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 <[email protected]> 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

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