XPost: comp.theory, sci.logic   
   From: flibble@red-dwarf.jmc.corp   
      
   On Mon, 28 Jul 2025 21:56:16 -0400, Richard Damon wrote:   
      
   > On 7/28/25 7:58 PM, olcott wrote:   
   >> On 7/28/2025 6:49 PM, Richard Damon wrote:   
   >>> On 7/28/25 7:20 PM, olcott wrote:   
   >>>> On 7/28/2025 5:57 PM, Richard Damon wrote:   
   >>>>> On 7/28/25 9:54 AM, olcott wrote:   
   >>>>>> On 7/28/2025 8:21 AM, joes wrote:   
   >>>>>>> Am Mon, 28 Jul 2025 07:11:11 -0500 schrieb olcott:   
   >>>>>>>> On 7/28/2025 2:30 AM, joes wrote:   
   >>>>>>>>> Am Sun, 27 Jul 2025 21:58:05 -0500 schrieb olcott:   
   >>>>>>>>>> On 7/27/2025 9:48 PM, Richard Damon wrote:   
   >>>>>>>>>>> On 7/27/25 8:20 PM, olcott wrote:   
   >>>>>>>   
   >>>>>>>>>>>> When DDD is emulated by HHH1 it need not emulate itself at   
   >>>>>>>>>>>> all.   
   >>>>>>>>>>> But "itself" doesn't matter to x86 instructions,   
   >>>>>>>>>> By itself I mean the exact same machine code bytes at the exact   
   >>>>>>>>>> same machine address.   
   >>>>>>>>> Yeah, so when you change HHH to abort later, you also change   
   >>>>>>>>> DDD.   
   >>>>>>>> HHH is never changed.   
   >>>>>>   
   >>>>>>> It is changed in the hypothetical unaborted simulation. HHH is   
   >>>>>>> reporting on UTM(HHH', DDD) where HHH' calls UTM(DDD), and not on   
   >>>>>>> the halting DDD,   
   >>>>>>> and definitely not on HHH(DDD), itself.   
   >>>>>>>   
   >>>>>>>   
   >>>>>> All halt deciders are required to predict the behavior of their   
   >>>>>> input. HHH does correctly predict that DDD correctly simulated by   
   >>>>>> HHH cannot possibly reach its own simulated "return" instruction   
   >>>>>> final halt state.   
   >>>>>>   
   >>>>>>   
   >>>>> How is it a "correct prediction" if it sees something different than   
   >>>>> what that DDD does.   
   >>>>>   
   >>>>>   
   >>>> What DDD does is keep calling HHH(DDD) in recursive simulation until   
   >>>> HHH kills this whole process.   
   >>>   
   >>> But the behavior of the program continues past that (something you   
   >>> don't seem to understand) and that behavior will also have its HHH   
   >>> terminate the DDD it is simulating and return 0 to DDD and then Halt.   
   >>>   
   >>> Your problem is you don't understand that the simulating HHH doesn't   
   >>> define the behavior of DDD, it is the execution of DDD that defines   
   >>> what a correct simulation of it is.   
   >>>   
   >>>   
   >>>>> Remember, to have simulated that DDD, it must have include the code   
   >>>>> of the HHH that it was based on, which is the HHH that made the   
   >>>>> prediction, and thus returns 0, so DDD will halt.   
   >>>>>   
   >>>>>   
   >>>> We are not asking: Does DDD() halt.   
   >>>> That is (as it turns out) an incorrect question.   
   >>>   
   >>> No, that is EXACTLY the question.   
   >>>   
   >>> I guess you are just admitting that you whole world is based on LYING   
   >>> about what things are supposed to be.   
   >>>   
   >>>   
   >>>> Turing machines cannot directly report on the behavior of other   
   >>>> Turing machines they can at best indirectly report on the behavior of   
   >>>> Turing machines through the proxy of finite string machine   
   >>>> descriptions such as ⟨M⟩.   
   >>>   
   >>> Right, and HHH was given the equivalenet of (M) by being given the   
   >>> code of *ALL* of DDD   
   >>>   
   >>> I guess you don't understand that fact, even though you CLAIM the   
   >>> input is the proper representation of DDD.   
   >>>   
   >>>   
   >>>> Thus the behavior specified by the input finite string overrules and   
   >>>> supersedes the behavior of the direct execution.   
   >>>   
   >>> No, it is DEFINED to be the behavior of the direct execution of the   
   >>> program it represent.   
   >>>   
   >>>   
   >> *That has always been the fatal flaw of all of the proofs*   
   >   
   > No, your failure to follow the rules is what makes you just a liar.   
      
   Yet another ad hominem attack!   
      
   /Flibble   
      
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