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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Simulating termination analyzers for dummies
Date: Tue, 18 Jun 2024 22:16:35 -0400
Organization: i2pn2 (i2pn.org)
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On 6/18/24 12:30 PM, olcott wrote:
> On 6/18/2024 11:21 AM, Mikko wrote:
>> On 2024-06-17 03:33:50 +0000, olcott said:
>>
>>> To understand this analysis requires a sufficient knowledge of
>>> the C programming language and what an x86 emulator does.
>>>
>>> Unless every single detail is made 100% explicit false assumptions
>>> always slip though the cracks. This is why it must be examined at
>>> the C level before it is examined at the Turing Machine level.
>>>
>>> typedef void (*ptr)();
>>> int H0(ptr P);
>>>
>>> void Infinite_Loop()
>>> {
>>>    HERE: goto HERE;
>>> }
>>>
>>> void Infinite_Recursion()
>>> {
>>>    Infinite_Recursion();
>>> }
>>>
>>> void DDD()
>>> {
>>>    H0(DDD);
>>>    return;
>>> }
>>>
>>> int main()
>>> {
>>>    H0(Infinite_Loop);
>>>    H0(Infinite_Recursion);
>>>    H0(DDD);
>>> }
>>>
>>> Every C programmer that knows what an x86 emulator is knows that when H0
>>> emulates the machine language of Infinite_Loop, Infinite_Recursion, and
>>> DDD that it must abort these emulations so that itself can terminate
>>> normally.
>>>
>>> When this is construed as non-halting criteria then simulating
>>> termination analyzer H0 is correct to reject these inputs as non-
>>> halting.
>>
>> The subject line is incorrect. The OP of "Simulating termination 
>> analyzers
>> for dummies" should tell what a "simulating termination analyzer" is.
>> The OP of this thread does not.
>>
> 
> I state the prerequisites if you don't have them
> then you cannot understand. If you have them then
> what I say is self-evidently true.
> 

Since you fail to have them, so you can't talk about it either.

An aborted simulation never, by itself, tells us about the halting 
behavior of the machine simulated.

 From Infinite_Loop, and Infinite_Recursion, there are valid induction 
arguements that can be used to prove tha the actual behavior of the 
machines is non-halting.

That doesn't work for DDD, as the PROGRAM DDD includes the code for H0, 
and thus the valid induction proof can't change it. Thus when you look 
farther, we see that H1, which simulate DDD calling H0 far enogh to see 
H0 make it decision, see that it will halt.

Your problem is you don't understand what a "program" is, so you don't 
understand what the input actually needs to be.

Thus, your FAIL and by continuing to repeat the error after it being 
pointed out, it turns into a LIE>