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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Can D simulated by H terminate normally? ---
Date: Fri, 3 May 2024 07:17:20 -0500
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On 5/2/2024 8:48 PM, Richard Damon wrote:
> On 5/2/24 2:48 PM, olcott wrote:
>> On 5/2/2024 6:04 AM, Richard Damon wrote:
>>> On 5/2/24 12:24 AM, olcott wrote:
>>>> On 5/1/2024 7:28 PM, Richard Damon wrote:
>>>>> On 5/1/24 11:51 AM, olcott wrote:
>>>>>> Every D simulated by H that cannot possibly stop running unless
>>>>>> aborted by H does specify non-terminating behavior to H. When
>>>>>> H aborts this simulation that does not count as D halting.
>>>>>
>>>>> Which is just meaningless gobbledygook by your definitions.
>>>>>
>>>>> It means that
>>>>>
>>>>> int H(ptr m, ptr d) {
>>>>>     return 0;
>>>>> }
>>>>>
>>>>
>>>> Your H is not simulating D at all thus not
>>>> "Every D simulated by H" quoted above
>>>
>>>
>>> Yes it is, it is just aborting the simulation before it started.
>>>
>>
>> "Interpreting" {D is simulated by H} as {D is NEVER simulated by H}
>> does not seem honest to me.
>>
> 
> But ALL steps simulated by H were correctly simulated, weren't they?
> 

YOU DIDN'T MEET THIS SPEC AND YOU KNOW IT AND WEASEL WORDS WILL NOT HELP

(a) It is a verified fact that D(D) simulated by H cannot
possibly reach past line 03 of D(D) simulated by H whether
H aborts its simulation or not.

> 
> Just shows you don't understand the essential nature of the problem, 
> perhaps taken slightly too extreme.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer