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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2
Date: Sat, 27 Jul 2024 10:21:05 +0300
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On 2024-07-26 14:08:11 +0000, olcott said:

> On 7/26/2024 3:45 AM, Mikko wrote:
>> On 2024-07-24 13:33:55 +0000, olcott said:
>> 
>>> On 7/24/2024 3:57 AM, Mikko wrote:
>>>> On 2024-07-23 13:31:35 +0000, olcott said:
>>>> 
>>>>> On 7/23/2024 1:32 AM, 0 wrote:
>>>>>> On 2024-07-22 13:46:21 +0000, olcott said:
>>>>>> 
>>>>>>> On 7/22/2024 2:57 AM, Mikko wrote:
>>>>>>>> On 2024-07-21 13:34:40 +0000, olcott said:
>>>>>>>> 
>>>>>>>>> On 7/21/2024 4:34 AM, Mikko wrote:
>>>>>>>>>> On 2024-07-20 13:11:03 +0000, olcott said:
>>>>>>>>>> 
>>>>>>>>>>> On 7/20/2024 3:21 AM, Mikko wrote:
>>>>>>>>>>>> On 2024-07-19 14:08:24 +0000, olcott said:
>>>>>>>>>>>> 
>>>>>>>>>>>>> When we use your incorrect reasoning we would conclude
>>>>>>>>>>>>> that Infinite_Loop() is not an infinite loop because it
>>>>>>>>>>>>> only repeats until aborted and is aborted.
>>>>>>>>>>>> 
>>>>>>>>>>>> You and your HHH can reason or at least conclude correctly about
>>>>>>>>>>>> Infinite_Loop but not about DDD. Possibly because it prefers to
>>>>>>>>>>>> say "no", which is correct about Infinte_loop but not about DDD.
>>>>>>>>>>>> 
>>>>>>>>>>> 
>>>>>>>>>>> *Because this is true I don't understand how you are not simply lying*
>>>>>>>>>>> int main
>>>>>>>>>>> {
>>>>>>>>>>>    DDD();
>>>>>>>>>>> }
>>>>>>>>>>> 
>>>>>>>>>>> Calls HHH(DDD) that must abort the emulation of its input
>>>>>>>>>>> or {HHH, emulated DDD and executed DDD} never stop running.
>>>>>>>>>> 
>>>>>>>>>> You are the lying one.
>>>>>>>>>> 
>>>>>>>>>> If HHH(DDD) abrots its simulation and returns true it is correct as a
>>>>>>>>>> halt decider for DDD really halts.
>>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> (b) We know that a decider is not allowed to report on the behavior
>>>>>>>>> computation that itself is contained within.
>>>>>>>> 
>>>>>>>> No, we don't. There is no such prohibition.
>>>>>>>> 
>>>>>>> 
>>>>>>> Turing machines never take actual Turing machines as inputs.
>>>>>>> They only take finite strings as inputs and an actual executing
>>>>>>> Turing machine is not itself a finite string.
>>>>>> 
>>>>>> The definition of a Turing machine does not say that a Turing machine
>>>>>> is not a finite string. It is an abstract mathematical object without
>>>>>> a specification of its exact nature. It could be a set or a finite
>>>>>> string. Its exact nature is not relevant to the theory of computation,
>>>>>> which only cares about certain properties of Turing machines.
>>>>>> 
>>>>>>> Therefore It is not allowed to report on its own behavior.
>>>>>> 
>>>>>> Anyway, that does not follow. The theory of Turing machines does not
>>>>>> prohibit anything.
>>>>>> 
>>>>>>> Another different TM can take the TM description of this
>>>>>>> machine and thus accurately report on its actual behavior.
>>>>>> 
>>>>>> If a Turing machine can take a description of a TM as its input
>>>>>> or as a part of its input it can also take its own description.
>>>>>> Every Turing machine can be given its own description as input
>>>>>> but a Turing machine may interprete it as something else.
>>>>>> 
>>>>> In this case we have two x86utm machines that are identical
>>>>> except that DDD calls HHH and DDD does not call HHH1.
>>>> 
>>>> That DDD calls HHH and DDD does not call HHH1 is not a difference
>>>> between two unnamed turing machines.
>>>> 
>>> 
>>> The same thing happens at the Peter Linz Turing Machine level
>>> I will provide that more difficult example if and only if you
>>> prove that you understand this one.
>> 
>> However, Peter Linz does not call taht same thing a difference.
> 
> We can call everything "late for dinner" with a unique integer
> index and the properties that I assert exist still exist.

That you can say all sorts stupid things does not mean that it be a
good idea to do so.

Some of the properties you assert exsit actually do exist, some don't.

> When Ĥ is applied to ⟨Ĥ⟩
> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

That does not specify a time or a situation.

> (a) Ĥ copies its input ⟨Ĥ⟩
> (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩

Its not really an invocation. Ĥ imitates H with ⟨Ĥ⟩ ⟨Ĥ⟩.

> (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩

Imitating H, Ĥ simulates ⟨Ĥ⟩ ⟨Ĥ⟩.

> (d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩

Ĥ, as a part of the simulation of what its input specifecs, simulates
the specifiec copying of the simulated input ⟨Ĥ⟩.

> (e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩

Again, this is not really an invocation. Imitating H, Ĥ simulates
the simulation of H with ⟨Ĥ⟩ ⟨Ĥ⟩.

> (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
> (g) goto (d) with one more level of simulation
> 
> Two complete simulations show a pair of identical TMD's are
> simulating a pair of identical inputs. We can see this thus
> proving recursive simulation.

Detection that the simulated simulation is identical to the simulating
simulation is not trivial. In particular, it requires that the
simulator is able to identify a part of the simulated process as a
simulation.

> When embedded_H is accountable for the behavior of its
> input and not accountable for the behavior of the
> computation that it is contained within then embedded_H
> is necessarily correct to transition to its own Ĥ.qn state.

Embedded_H is not accountable for anything. It just is there and does
whatever it does. There is no requirements on Ĥ so whatever it does
it can do no wrong.

H is required to predict whether Ĥ ⟨Ĥ⟩, it executed, will terminate.
H is not a halt decider if it says "yes" and Ĥ ⟨Ĥ⟩ does not terminate,
nor if it says "no" and terminates.

-- 
Mikko