XPost: sci.math, comp.theory
From: polcott333@gmail.com
On 11/28/2025 2:29 AM, Mikko wrote:
> olcott kirjoitti 27.11.2025 klo 17.17:
>> On 11/27/2025 1:40 AM, Mikko wrote:
>>> olcott kirjoitti 26.11.2025 klo 17.15:
>>>> On 11/26/2025 3:22 AM, Mikko wrote:
>>>>> olcott kirjoitti 26.11.2025 klo 5.17:
>>>>>> On 11/25/2025 9:09 PM, Kaz Kylheku wrote:
>>>>>>> On 2025-11-26, olcott wrote:
>>>>>>>> On 11/25/2025 8:36 PM, Kaz Kylheku wrote:
>>>>>>>>> On 2025-11-26, olcott wrote:
>>>>>>>>>> None of them ever had the slightest clue about Montague
>>>>>>>>>> Grammar. Except for one they all had very severe math
>>>>>>>>>> phobia.
>>>>>>>>>
>>>>>>>>> So do you; you are terribly afraid of the mathematical idea that
>>>>>>>>> simulations that are paused still exist and have future
>>>>>>>>> states.
>>>>>>>>>
>>>>>>>>
>>>>>>>> I am not going to discuss your psychotic nonsense.
>>>>>>>
>>>>>>> In all honesty, you and your therapist /should/ be laser focused
>>>>>>> on your
>>>>>>> own psychotic nonsense.
>>>>>>>
>>>>>>>> You already agreed that I am correct so this subject
>>>>>>>> is closed.
>>>>>>>
>>>>>>> Whaaat ...
>>>>>>>
>>>>>>>> news://news.eternal-september.org/20251104183329.967@kylheku.com
>>>>>>>> On 11/4/2025 8:43 PM, Kaz Kylheku wrote:
>>>>>>>>> On 2025-11-05, olcott wrote:
>>>>>>>>>>
>>>>>>>>>> The whole point is that D simulated by H
>>>>>>>>>> cannot possbly reach its own simulated
>>>>>>>>>> "return" statement no matter what H does.
>>>>>>>>>
>>>>>>>>> Yes; this doesn't happen while H is running.
>>>>>>>>>
>>>>>>>>> So while H does /something/, no matter what H does,
>>>>>>>>> that D simulation won't reach the return statement.
>>>>>>>
>>>>>>> But we know that. If H is nonreturning, of course D is.
>>>>>>> Since D calls H(D), D is suspended until H(D) returns,
>>>>>>> which means forever if H(D) is nonterminating.
>>>>>>>
>>>>>>> I have no idea what you are trying to milk out of this;
>>>>>>> it is completely uncontroversial.
>>>>>>
>>>>>> I really did figure out how to determine the
>>>>>> correct halt status that the halting problem's
>>>>>> counter-example input specifies to it decider.
>>>>>
>>>>> The basic halting problem is about Turing machines. A Turing machine
>>>>> specifies only one bhavour. It does not specify anything else to the
>>>>> decider. An ambiguous program is outside of the domain of the halting
>>>>> problem.
>>>>
>>>> That is inaccurate.
>>>
>>> No, it is not. Of course there are many ways to formulate the problem
>>> but what I said is true about the basic formulation. All formulations
>>> restrict the domain to unambiguous specifications.
>>>
>>
>> It is a perfectly unambiguous ultimately
>> self-contradictory specification.
>
> If the specification of D is perfectly unambiguous there is no point
> to say "specifies to its decider".
That DD simulated by HHH never stops running
unless aborted by HHH proves that the input
to HHH(DD) specifies non halting behavior.
The caller of a function is never an argument to
this same function. The DD executed in main that
calls HHH(DD) is not the same DD as the one that
HHH simulates.
> If D specifies a halting behaviour
> it specifies it both to its normal execution environment and to every
> decider, otherwise it specifies non-halting behaviour both to its
> normal execution environment and its decider. If it specifies one
> behaviour to one and a diffetent behaviour to the other then it is
> ambiguous. Becauyse that is what the words mean.
>
--
Copyright 2025 Olcott
My 28 year goal has been to make
"true on the basis of meaning" computable.
This required establishing a new foundation
for correct reasoning.
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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