Re: DD specifies non-terminating behavior to HHH --- COMPLETE PROOF

On 2/25/2025 10:38 PM, olcott wrote:

On 2/25/2025 8:52 PM, dbush wrote:

On 2/25/2025 6:14 PM, olcott wrote:

On 2/25/2025 2:46 PM, olcott wrote:

On 2/25/2025 12:49 PM, dbush wrote:

On 2/25/2025 1:01 PM, olcott wrote:

On 2/25/2025 10:13 AM, Mikko wrote:

Althogh the subject line has the words "COMPLETE PROOF" there is no
proof or pointer to proof below.
>

>
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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);
}
>
The above does specify that DD simulated by HHH
cannot possibly terminate normally by reaching its
own "return" instruction.
>
That this may be beyond your technical skill level.
is less than no rebuttal at all.
>
Ignoring the code in main() seemed dishonest.
>

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int no_numbers_greater_than_10();
>
int F(uintptr_t p);
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int no_numbers_greater_than_10()
{
   return F((uintptr_t)no_numbers_greater_than_10);
}
>
int main()
{
   F((uintptr_t)no_numbers_greater_than_10);
   return 0;
}
>
>
The above does specify that no_numbers_greater_than_10 simulated by F
cannot possibly terminate normally by reaching its
own "return" instruction.
>
That this may be beyond your technical skill level
is less than no rebuttal at all.

>
Finally you made something that was not wrong in several different ways.
So what is your point?
>

>
Yes.
>

>
Good.  So now looking again at the code, this time showing the implementation of F:
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int no_numbers_greater_than_10();
>
int F(uintptr_t p)
{
   uintptr_t ptr = (uintptr_t)no_numbers_greater_than_10;
   uintptr_t i = p ^ ptr;i

 I am never going to attempt to deal with this convoluted bullshit.

It's XORing the input value with the address of no_numbers_greater_than_10.  It allows F to both accept the address of no_numbers_greater_than_10 as a parameter as well an arbitrary number.
By doing allowing F to take the address of a function, F can be construed as a simulator by performing a calculation on that address to start emulating the first instruction, and breaking out of the recursive call can be construed as aborting that simulation.
The point is, no_numbers_greater_than_10 has been proven not to halt, as you have agreed, and since it doesn't halt that proves that there are no natural numbers greater than 10.
Agreed?

 

   if (i > 10)
     return 0;
   else
     return F((i+1) ^ ptr);
}
>
int no_numbers_greater_than_10()
{
   return F((uintptr_t)no_numbers_greater_than_10);
}
>
int main()
{
   F((uintptr_t)no_numbers_greater_than_10);
   return 0;
}
>
no_numbers_greater_than_10 determines whether there exists a number greater than 10 by testing all numbers.  If if finds one, it aborts and halts returning 0.  If it does not, it gets stuck in infinite recursion and does not halt.
>
Since it was correctly determined, and you agreed, that no_numbers_greater_than_10 does not halt, we can conclude that there are no natural numbers greater than 10.
>
Agreed?