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From: joes <noreply@example.com>
Newsgroups: comp.theory
Subject: Re: D simulated by H unproved for THREE YEARS ---
Date: Mon, 10 Jun 2024 17:06:05 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Mon, 10 Jun 2024 10:36:44 -0500 schrieb olcott:
> On 6/10/2024 10:25 AM, joes wrote:
>> Am Mon, 10 Jun 2024 09:36:09 -0500 schrieb olcott:
>>> On 6/10/2024 3:35 AM, joes wrote:
>>>> Am Sun, 09 Jun 2024 22:54:52 -0500 schrieb olcott:
>>>>> On 5/29/2021 2:26 PM, olcott wrote:
>> 
>>>>> THE ONLY POSSIBLE WAY for D simulated by H to have the same behavior
>>>>> as the directly executed D(D) is for the instructions of D to be
>>>>> incorrectly simulated by H (details provided below).
>> The only correct simulation must simulate incorrectly? Wat.
> Try carefully studying those words again and again until you see how
> your paraphrase is wrong.
Perhaps you can paraphrase it better?
A simulator MUST have the same behaviour.

>>> H does not ignore that instruction and simulates itself simulating D.
>>> The directly executed D(D) reaps the benefit of D simulated by H
>>> proving that *its input never halts*
If that simulation is right, D(D) never halts.

>>> Because the H(D,D) that D(D) calls recognizes the its input DOES NOT
>>> HALT, it aborts the simulation of this input causing the directly
>>> executed D(D) to halt.
Simulating something most definitely does NOT cause any change in 
its behaviour.
If the abortion causes D to halt, how could H detect D not to?
If D halts, it does so whether it is simulated or not.
>> How can H report "non-halting" when D(D) halts?

>>> I proved that D simulated by H can only have the same behavior as the
>>> directly executed D(D) when D is simulated by H incorrectly.
>> You've got it the wrong way around. A simulation must have the same
>> behaviour.
> I proved otherwise. Maybe the proof is simply over-your-head?
Your simulator does not simulate if it has different behaviour.

-- 
joes