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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Because I have repeated this same point 500 times in the last three years...
Date: Sat, 27 Jul 2024 10:42:47 +0300
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On 2024-07-26 13:17:19 +0000, olcott said:

> On 7/26/2024 3:42 AM, Mikko wrote:
>> On 2024-07-26 01:49:46 +0000, olcott said:
>> 
>>> If you understand the x86 language and can't tell how DDD
>>> emulated by HHH differs from DDD emulated by HHH1 by the
>>> following then you are probably lying about understanding
>>> the x86 language.

Any emulation of DDD that differs from DDD is wrong.

If the emulation of DDD by HHH differs from the emulation of DDD by HHH1
then at least one of them differes from DDD and is wrong.

>>> *I did annotate it a little better this time*
>>> 
>>> typedef void (*ptr)();
>>> int HHH(ptr P);
>>> int HHH1(ptr P);
>>> 
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>> }
>>> 
>>> int main()
>>> {
>>>    HHH1(DDD);
>>> }
>>> 
>>> *You really don't need to know one damn thing else besides this*
>>> *You really don't need to know one damn thing else besides this*
>>> *You really don't need to know one damn thing else besides this*
>>> 
>>> All that you have to know is that HHH and HHH1 are x86 emulators
>>> and that HHH sees that same repeated state (first four lines of DDD)
>>> that anyone knowing the x86 language can see.

No knowledge of x86 is necessary as no x86 code is shown above.
From C semantics it is obvious that DDD returns if and only if
HHH(DDD) returns.

Whther DDD returns can be checked with

int main()
{
   DDD();
}

If this main returns it proves that there is no non-halting pattern in DDD.
If HHH finds one there it is wrong.

-- 
Mikko