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

On 2/8/25 9:55 AM, olcott wrote:

On 2/8/2025 4:25 AM, Mikko wrote:

On 2025-02-07 23:13:04 +0000, olcott said:
>

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

>
Wrong, they understand that nothing below exludes the possibility that
HHH is a program that can correctly simulate DD to its "if" statement.

 Show the execution trace of that.
 

The code of HHH might exlude that but that is not sohwn below.
>

The finite string DD specifies non-terminating recursive
simulation to simulating termination analyzer HHH.

>
No, it does not. DD as quoted below pecifies nothing about the behaviour
of HHH, only its argument types and return type.
>

  >>    int Halt_Status = HHH(DD); // line 3 of DD
Requires HHH to simulate itself simulating DD recursively.

No, it requires HHH to correctly determine what that call will do./
If HHH was just a correct simulator, you are correct, but a correct simulator fails to be a correct halt decider.
Your problem is you have built your logic on a contradiction, and thus it has exploded.

 

This makes HHH necessarily correct to reject its input as
non-halting.

>
No, it does not. It is not correct to reject the input to HHH as non- halting
unless it really is non-halting.
>

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://github.com/plolcott/x86utm/blob/master/Halt7.c
has fully operational HHH and DD

>
No, it has not. There is no DD there.
>

 line 1354 through 1360
 

WHich isnt a FULL program, so not a valid input to a halt decider.
Sorrtg, you are just proving your stupidity.