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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Pathological self-reference changes the semantics of the same
 finite string.
Date: Sat, 31 Aug 2024 17:24:38 +0200
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Op 31.aug.2024 om 14:26 schreef olcott:
> On 8/30/2024 8:22 AM, Mikko wrote:
>> On 2024-08-30 12:57:49 +0000, olcott said:
>>
>>> On 8/30/2024 3:11 AM, Mikko wrote:
>>>> On 2024-08-29 17:53:44 +0000, olcott said:
>>>>
>>>>> I just proved that the basic notion of finite strings
>>>>> having unique meanings independently of their context
>>>>> is incorrect.
>>>>
>>>> The context is the halting problem.
>>>
>>> The behavior of
>>> the directly executed DDD and executed HHH
>>> is different from the behavior of
>>> the emulated DDD and the emulated HHH
>>
>> The correct behaviour is the computation that the user wants to
>> ask about. If the input string specifies a different behaviour
>> then the input string is worng, not the behaviour.
>>
> 
> int sum(int x, int y) { return x + y; }
> And in the exact same way Bill wants to get the
> sum of 5+6 from sum(3,2).
> 
> HHH must use its actual input as its basis
> and it not allowed to use anything else.

HHH is given a finite string of a halting program.
But olcott wants it to process a modified input, where the abort code 
has been removed.
That is not allowed. HHH should process its input, not a non-input.
Just as sum(3,2) is not allowed to modify the 3 into a 4 before it adds 
the two numbers. But olcott seems to think that it is correct to write 
sum as
int sum(int x, int y) { return (x+1) + y; }

> 
> DDD emulated by HHH according to the semantics of the x86
> language cannot possibly stop running unless aborted and
> cannot possibly reach its only final halt state no matter
> what HHH does. 

Exactly, showing that this is an incorrect simulation, no matter what 
HHH does.


 > Therefore DDD never halts even if everyone> in the universe including 
myself disagrees.

No. This proves that HHH is unable to see the halting behaviour of DDD 
and makes a wild guess. An incorrect guess in this case.

No matter how much olcott wants it to be correct, or how many times 
olcott repeats that it is correct, it does not change the fact that such 
a simulation is incorrect, because it is unable to reach the end of a 
halting program.