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From: olcott <polcott2@gmail.com>
Newsgroups: comp.theory,sci.logic
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On 3/9/2024 7:40 PM, immibis wrote:
> On 10/03/24 02:37, olcott wrote:
>> On 3/9/2024 7:32 PM, immibis wrote:
>>> On 10/03/24 02:29, olcott wrote:
>>>> On 3/9/2024 7:24 PM, immibis wrote:
>>>>> On 10/03/24 01:30, olcott wrote:
>>>>>> On 3/9/2024 6:24 PM, immibis wrote:
>>>>>>> On 10/03/24 01:22, olcott wrote:
>>>>>>>> On 3/9/2024 5:57 PM, immibis wrote:
>>>>>>>>> On 10/03/24 00:26, olcott wrote:
>>>>>>>>>> On 3/9/2024 5:10 PM, immibis wrote:
>>>>>>>>>>> On 9/03/24 23:22, olcott wrote:
>>>>>>>>>>>> On 3/9/2024 3:50 PM, immibis wrote:
>>>>>>>>>>>>> On 9/03/24 22:34, olcott wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows what
>>>>>>>>>>>>>> wrong answer to provide?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective 
>>>>>>>>>>>>> criteria that H ⟨Ĥ⟩ uses.
>>>>>>>>>>>>
>>>>>>>>>>>> Simulating halt deciders must make sure that they themselves
>>>>>>>>>>>> do not get stuck in infinite execution. This means that they
>>>>>>>>>>>> must abort every simulation that cannot possibly otherwise 
>>>>>>>>>>>> halt.
>>>>>>>>>>>>
>>>>>>>>>>>> This requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
>>>>>>>>>>>> require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ 
>>>>>>>>>>>> aborts
>>>>>>>>>>>> its simulation.
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H 
>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>>>>>>>> does not simulate itself in recursive simulation.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective 
>>>>>>>>>>> criteria that H ⟨Ĥ⟩ uses.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> *Only because Ĥ.H is embedded within Ĥ and H is not*
>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
>>>>>>>>>> H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use 
>>>>>>>>> the exact same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.
>>>>>>>>
>>>>>>>> The above is true no matter what criteria that is used
>>>>>>>> as long as H is a simulating halt decider.
>>>>>>>>
>>>>>>>
>>>>>>> Objective criteria cannot vary based on who the subject is. They 
>>>>>>> are objective. The answer to different people is the same answer 
>>>>>>> if the criteria are objective.
>>>>>>
>>>>>> It is objectively true that Ĥ.H can get stuck in recursive
>>>>>> simulation because Ĥ copies its input thus never runs
>>>>>> out of params.
>>>>>>
>>>>>> It is objectively true that Ĥ cannot possibly get stuck
>>>>>> in recursive because H does not copy its input thus runs
>>>>>> out of params.
>>>>>>
>>>>>
>>>>> Wrong. Dead wrong. Stupidly wrong. So wrong that a dead monkey 
>>>>> could do better. Write the Olcott machine (not x86utm) code for Ĥ 
>>>>> and I would show you.
>>>>
>>>> *In other words you are denying these verified facts*
>>>> *In other words you are denying these verified facts*
>>>> *In other words you are denying these verified facts*
>>>>
>>>> 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
>>>
>>> That's not a verified fact, that's just something you want to be true.
>>>
>>> ∞ means infinite loop. Infinite loop doesn't halt. You see how stupid 
>>> it is, to say that an infinite loop halts?
>>>
>>>> Execution trace of Ĥ applied to ⟨Ĥ⟩
>>>> (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
>>>
>>> Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ BECAUSE IT IS PRECISELY 
>>> IDENTICAL TO STEPS B AND C:
>>>  > (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
>>>  > (c) which begins at Ĥ's own simulated ⟨Ĥ.q0⟩ to repeat the process
>>>
>>
>> *Yes and the key step of copying its input is left out so*
>> *H ⟨Ĥ⟩ ⟨Ĥ⟩ runs out of params and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ never runs out of params*
>>
> 
> that isn't how any of this works. Do you even know what words mean?

(b) and (c) are not the same as (1) and (2)
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

This means that Turing machine H ⟨Ĥ⟩ ⟨Ĥ⟩ can see one more execution
trace of Ĥ ⟨Ĥ⟩ than its simulated Turing machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see.

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