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

On 5/8/2024 8:32 PM, immibis wrote:

On 7/05/24 05:00, 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).
>

 Does the one that simulates ∞ steps terminate?

Message-ID: <v1h8qu$flc1$3@i2pn2.org>
On 5/8/2024 8:27 PM, Richard Damon wrote:
 > On 5/8/24 3:36 PM, olcott wrote:
 >> 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 }
 >>
 >> Any H/D pair matching the above template where D(D)
 >> is simulated by the same H(D,D) that it calls cannot
 >> possibly reach past its own line 03. Simple software
 >> engineering verified fact.
 >>
 >
 > Proven wrong, and not refuted, so just shows that you are
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer