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Subject: Re: Who here is too stupid to know that DDD correctly simulated by
 HHH cannot possibly reach its own return instruction?
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On 8/5/2024 2:44 AM, Mikko wrote:
> On 2024-08-04 13:11:56 +0000, olcott said:
> 
>> On 8/4/2024 1:26 AM, Fred. Zwarts wrote:
>>> Op 03.aug.2024 om 17:20 schreef olcott:>>
>>>> When you try to show how DDD simulated by HHH does
>>>> reach its "return" instruction you must necessarily
>>>> must fail unless you cheat by disagreeing with the
>>>> semantics of C. That you fail to have a sufficient
>>>> understanding of the semantics of C is less than no
>>>> rebuttal what-so-ever.
>>>
>>> Fortunately that is not what I try, because I understand that HHH 
>>> cannot possibly simulate itself correctly.
>>>
>>
>> void DDD()
>> {
>>    HHH(DDD);
>>    return;
>> }
>>
>> In other words when HHH simulates itself simulating DDD it
>> is supposed to do something other than simulating itself
>> simulating DDD ???  Do you expect it to make a cup of coffee?
> 
> In another message you have said that when HHH simulates itself
> simulating DDD is does not simulate itself simulating itself
> simulating DDD. You have not told whether it makes a cup of coffee.
> Neither action can be seen in the traces you have shown.
> 

HHH and HH and the original H have proved that they simulate
themselves simulating DDD, DD and P for three years now.

They did this by deriving the correct execution trace that
simulating themselves simulating their input would derive.

Maybe all of my reviewers have been saying that I am wrong
about this on the basis of pure bluster in that they are
totally confused by assembly language and don't have the
slightest clue what it means.

-- 
Copyright 2024 Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer