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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Halt Deciders must be computable functions --- dbush was always
 wrong
Date: Mon, 24 Mar 2025 15:34:24 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Mon, 24 Mar 2025 09:49:34 -0500 schrieb olcott:
> On 3/24/2025 6:23 AM, Richard Damon wrote:
>> On 3/23/25 8:59 PM, olcott wrote:
>>> On 3/23/2025 7:01 PM, joes wrote:
>>>> Am Sun, 23 Mar 2025 15:08:25 -0500 schrieb olcott:
>>>>
>>>>> The behavior of a directly executing Turing Machine cannot be
>>>>> computed because a directly executing Turing machine cannot be the
>>>>> input to any computable function.
>>>> Lol. This is such a ridiculously silly objection. Of course a TM is
>>>> nothing but a finite string (or can be encoded as such). TMs are most
>>>> definitely computable - UTMs are possible.
>>>>
>>> It is enormously more nuanced than that.
>>> https://www.liarparadox.org/Linz_Proof.pdf
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> 
>> Incorrect, UTM was never there;, so you are just showing that your
>> logic was a lie.
>> 
>> 
>>> When we define the Peter Linz Ĥ with a UTM that simulates a finite
>>> number of "moves"
>>> before transitioning to Ĥ.qn then Ĥ reaches Ĥ.qn and the simulated ⟨Ĥ⟩
>>> cannot possibly reach ⟨Ĥ.qn⟩.
>>>
>> Which is impossible, as such a thing is not a UTM.
>> 
> Saying that a UTM that simulates a finite number of states is not a UTM
> is like saying that a red car is not a car.
The other way around: a limited TM is not universal.

>> You are just showing that your logic is based on the presumption of the
>> impossible and lies to fill the holes.
> It is not impossible for a UTM to simulate a finite number of states.
Neither is it identical to the direct execution (which doesn't stop).

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.