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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: Computable Functions --- finite string transformation rules
Date: Fri, 25 Apr 2025 08:02:37 -0400
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In-Reply-To: <vuf4gv$38ei5$1@dont-email.me>

On 4/25/2025 12:53 AM, olcott wrote:
> On 4/24/2025 10:00 PM, dbush wrote:
>> On 4/24/2025 10:50 PM, olcott wrote:
>>> On 4/24/2025 6:07 PM, Richard Damon wrote:
>>>> On 4/24/25 3:41 PM, olcott wrote:
>>>>> On 4/24/2025 2:12 PM, Fred. Zwarts wrote:
>>>>>> Op 24.apr.2025 om 19:13 schreef olcott:
>>>>>>>
>>>>>>> HHH correctly determines through mathematical induction that
>>>>>>> DD emulated by HHH (according to the finite string transformations
>>>>>>> specified by the x86 language) cannot possibly reach its final
>>>>>>> halt state in an infinite number of steps.
>>>>>
>>>>>> No, HHH has a bug which makes that it fails to see that there is 
>>>>>> only a finite recursion, 
>>>>>
>>>>> *You are technically incompetent on this point*
>>>>> When the finite string transformation rules of the
>>>>> x86 language are applied to the input to HHH(DD)
>>>>> THIS DD CANNOT POSSIBLY REACH ITS FINAL HALT STATE
>>>>> not even after an infinite number of emulated steps.
>>>>
>>>
>>>
>>>> When the defined finite string trasnsformation rules, thos of the 
>>>> x86 language, are applied to this input, completed with the 
>>>> definitions from Halt7.c as stipulated, we see that DD calls 
>>>> HHH(DD), that it will spend some time emulating DDm then it will
>>>
>>> Correctly determine that DD emulated by HHH can never possibly
>>> reach its final halt state even after an infinite number of
>>> steps are emulated.
>>
>> Category error.  The fixed code of algorithm HHH, which is part of the 
>> input as you agreed, emulates for a fixed number of steps.  Therefore 
>> there is no infinite number of steps emulated by algorithm HHH.
>>
> 
> You are flat out stupid about hypothetical possibilities.
> Of every possible HHH/DD pair where DD calls HHH(DD) and
> DD is emulated by HHH according to the finite string transformation
> rules of the x86 language no DD ever reaches its own final halt state.
> 

In other words, you're hypothesizing changing the input.

Changing the input, hypothetically or otherwise, is not allowed.


>> However, UTM(DD) emulates the same input which includes HHH as part if 
>> it for a finite number of steps and reaches a final state.  The 
>> instructions emulated by UTM are exactly the same as those emulated by 
>> HHH up to the point that it aborts, showing that the fixed code of HHH 
>> is incorrect in showing non-halting.
> 
>