Re: Defining a correct simulating halt decider

On 9/3/24 9:00 AM, olcott wrote:

On 9/3/2024 5:25 AM, Mikko wrote:

On 2024-09-02 16:38:03 +0000, olcott said:
>

A halt decider is a Turing machine that computes
the mapping from its finite string input to the
behavior that this finite string specifies.

>
A halt decider needn't compute the full behaviour, only whether
that behaviour is finite or infinite.
>

 void DDD()
{
   HHH(DDD);
   return;
}
 New slave_stack at:1038c4
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
 Hence  HHH(DDD)==0 is correct
 

But the behavior of DDD is to Halt, so it is incorrect, as you just defined that HHH(DDD) will return 0, and thus ALL copies of HHH(DDD) return 0, or it isn't the "pure function" that deciders are required to be.
The above is NOT a "Correct Emulation" of the input, just proof that you are a LIAR. when you explain your logic, you admit that HHH just needs to ASSUME (incorrectly) that the HHH that DDD calls will not return, or it INCORRECTLY says that there was no conditional instruction in the path of the correctly emulated DDD, becuase the above is not the correct emulation, which would show the code of HHH, which is part of that DDD, and has the conditional instructions that invalidates that rule.
You are just proving yourself to be a LIAR.