Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

On 2/7/25 6:13 PM, olcott wrote:

Experts in the C programming language will know that DD
correctly simulated by HHH cannot possibly reach its own
"if" statement.
 

But that isn't the Halting Question, and HHH doesn't correctly simulate DD and return an answer, so you claim is based on two LIES.

The finite string DD specifies non-terminating recursive
simulation to simulating termination analyzer HHH. This
makes HHH necessarily correct to reject its input as
non-halting.

Nope, the "finite string DD" is just an invalid program representation, as it isn't the representation of a complete program.
And thus your claim is based on another lie.

 typedef void (*ptr)();
int HHH(ptr P);
 int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 int main()
{
   HHH(DD);
}
 https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
 https://github.com/plolcott/x86utm/blob/master/Halt7.c
has fully operational HHH and DD
 The halting problem has always been a mathematical mapping
from finite strings to behaviors. That people do not comprehend
this shows the shallowness of the depth of the learned-by-rote
(lack of) understanding.
 

And the string needs to be the representation of a FULL PROGRAM.
And the mapping is whether to whether or not said program will halt when run.
So, all you are doing is proving you don't know what you talk about, and are just a pathological lying idiot.
YOU are the one with a lack of understanding, because you have a never-learned-so-ignorant understanding of the topic.
You don't seem to understand that logic system have something call RULES and DEFINITIONS, and you ignore them at your peril, because doing so makes your logic just a LIE.