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

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.