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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Who here is too stupid to know that DDD correctly simulated by HHH cannot possibly reach its own return instruction?
Date: Wed, 7 Aug 2024 10:22:02 +0300
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On 2024-08-05 15:00:12 +0000, olcott said:

> 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.

Your trace don't show siulation of exectuion differently from
simulation of simulation of execution.

-- 
Mikko