XPost: comp.theory, sci.logic   
   From: F.Zwarts@HetNet.nl   
      
   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. We know that every DDD would halt when simulated   
   correctly. Each HHH misses the last part of the behaviour of the input,   
   because they all abort one cycle too soon.   
   You like to ignore it and continue to double-talk about the properties   
   of an empty set of HHH that correctly simulate itself.   
      
   No matter how much you want it to be correct, or how many times you   
   repeat that it is correct, it does not change the fact that none of   
   these simulations is correct, because none of them is able to reach the end.   
   For each HHH in this set we see that HHH cannot possibly simulate itself   
   correctly.   
      
   DDD has nothing to do with it. It is easy to eliminate DDD:   
      
           int main() {   
             return HHH(main);   
           }   
      
   This has the same problem. This proves that the problem is not in DDD,   
   but in HHH, which halts when it aborts the simulation, but it decides   
   that the simulation of itself does not halt.   
      
   HHH is unable to decide about finite recursions.   
      
   void Finite_Recursion (int N) {   
      if (N > 0) Finite_Recursion (N - 1);   
   }   
      
   It decides after N recursions that there is an infinite recursion, which   
   is incorrect.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)