Re: Disagreeing with tautologies is always incorrect

On 5/29/25 11:37 AM, olcott wrote:

HHH is a simulating termination analyzer that uses
an x86 emulator to emulate its input. HHH is capable
of emulating itself emulating DDD.

So, since HHH hand DDD have been admitted to not be programs, they can't do ANYTHING or be emulated.

 HHH is executed within the x86utm operating system
that enables any C function to execute another C
function in debug step mode.

But you can only do that to actual computations, that is C functions that include all the code they reference.
Since DDD doesn't that just makes the work that HHH does incorrect.

 *Here is the fully operational code*
https://github.com/plolcott/x86utm/blob/master/Halt7.c

Right, which as explaieed many times, just shows that you have been LYING for years about your claims that HHH does what you claim.

 void DDD()
{
   HHH(DDD);
   return;
}
 _DDD()
[00002192] 55             push ebp
[00002193] 8bec           mov ebp,esp
[00002195] 6892210000     push 00002192
[0000219a] e833f4ffff     call 000015d2  // call HHH
[0000219f] 83c404         add esp,+04
[000021a2] 5d             pop ebp
[000021a3] c3             ret
Size in bytes:(0018) [000021a3]

 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     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
 

Which since those only apply to PROGRAMS< of which neither your H or D are, it isn't applicatable.

It is a tautology that any input D to termination
analyzer H that *would never stop running unless aborted*
DOES SPECIFY NON-TERMINATING BEHAVIOR.

Right, once H and D are fully programs. If H does continue simulating its input until it can actually PROVE that its input will be non-halting, it can never abort. Because, if H ever aborts and returns 0, its input WILL stop running and reach its return statement, just after the point that H stopped looking at its input.
If you replace the H with your hypothetical H, but keep the same input (which has been made into a full program by including all the code of the actual H, the one that does abort and return 0) then they hypothetical H will see the simulation of the input reach a final state.
Thus, no H that aborts and returns 0 can ever have corrected determined its input will not halt.
Your logic is just based on you lying about what you are doing, and starting with a category error.

 Simulating Termination Analyzer H is Not Fooled by Pathological Input D
https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D