Re: A simulating halt decider applied to the The Peter Linz Turing Machine description ⟨Ĥ⟩

On 5/26/24 9:03 PM, olcott wrote:

On 5/26/2024 7:44 PM, Richard Damon wrote:

On 5/26/24 8:21 PM, olcott wrote:

On 5/26/2024 7:15 PM, Richard Damon wrote:

On 5/26/24 7:45 PM, olcott wrote:

On 5/26/2024 6:07 PM, Richard Damon wrote:

On 5/26/24 6:47 PM, olcott wrote:

On 5/26/2024 3:20 PM, Richard Damon wrote:

On 5/26/24 3:14 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
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When we see that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H in an
infinite number of steps cannot possibly reach its own simulated
final state of ⟨Ĥ.qn⟩ and halt then we correctly deduce that the
same thing applies when simulating halt decider embedded_H correctly
simulates less than an infinite number of steps of ⟨Ĥ⟩ ⟨Ĥ⟩.
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Nope.
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Since we are talking about Turing Machines, your stipulated POOP definitions go away,

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https://www.liarparadox.org/Linz_Proof.pdf
*Simplified the notation for Ĥ on the top of page three*
and put back in the qy state shown in figure 12.2
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When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
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   Ĥ copies its own Turing machine description: ⟨Ĥ⟩
   then invokes embedded_H that simulates ⟨Ĥ⟩ with ⟨Ĥ⟩ as input.
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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.

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Nope, since we are in Turing Machines, the term "Correctly Simulated" means, and can ONLY mean, the resuts of a UTM simulation, which BY DEFINITION is nopt aborted.

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You always seem to make sure respond to a different set of words
than the words I actually said. This could be an honest mistake.
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*I SAID A CORRECT SIMULATION OF A FINITE NUMBER OF STEPS*

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No you didn't, not the last time.
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You said (H^) H^) correctly simulated by embbeded_H cannot ..."
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If embedded_H does a "Correct Simulation", then BY DEFINITION, it never aborts.
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That it doesn't reach a final state in a finte number of steps, and thus, that "Correct Simulation" was non-halting.
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(and your earlier statement tried to assert behavior of THIS H^ based on the behaviof or a DIFFERENT H^ built on a diffferent embedded_H with differet behaivor which is just unsound logic, as the two machines are essentially unrelated as far as this behavior)
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*WHEN I EXPLICITLY STATE A FINITE NUMBER OF STEPS THEN YOU ARE*
*FLAT OUT WRONG TO SIMPLY ASSUME AN INFINITE NUMBER OF STEPS*

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Nope, you said it didn't reach a final state in a finite number of steps, i.e the simulation is shown to be non-halting.

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*If you need to, reread that many times*
 >>>> 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.

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Right, so if you claim embedded_H is actually DOING a "Correct Simulation", then BY the DEFINITION of COMPUTATION THEORY, that is an non-aborted simulation.
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 CORRECT SIMULATION OF A FINITE NUMBER OF STEPS
CORRECT SIMULATION OF A FINITE NUMBER OF STEPS
CORRECT SIMULATION OF A FINITE NUMBER OF STEPS

Not. what you said. You do like to change meaning with a "paraphrase" which is actually just a way to LIE.
Typical method of deliberate liars.
your clause about a finite sequence of steps was about reaching a final state, NOT about the actual simulation, which was just defined a "Correct" which means like a UTM which means it doesn't abort until it reaches the final state (which you admit it won't, so the simulator is stuck being non-halting, and thus not a decider).

 UNLESS YOU CAN PROVE THAT A UTM CANNOT POSSIBLY BE ADAPTED TO COUNT
THE NUMBER OF STEPS AND THEN STOP I AM CORRECT AND YOU ARE DISHONEST
 

It can be, but the isn't a UTM, BY DEFINITION.
Look up the definition of a Universal Turing Machine.
https://en.wikipedia.org/wiki/Universal_Turing_machine
In short, the Universal Turing Machine, when given the proper description of the Machine it emulate, computes the same result as the machine described.
This can not be done if the UTM aborts after a finite number of steps without reaching the final state.
And also, there is nothing in the definition that the UTM actually step through the same virtual sequence of states of the machine it is emulating. Yes, that is a natural way to do it, but isn't required, because everything in this area of computation theory is about input to output mapping, and NOT about how it was done, except for the requirement that it was done with a finite algorithm processing the data given to it.
You are just proving your ignorance of the subject.