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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: HHH maps its input to the behavior specified by it --- never reaches its halt state
Date: Fri, 9 Aug 2024 12:03:17 +0300
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On 2024-08-08 13:18:34 +0000, olcott said:

> On 8/8/2024 3:13 AM, Mikko wrote:
>> On 2024-08-08 04:41:11 +0000, olcott said:
>> 
>>> On 8/7/2024 8:22 PM, Richard Damon wrote:
>>>> On 8/7/24 9:12 PM, olcott wrote:
>>>>> On 8/7/2024 8:03 PM, Richard Damon wrote:
>>>>>> On 8/7/24 2:14 PM, olcott wrote:
>>>>>>> On 8/7/2024 1:02 PM, joes wrote:
>>>>>>>> Am Wed, 07 Aug 2024 08:54:41 -0500 schrieb olcott:
>>>>>>>>> On 8/7/2024 2:29 AM, Mikko wrote:
>>>>>>>>>> On 2024-08-05 13:49:44 +0000, olcott said:
>>>>>>>> 
>>>>>>>>>> I know what it means. But the inflected form "emulated" does not mean
>>>>>>>>>> what you apparently think it means. You seem to think that "DDD
>>>>>>>>>> emulated by HHH" means whatever HHH thinks DDD means but it does not.
>>>>>>>>>> DDD means what it means whether HHH emulates it or not.
>>>>>>>>>> 
>>>>>>>>> In other words when DDD is defined to have a pathological relationship
>>>>>>>>> to HHH we can just close our eyes and ignore it and pretend that it
>>>>>>>>> doesn't exist?
>>>>>>>> It doesn't change anything about DDD. HHH was supposed to decide anything
>>>>>>>> and can't fulfill that promise. That doesn't mean that DDD is somehow
>>>>>>>> faulty, it's just a counterexample.
>>>>>>>> 
>>>>>>> 
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>    HHH(DDD);
>>>>>>>    return;
>>>>>>> }
>>>>>>> 
>>>>>>> *HHH is required to report on the behavior of DDD*
>>>>>>> Anyone that does not understand that HHH meets this criteria
>>>>>>> has insufficient understanding.
>>>>>> 
>>>>>> But it doesn't, as a correct simulation of a DDD that calls an HHH that 
>>>>>> returns will stop running,
>>>>> 
>>>>> I really think that you must be a liar here because
>>>>> you have known this for years:
>>>>> 
>>>>> On 8/2/2024 11:32 PM, Jeff Barnett wrote:
>>>>>  > ...In some formulations, there are specific states
>>>>>  >    defined as "halting states" and the machine only
>>>>>  >    halts if either the start state is a halt state...
>>>>> 
>>>>>  > ...these and many other definitions all have
>>>>>  >    equivalent computing prowess...
>>>>> 
>>>>> Anyone that knows C knows that DDD correctly simulated
>>>>> by any HHH cannot possibly reach its "return" {halt state}.
>>>>> 
>>>> 
>>>> But the problem is that you HHH ODESN'T correctly emulate the DDD it is 
>>>> given, because it aborts its emulation.
>>>> 
>>> 
>> 
>> Every one can see that Olcott is trying to get way
>> with ad-hominem instead of staying on topic.
>> 
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> Each HHH of every HHH that can possibly exist definitely
> *emulates zero to infinity instructions correctly* In
> none of these cases does the emulated DDD ever reach
> its "return" instruction halt state.

The ranges of "each HHH" and "every HHH" are not defined above
so that does not really mean anything.

-- 
Mikko