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From: Richard Damon <richard@damon-family.org>
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 23:05:02 -0400
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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, and in doing so change 
the contents of the input (which actually reference the embedded_H that 
actually is, not the hypothetical) you are just lying.

Now, if you want to look at the case where H and embedded_H are ACTUALLY 
UTMs, we can do this, and see that H will never answer and thus fails to 
be a decider.

> 
> ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by UTM embedded_H cannot possibly reach its
> own simulated final state of ⟨Ĥ.qn⟩ in an infinite number of steps.
> 
> This conclusively proves that ⟨Ĥ⟩ will not reach ⟨Ĥ.qn⟩ is less than
> an infinite number of steps. A decider is not allowed to simulate
> an infinite number of steps.
> 

Note, the CORRECT way to do that hypothetical, is to say let us assume 
that we give this input to a UTM, but not change the input, so it still 
acts as the actual H / embedded_H, and then we see that if H / 
embedded_H will ever decide to end its partial simulation, and go to qn, 
then H^ will go go H^.qn and halt, so the UTM simulation of this input 
will ahalt.

There is n way to actually define an input machine H^ that will change 
itself when we hypotosize the decider being something different and 
giving the input to the original machine to it.

Note, the fact that the decider doesn't get the right answer in a finite 
number of steps is NOT grounds for it to give the wrong answer. The 
problem was NEVER can H simulate the input to the final state, but 
ALWAYS does the machine itself get to the final state (which can by the 
definition of a UTM, be shown by an actual UTM simulation of the input, 
but NOT a 'partial simulation'). If it takes an infinite number of steps 
to show that the input is non-halting, then the decider is put on the 
spot, as it must due so, but can't, which is one cause of problems being 
non-computable.

Also, when we look at the actual machine, which is the actual goal of 
the problem refered to, we see, and Peter Olcott has agreed, that the 
machine H^ (H^) will go to H^.qn, despite his claim above otherwise (so 
he is disagreeing with himself).


Peter Olcott is just proving his total lack of understanding of the 
field, and is reckless disregard for the truth about it, and that he is 
just willing to state lies to try to prove his false thesis.