Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail
From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: Formal systems that cannot possibly be incomplete except for
 unknowns and unknowable
Date: Wed, 7 May 2025 20:44:14 -0400
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In-Reply-To: <vvgtbe$1b0li$1@dont-email.me>

On 5/7/2025 8:19 PM, olcott wrote:
> On 5/7/2025 7:15 PM, dbush wrote:
>> On 5/7/2025 7:40 PM, olcott wrote:
>>> On 5/7/2025 6:31 PM, dbush wrote:
>>>> On 5/7/2025 7:26 PM, olcott wrote:
>>>>>
>>>>> When N instructions of DD are emulated by HHH
>>>>> according to the rules of the x86 language then
>>>>
>>>> The subject was "DD emulated by HHH", not "N instructions of DD 
>>>> emulated by HHH".
>>>>
>>>> Changing the subject is the dishonest dodge of the strawman deception.
>>>
>>>
>>> That you and Richard construe anything less than an
>>> infinite number of steps of DD emulated by HHH
>>> (according to the rules of the x86 language)
>>> as an incorrect emulation IS MORONICALLY STUPID.
>>>
>>
>> The fixed immutable code of HHH simulates a fixed number X of 
>> instructions of DD, the last of which was simulated incorrectly.  Any 
>> number other than X is not what HHH simulates and is therefore 
>> irrelevant to HHH.
>>
>> UTM simulates X+Y instruction of DD correctly and reaches a final state.
>>
> 
> I will make it easier to understand.
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> Can DDD simulated by HHH reach its own "return" instruction?
> 

Category error. There is no "can" as algorithm HHH is fixed and 
immutable, as is algorithm DDD.

Algorithm HHH *does not* simulate algorithm DDD to the end but instead 
aborts in violation of the x86 language.

Algorithm UTM *does* simulate algorithm DDD to the end.