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|    comp.ai.philosophy    |    Perhaps we should ask SkyNet about this    |    59,235 messages    |
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|    Message 57,279 of 59,235    |
|    olcott to Richard Damon    |
|    Re: People are still trying to get away     |
|    29 Jun 24 13:06:49    |
   
   XPost: comp.theory, sci.logic   
   From: polcott333@gmail.com   
      
   On 6/29/2024 12:59 PM, Richard Damon wrote:   
   > On 6/29/24 1:17 PM, olcott wrote:   
   >> On 6/29/2024 11:45 AM, Richard Damon wrote:   
   >>> On 6/29/24 12:09 PM, olcott wrote:   
   >>>> People are still trying to get away with disagreeing with   
   >>>> the semantics of the x86 language. That is isomorphic to   
   >>>> trying to get away with disagreeing with arithmetic.   
   >>>   
   >>> Nope, we are not disagreeing with the semantics of the x86 language,   
   >>> we are disagreeing with your misunderstanding of how it works.   
   >>>   
   >>>>   
   >>>> typedef void (*ptr)();   
   >>>> int H0(ptr P);   
   >>>>   
   >>>> void Infinite_Loop()   
   >>>> {   
   >>>> HERE: goto HERE;   
   >>>> }   
   >>>>   
   >>>> void Infinite_Recursion()   
   >>>> {   
   >>>> Infinite_Recursion();   
   >>>> }   
   >>>>   
   >>>> void DDD()   
   >>>> {   
   >>>> H0(DDD);   
   >>>> }   
   >>>>   
   >>>> int main()   
   >>>> {   
   >>>> H0(Infinite_Loop);   
   >>>> H0(Infinite_Recursion);   
   >>>> H0(DDD);   
   >>>> }   
   >>>>   
   >>>> Every C programmer that knows what an x86 emulator is knows   
   >>>> that when H0 emulates the machine language of Infinite_Loop,   
   >>>> Infinite_Recursion, and DDD that it must abort these emulations   
   >>>> so that itself can terminate normally.   
   >>>   
   >>> No the x86 language "knows" NOTHING about H0 being a x86 emulator. It   
   >>> is just a function that maybe happens to be a partial x86 emulator,   
   >>> but that is NOT a fundamental result of it being H0.   
   >>>   
   >>>>   
   >>>> When this is construed as non-halting criteria then simulating   
   >>>> termination analyzer H0 is correct to reject these inputs as   
   >>>> non-halting by returning 0 to its caller.   
   >>>   
   >>> It is construed as non-halting BECAUSE it has been shown that your H0   
   >>> *WILL* terminate its PARTIAL emulation of the code it is emulating   
   >>> and returning.   
   >>>   
   >>>>   
   >>>> Simulating termination analyzers must report on the behavior   
   >>>> that their finite string input specifies thus H0 must report   
   >>>> that DDD correctly emulated by H0 remains stuck in recursive   
   >>>> simulation.   
   >>>   
   >>> Right, so H0 is REQUIRED to return, and thus if the termination   
   >>> analyser knows that H0 is a termination analyzer it knows that the   
   >>> call to H0 MUST return, and thus DDD must be a terminating program.   
   >>>   
   >>> An H0 that doesn't know this, and can't figure out that H0 will   
   >>> return, but just keeps emulating H0 emulating its input will just   
   >>> fail to meet its own requirement to return.   
   >>>   
   >>>>   
   >>>>
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