Re: Every D(D) simulated by H presents non-halting behavior to H +++

On 7/05/24 05:07, Richard Damon wrote:

On 5/6/24 11:00 PM, olcott wrote:

On 5/6/2024 8:39 PM, immibis wrote:

On 5/05/24 21:40, olcott wrote:

If you are claiming that you have some top secret proof that shows
the above execution trace is incorrect I am taking this as the empty
claims of evidence of election fraud that no one has ever seen.

>
The simulated execution trace is proven incorrect because it is different from the actual execution trace.
>
This is not top secret - it is very obvious.
>

>
00 int H(ptr x, ptr x)  // ptr is pointer to int function
01 int D(ptr x)
02 {
03   int Halt_Status = H(x, x);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 int main()
10 {
11   H(D,D);
12 }
>
Every H/D pair in the universe where D(D) is simulated by the
same H(D,D) that D(D) calls: AS IN THE ABOVE TEMPLATE
Involves 1 to ∞ steps of D and also includes zero to ∞
recursive simulations where H simulates itself simulating D(D).
>
None of these simulated inputs can possibly reach past their
own line 03, thus none of them teach their own line 06 and halt.
>
Once you accept the software engineering of this we can get
into the much more difficult computer science of this.
>
If you cannot understand the software engineering of this
then there is no sense moving on to something more difficult.
>

 Except I have described how to design a set of machines H that can simulate this input to the finis input to the final line 06.

The simulated execution trace is proven incorrect because it is different from the actual execution trace.
This is not top secret - it is very obvious.