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On 3/9/2024 3:59 PM, immibis wrote:Nope, proven elsewhereOn 9/03/24 22:55, olcott wrote:This has nothing to do with the Linz proof it only pertains to whetherOn 3/9/2024 3:17 PM, Richard Damon wrote:>On 3/9/24 10:33 AM, olcott wrote:>*Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior*>
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Specifications, not actual behavior until the existance of such an H is shown.
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IF taken as actual behavior, then it is conditional on such an H existing.
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Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
*This proves that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation*
It NEEDS to in order to meet its specification
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It DOESN'T unless its algorithm says it does,
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If it just fails to answer, then it has failed to be a correct Halt Decider.
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The fact that you reach this conflict in actions, is the reason Halt Deciding is uncomputable.
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*This is a verified fact*
When simulating halt deciders always report on the behavior of
their simulated input from their own POV then when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
transitions to Ĥ.Hqn it is correct from its own POV.
In other words, you are admitting to changing the question, and thus LYING that you are working on the actual original problem.
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*This is a verified fact*
When that occurs then H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to H.qy from
its own POV.
Which just means you are LYING that this apply to an actual Halt Decider per the Halting Theory,
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When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ report on the basis of their own
POV then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ reports incorrectly about the behavior of
Ĥ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ reports the behavior of Ĥ ⟨Ĥ⟩ correctly.
And thus you are admitting that the H in H^.H is WRONG and thus H is not a correct Halt Decider, because it gets some cases wrong.
It is a verified fact that when H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ meet this criteria that H gets the right answer and Ĥ.H gets the wrong answer.
It is a verified fact that it's impossible for H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to both meet this criteria if you honestly followed the Linz proof (for Turing machines).
or not Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can meet this criteria as a Turing machine. If it can
then there is no need for Olcott machines.
It is a verified fact that it's impossible for H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> to both meet this criteria if you honestly followed the modified Linz proof (for Olcott machines).It is still the case that the Linz Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ will
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(a) Transition to Ĥ.Hqn and halt
(b) Transition to Ĥ.Hqy and loop
(c) Loop without transitioning to Ĥ.Hqn or Ĥ.Hqy
Because Linz H only contradicts itself that means that it
does not contradict Linz H.
Thus the only reason why anyone is saying that Linz H cannot get
the right answer is because they assume that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and Linz
H ⟨Ĥ⟩ ⟨Ĥ⟩ are the same computation when I proved otherwise:
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
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