On 2025-08-30 15:08:14 +0000, olcott said:
On 8/30/2025 1:06 AM, Kaz Kylheku wrote:
On 2025-08-29, olcott <polcott333@gmail.com> wrote:
When I say something 500 times that does not
count as I said it zero times.
DD correctly simulated by HHH specifies the
non-halting behavior pattern of recursive
simulation.
Using a bit of logical formalism:
correctly_simulated(DD) -> ~ halts(DD)
Given P -> Q, we can derive the contrapositive ~Q -> ~P.
In this case:
~ ~ halts(DD) -> ~ correctly_simulated(DD)
Cancel double negative:
halts(DD) -> ~ correctly_simulated(DD)
If HHH can recognize this repeating state as
a pure function of its inputs then HHH(DD)
returns 0 for non-halting and DD() halts.
From this paragraph we can extract:
returns_zero(HHH(DD)) -> halts (DD)
OK, so we have these propositions together:
1. returns_zero(HHH(DD)) -> halts (DD) ;; from second paragraph
2. halts(DD) -> ~ correctly_simulated(DD) ;; from first paragraph
By the transitive property of the arrow: P -> Q ^ Q -> R => P -> R,
therefore we have:
3. returns_zero(HHH(DD)) -> ~ correctly_simulated(DD)
You are logically arguing that if HHH(DD) returns
zero, then DD is not correctly simulated (because it halts).
We just use your idea of establishing
naming conventions:
HHH(DD).exe returns 0 because DD.sim1
cannot possibly halt then DD.exe halts.
The code of DD.exe is the same as the code of DD.siml so they specify
the same behaviour. Let N be the number of instructions executed when
DD.exe is executed. It is not possible to simulate correctly nore than
N steps of DD.siml.
DD.exe is outside of the scope of HHH.exe.
Likewise, DD.siml is outside of the scope of the halting problem.
--
Mikko
--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)