-Win32 v1.05   
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