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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How Partial Simulations correctly determine non-halting ---Ben's 10/2022 analysis
Date: Sun, 9 Jun 2024 18:13:11 +0300
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On 2024-06-09 14:03:08 +0000, olcott said:

> On 6/9/2024 3:52 AM, Mikko wrote:
>> On 2024-06-08 13:46:07 +0000, joes said:
>> 
>>> Am Sat, 08 Jun 2024 08:04:14 -0500 schrieb olcott:
>>>> On 6/8/2024 1:42 AM, Mikko wrote:
>>>>> On 2024-06-07 22:26:05 +0000, olcott said:
>>>>>> On 6/7/2024 4:00 PM, joes wrote:
>>>>>>> Am Fri, 07 Jun 2024 09:47:35 -0500 schrieb olcott:
>>>>>>>> On 6/7/2024 1:30 AM, Mikko wrote:
>>>>>>>>> On 2024-06-06 15:31:36 +0000, olcott said:
>>>>>>>>>> On 6/6/2024 10:14 AM, Mikko wrote:
>>>>>>>>>>> On 2024-06-06 13:53:58 +0000, olcott said:
>>>>>>>>>>>> On 6/6/2024 5:15 AM, Mikko wrote:
>>>>>>>>>>>>> On 2024-06-05 13:29:28 +0000, olcott said:
>>>>>>>>>>>>>> On 6/5/2024 2:37 AM, Mikko wrote:
>>>>>>>>>>>>>>> On 2024-06-04 18:02:03 +0000, olcott said:
>>> 
>>>>> Any finite string can be an input to some Turing machine.
>>>>> Can you prove that a Turing machine is not a finite string?
>>>> By definition Turing Machines are not finite strings in the conventional
>>>> model. In my x86utm model of computation x86 machine language <is> the
>>>> input to another function written in the x86 language.
>>> In your model, the machine code is also finite.
>>> 
>>>>> Your own attempts of a conter-proof are not about Turing machines but C
>>>>> programs. C programs are finite strings, so a C program is a valid
>>>>> input to a C program (and a Turing machine, too).
>>>> They are about Turing Machines yet cannot be sufficiently understood
>>>> with less than the 100% compete precision of the x86 language. They
>>>> x86utm model is required to prove that false assumptions about the
>>>> nature of correct simulation are false assumptions.
>>> You can't hide behind an x86 implementation. The same arguments hold.
>>> Which assumptions are false?
>>> 
>>>> I have all of the details of the machine code and C code for HH and DDD.
>>>> I can't to the same thing for embedded_H and ⟨Ĥ⟩ ⟨Ĥ⟩ so we have to learn
>>>> by analogy.
>>> Why can't you do that? The simulator can simulate itself.
>>> 
>>>>>>>> This means that H is not allowed to report on the behavior of the
>>>>>>>> directly executed P(P).
>>>>>>> So on which program does it report then?
>>>> DD(DD) is just like infinite recursion that gets terminated at its
>>>> second recursive call. DD(DD) halts only because HH(DD,DD)
>>>> correctly determines that its input DOES NOT HALT.
>>>> If HH(DD,DD) did not correctly determine that its input DOES NOT HALT
>>>> then DD(DD) would never halt.
>>> That doesn't make sense. The function halts because a simulator says it
>>> doesn't?
>> 
>> You missed an important point: The function halts because a simulator
>> CORRECTLY says it doesn't.
>> 
> 
> The first recursive call halts because the second recursive
> call has been aborted. That the second recursive call had to
> be aborted proves that the entire computation is non-halting.
> 
> DD(DD) is just like infinite recursion that gets terminated at
> its second recursive call. DD(DD) halts only because HH(DD,DD)
> correctly determines that its input DOES NOT HALT.
> 
> If HH(DD,DD) did not correctly determine that its input
> DOES NOT HALT then DD(DD) would never halt.

Nice to see that you don't disagree with my silly remark.

-- 
Mikko