XPost: comp.theory, sci.logic
From: polcott333@gmail.com
On 7/29/2025 8:48 PM, Richard Damon wrote:
> On 7/29/25 9:36 PM, olcott wrote:
>> On 7/29/2025 8:28 PM, Richard Damon wrote:
>>> On 7/29/25 5:34 PM, olcott wrote:
>>>> On 7/29/2025 3:37 AM, Fred. Zwarts wrote:
>>>>> Many times already olcott has claimed to have a proof.
>>>>> Each time again it turned out that there was no proof.
>>>>>
>>>>> Op 28.jul.2025 om 18:13 schreef olcott:
>>>>>>
>>>>>> 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.
>>>>>> >>>>> 10/13/2022>
>>>>>>
>>>>>> To make this easier to understand we replace line three
>>>>>> with the more conventional terminology this line:
>>>>>> "cannot possibly reach its own simulated final halt state then"
>>>>>
>>>>> Which changes the meaning of the sentence, so you can no longer
>>>>> claim that Sipser agreed to it.
>>>>>
>>>>
>>>> Yes
>>>>
>>>>> In addition not-reaching the halt state can have more reasons than
>>>>> a non-halting sequence of instructions. E.g.: a computer being
>>>>> switched off, a simulation aborted.
>>>>
>>>> This is you not paying close enough attention to the exact
>>>> words that I exactly said.
>>>>
>>>> I did not say: DOES NOT REACH ITS FINAL STATE (might have been aborted)
>>>> I said: CANNOT POSSIBLY REACH ITS FINAL STATE (aborted or not)
>>>>
>>>>>>
>>>>>> On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
>>>>>>>
>>>>>>> I don't think that is the shell game. PO really /has/ an H (it's
>>>>>>> trivial to do for this one case) that correctly determines that P(P)
>>>>>>> *would* never stop running *unless* aborted. He knows and
>>>>>>> accepts that
>>>>>>> P(P) actually does stop. The wrong answer is justified by what
>>>>>>> would
>>>>>>> happen if H (and hence a different P) where not what they
>>>>>>> actually are.
>>>>>>>
>>>>>>
>>>>>> Saying that H is required report on the behavior of
>>>>>> machine M is a category error.
>>>>>>
>>>>>> Turing machines cannot directly report on the behavior
>>>>>> of other Turing machines they can at best indirectly
>>>>>> report on the behavior of Turing machines through the
>>>>>> proxy of finite string machine descriptions such as ⟨M⟩.
>>>>>>
>>>>>> Thus the behavior specified by the input finite string
>>>>>> overrules and supersedes the behavior of the direct
>>>>>> execution.
>>>>>
>>>>> No, because there is no difference between the specification of the
>>>>> input and the direct execution.
>>>> *I have proven this to be counter-factual*
>>>>
>>>> HHH(DDD) must simulate itself simulating DDD because DDD calls HHH(DDD)
>>>>
>>>> HHH1(DDD) must NOT simulate itself simulating DDD because DDD DOES
>>>> NOT CALL HHH1(DDD) --- [same behavior as direct execution]
>>>>
>>>
>>> The problem is since you HHH *DOES* abort,
>> void Infinite_Recursion()
>> {
>> Infinite_Recursion();
>> return;
>> }
>>
>> void Infinite_Loop()
>> {
>> HERE: goto HERE;
>> return;
>> }
>>
>> As it always does as soon as it correctly determines
>> that there exists no N steps of correct simulation
>> that can possibly reach the "return" instruction final
>> halt state.
>>
>
> But their is a N number of steps of the correct simulation of DDD that
> cause it to halt,
Again you are a fucking liar. You know that halting is
reaching a final halt state and nothing else is halting.
When 0 to infinity steps of DDD are correctly simulated
by HHH no simulated DDD ever halts.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
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