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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: =?UTF-8?Q?Re=3A_A_simulating_halt_decider_applied_to_the_The_Peter_?=
 =?UTF-8?Q?Linz_Turing_Machine_description_=E2=9F=A8=C4=A4=E2=9F=A9?=
Date: Sun, 26 May 2024 22:17:24 -0500
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On 5/26/2024 10:05 PM, Richard Damon wrote:
> On 5/26/24 10:43 PM, olcott wrote:
>> On 5/26/2024 9:06 PM, olcott wrote:
>>> When Ĥ is applied to ⟨Ĥ⟩
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>>   Ĥ copies its own Turing machine description: ⟨Ĥ⟩
>>>   then invokes embedded_H that simulates ⟨Ĥ⟩ with ⟨Ĥ⟩ as input.
>>>
>>> It is an easily verified fact that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by
>>> embedded_H cannot possibly reach its own simulated final state of
>>> ⟨Ĥ.qn⟩ in any finite sequence of steps.
>>
>> *To other reviewers that are not dishonest*
>> The complete proof of the above statement is that when we hypothesize
>> that embedded_H is a UTM we can see that:
> 
> i.e. when we assume it is something it isn't, i.e we LIE to ourselves.
> 
> If you assume embedded_H is something it isn't, 

Not at all.
*It looks like you may be utterly clueless about what-if scenarios*

What-if embedded_H was a UTM would ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated
by embedded_H reach its own simulated final state of ⟨Ĥ.qn⟩ ?
(a) YES
(b) NO
(c) DISHONEST HONEST ATTEMPT TO CHANGE THE SUBJECT

*We can't bounce all around this point like a manic person on meth*
*We can't bounce all around this point like a manic person on meth*
*We can't bounce all around this point like a manic person on meth*


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