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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V2
Date: Sun, 16 Jun 2024 13:30:28 -0400
Organization: i2pn2 (i2pn.org)
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On 6/16/24 10:41 AM, olcott wrote:
> On 6/16/2024 9:16 AM, joes wrote:
>> Am Sun, 16 Jun 2024 07:44:41 -0500 schrieb olcott:
>>> On 6/16/2024 2:50 AM, Mikko wrote:
>>>> On 2024-06-15 13:14:57 +0000, olcott said:
>>>>> On 6/15/2024 7:19 AM, Mikko wrote:
>>>>>> On 2024-06-15 03:07:14 +0000, olcott said:
>>>>>>> On 6/13/2024 8:24 PM, Richard Damon wrote:
>>>>>>>   > On 6/13/24 11:32 AM, olcott wrote:
>>
>>>> Whenever a decider is run it answers the question it is made to answer.
>>> Not necessarily. Just because everyone falsely assumes that D correctly
>>> simulated by H must have the same behavior as the directly executed D(D)
>>> does not make this false assumption true.
> 
>> You still need to explain how you can call a simulation that differs from
>> the behaviour of its input "correct".
>>
> 
> I have proven it many times and this proof is simply over
> everyone's heads. When I ask what your C programming skill
> level is, this *is not* a rhetorical question.
> 
> 00   typedef void (*ptr)(); // pointer to void function
> 01
> 02   int H0(ptr P);
> 03
> 04   void DDD()
> 05   {
> 06     H0(DDD);
> 07     return;
> 08   }
> 09
> 10   int main()
> 11   {
> 12     H0(DDD);
> 13   }
> 
> Line 12 main()
>    invokes H0(DDD); that simulates DDD()
> 
> *REPEAT UNTIL outer H0 aborts*
>    Line 06 simulated DDD()
>    invokes simulated H0(DDD); that simulates DDD()

But the simulation you show doesn't show that, the simulation needs to 
show the instructions that H0 is using to do the simulation.

> 
> DDD correctly simulated by H0 never reaches its own "return"
> instruction and halts.
> 
>>> H is only asked about the behavior of D simulated by H and is
>>> not asked about the behavior of the directly executed D(D). >>
>> If H is a simulator, it must simulate the execution of D(D).
> 
> I have proven this to be a false assumption and people
> maintain this false assumption entirely on the basis
> that my proof is over their heads.

Nope, you prove that you LIE.

You have admitted that the simulation you (barely) looked at wasn't the 
simuation you claimed to have looked at, and even a moderate looking at 
it reveals it to not be the simulation you claimed (It starts at the 
wrong point, something never simulated by the simulator)

And you have admitted that you have just recently figured out how you 
might be able to get the trace you claimed, but have never actually 
shown such a trace.

So, you have not "proven" that they differ, and your claim is based on 
things that the simulation doesn't show.

> 
>> H does not compute the answer to "What does H say about its input?",
>> since it could answer anything then.
>> It makes no sense to call a wrong answer the correct answer to a 
>> different
>> question.
>>
>