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On 2/10/2025 2:35 AM, Mikko wrote:And that paper just proves that you don't understand what you are talking about.On 2025-02-09 15:24:53 +0000, olcott said:That you did not bother to look at my paper does not entail
>On 2/9/2025 4:08 AM, Mikko wrote:>On 2025-02-08 14:55:09 +0000, olcott said:>
>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.
Your request does not make sense. Non-existence of a exclusion does not
have an execution trace.
>>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 does not. I only requires that the execution of HHH with a function
pointer to DD must be started. OP does not show what happens next.
Within the context that HHH <is> a simulating termination
analyzer line 3 of DD proves that DD cannot possibly reach
its own "if" statement.
That is not the context of OP.
>
that the correct full context has not been provided. The
context has always been this paper for several years.
https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
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);
}
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