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From: joes <noreply@example.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Boilerplate Reply -- different simulation
Date: Sat, 22 Jun 2024 14:36:29 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sat, 22 Jun 2024 08:08:51 -0500 schrieb olcott:
> On 6/22/2024 3:47 AM, joes wrote:
>> Am Fri, 21 Jun 2024 23:18:50 -0500 schrieb olcott:
>>> On 6/21/2024 11:09 PM, joes wrote:
>>>> Am Fri, 21 Jun 2024 15:52:21 -0500 schrieb olcott:
>>>>> On 6/21/2024 3:00 PM, Richard Damon wrote:
>>>>>> On 6/21/24 3:45 PM, olcott wrote:
>>>>>>> On 6/21/2024 2:33 PM, Richard Damon wrote:
>>>>>>>> On 6/21/24 3:19 PM, olcott wrote:

>>>>> No you cannot show that the mapping for the input to H(D,D) maps to
>>>>> the behavior of D(D).
>>>> If it doesn't, H is not a simulator.
>>>> The input D(D) absolutely describes the behaviour of that machine.
>>>> H just can't map it.
>>>> Either H is not a decider or it returns.

>>>>> H(D,D) is not free to simply assume that the call from D(D) to
>>>>> H(D,D) will return.
>>>> Yes it is, because it is a decider. It (incorrectly) aborts
>>>> nonterminating inputs.
>>> The behavior of D correctly simulated by H1 is the same as the
>>> behavior of the directly executed D(D) because D does not call H1(D,D)
>>> in recursive simulation.
>> D1 however, which calls H1(D1, D1), can't be decided by H1.
> When you change the subject rather than address the point at hand I take
> this to mean that you do not want an honest dialogue.
Unlike you, I replied to the points at hand. See below.

>>> The behavior of D correctly simulated by H is NOT the same as the
>>> behavior of D correctly simulated by H1 because D DOES call H(D,D) in
>>> recursive simulation.
>> The simulation by H is then of course not correct.

>> What about the other points above?
Yeah, what about them?

-- 
Man kann mit dunklen Zahlen nicht rechnen. Für die eigentliche Mathematik 
sind sie vollkommen nutzlos. --Wolfgang Mückenheim