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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: =?utf-8?Q?Re:_When_H_=E2=9F=A8=C4=A4=E2=9F=A9_=E2=9F=A8=C4=A4=E2=9F=A9_reports_on_the_behavior_it_actually_sees_then_it_is_correct?=
Date: Tue, 12 Mar 2024 11:23:16 +0200
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On 2024-03-11 16:14:29 +0000, olcott said:

> On 3/11/2024 10:16 AM, Mikko wrote:
>> On 2024-03-11 14:38:58 +0000, olcott said:
>> 
>>> On 3/11/2024 5:12 AM, Mikko wrote:
>>>> On 2024-03-10 14:59:13 +0000, olcott said:
>>>> 
>>>>> On 3/10/2024 7:49 AM, Mikko wrote:
>>>>>> On 2024-03-09 18:22:53 +0000, Richard Damon said:
>>>>>> 
>>>>>>> On 3/9/24 9:39 AM, olcott wrote:
>>>>>>>> On 3/9/2024 11:12 AM, Richard Damon wrote:
>>>>>>>>> On 3/9/24 8:41 AM, olcott wrote:
>>>>>>>>>> On 3/9/2024 9:36 AM, immibis wrote:
>>>>>>>>>>> On 9/03/24 16:12, olcott wrote:
>>>>>>>>>>>> On 3/9/2024 7:47 AM, immibis wrote:
>>>>>>>>>>>>> On 8/03/24 22:34, olcott wrote:
>>>>>>>>>>>>>>> And since H^ can "lie" to that embedded H^.H about what its description 
>>>>>>>>>>>>>>> is, that H can't tell that it is part of an H^ computation that is 
>>>>>>>>>>>>>>> simulating an H^ computation.
>>>>>>>>>>>>>> 
>>>>>>>>>>>>>> That subject must be postponed until after the Olcott refutation
>>>>>>>>>>>>>> of the exact Linz proof is either fully accepted by three people
>>>>>>>>>>>>>> or actual errors or gaps are found that cannot be addressed or
>>>>>>>>>>>>>> corrected.
>>>>>>>>>>>>> 
>>>>>>>>>>>>> It's accepted that the Linz proof doen't work on Olcott machines 
>>>>>>>>>>>>> because the Linz proof is designed for Turing machines. But you can't 
>>>>>>>>>>>>> refute the Linz-immibis proof designed for Olcott machines, where H is 
>>>>>>>>>>>>> lied to about its own description.
>>>>>>>>>>>> 
>>>>>>>>>>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except halt or fail to halt
>>>>>>>>>>>> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> could see that.
>>>>>>>>>>>> 
>>>>>>>>>>> 
>>>>>>>>>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except for exactly the same 
>>>>>>>>>>> thing that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do.
>>>>>>>>>> 
>>>>>>>>>> It is easily proven that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of
>>>>>>>>>> its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input.
>>>>>>>>> 
>>>>>>>>> How?
>>>>>>>> 
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>>>>> 
>>>>>>> Comment are SPECIFICATION, not actual behavior until existance of a 
>>>>>>> conforming H is proven
>>>>>> 
>>>>>> To me they do not look like a specification but a false statement of
>>>>>> the actual behaviour (for some non-conforming H).
>>>>>> 
>>>>> 
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy  // Ĥ applied to ⟨Ĥ⟩ halts
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn  // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>>> 
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>>> 
>>>>> Unlike anyone else has ever done my simulating termination
>>>>> analyzers can always detect when their input will cause
>>>>> themselves to never terminate. I have demonstrated this
>>>>> for the Halting Problem's pathological input:
>>>>> 
>>>>> Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>> (1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
>>>>> (2) which begins at simulated ⟨Ĥ.q0⟩
>>>>> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
>>>>> (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
>>>>> (c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
>>>>> 
>>>>> The earliest point when Turing machine H can detect the repeating
>>>>> state of Ĥ ⟨Ĥ⟩ is when Ĥ first reaches (c) where it would begin
>>>>> simulating a copy of itself with a copy of its input.
>>>> 
>>>> Nice to see that you don't disagree with my opinion about the
>>>> apparent meahings.
>>>> 
>>> 
>>> *MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
>>> (He has neither reviewed nor agreed to anything else in this paper)
>>> (a) 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
>>> (b) H can abort its simulation of D and correctly report that D 
>>> specifies a non-halting sequence of configurations.
>>> 
>>> That goes directly against this definition
>>> H(D,D) must report on the behavior of D(D).
>>> H ⟨Ĥ⟩ ⟨Ĥ⟩ must report on the behavior of Ĥ ⟨Ĥ⟩.
>>> 
>>> The conventional definitions require the deciders to report
>>> on different behavior than the behavior they actually see.
>> 
>> If they don't actually see the behaviour they are reqquired to report on
>> the deciders are defective, not the requirements.
>> 
> 
> A specification is objective if the specified behavior does not
> depend on the agent that performs it, and subjective if it does.

Only if it doesn't depend on who compares the actual behaviour to
the specification, either.

> The Church-Turing Thesis applies to objective specifications,
> not to subjective ones.

As presented, it applies to both. Whether it really is true about
either kind is unknown.

> Because H(D,D) must abort the simulation of its input and H1(D,D)
> need not abort the simulation of its input this proves that the
> halting problem specification is subjective(Hehner).

Whether H(D,D) must abort its simulation is not part of the
specification of halting decider. Other specifications are
irrelevant. The specification of halting decider is objective:
in order to determine whether the behaviout is correct one
only needs to know:
- Did the candidate decider halt in its accepting state?
- Did the candidate decider halt in its reject state?
- Does the computation asked about halt?
The identity of the candidate decider need not be known.

-- 
Mikko