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On 10/11/2023 10:50 AM, polcott2 wrote:
> This text only news reader filters out all non-text
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> news.i2pn2.org

Original Linz Turing Machine H
H.q0 ⟨M⟩ w ⊢* H.qy // Turing Machine description M and finite string
w, accept state
H.q0 ⟨M⟩ w ⊢* Hqn // Turing Machine description M and finite string w,
reject state

--
Copyright 2023 Olcott

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

On 10/11/23 12:38 PM, olcott wrote:
> On 10/11/2023 10:50 AM, polcott2 wrote:
>> This text only news reader filters out all non-text
>>
>> news.i2pn2.org
>
>
> Original Linz Turing Machine H
> H.q0  ⟨M⟩ w ⊢* H.qy  // Turing Machine description M and finite string
> w, accept state
> H.q0  ⟨M⟩ w ⊢* Hqn  // Turing Machine description M and finite string w,
>  reject state
>

Nope, you are missing important parts of the description, namely the
conditions to get to those two states. Without state that, we can't tell
if H actually meets its requirements. You of course seem to omit them,
because you want to alter the requriments but still claim to be working
on the original problem, which just makes you a LIAR.

H, to be correct, must go to H.Qy if and only if the input (M) w
describes the machine M applied to the input w that will Halt in finite
steps, and to H.qn if M w will NEVER reach a finit state in any bounded
number of steps, and must do this in a finite number of steps.

Note, the condition is on the ACTUAL BEHAVIOR of the machine M that is
described by (M).

Note also, the definition of a "Correct Simulation" is a simulation that
exactly matches the behavior of the machine simulated, so if M applied
to w WILL halt if allowed to run to completion, then a "correct
simulation" of that input will also reach that final state, and if you
stop the simulation before then, it just is NOT a "Correct Simulation",
but a "Partial Simulation".

Since the "counter example" program will use its own copy of the program
H to determine what it will do, and that copy will behave in the exact
same manner at the program H you are going to claim to be correct, amd
if that copy goes to H.qn, then the counter example program halts, we
can show that for that input, H going to H.qn is INCORRECT, as the
DEFINITION of the Halting criteria says that H can go to H.qn ONLY IF
its input represents a NON-HALTING machine, which means it must continue
running for an unbounded number of steps, but we have an upper bound for
the number of steps based on the finite number of steps that H needs to
give its answer plus the finite number of steps to create the question
to give to H.

Thus, your claim that H going to H.qn is correct, is wrong. Given an H
that goes to H.qn, then we have just showed that the actual correct
answer, that ANY "Halt Decider" must give to be correct, is Halting, and
thus it needed to go to H.Qy.

So, you are proved to be a LIAR.