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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as
 non-halting V2
Date: Sun, 21 Jul 2024 08:34:40 -0500
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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. Deciders only take finite
string inputs. They do not take executing processes as inputs. Thus HHH
is not allowed to report on the behavior of this int main() { DDD(); }.

Even the Linz proof makes this same mistake

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

Although embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly correctly report on
its own behavior because its input does the opposite of whatever
it reports embedded_H is only allowed to report on the behavior
that its input specifies.

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.

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then

     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

Since we ourselves can directly see that UTM based embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
must abort the simulation of its input otherwise this input would
never stop running we know that the criteria have been met.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer