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On 3/9/24 2:22 PM, olcott wrote:It is objectively true that Ĥ.H can get stuck in recursiveOn 3/9/2024 3:50 PM, immibis wrote:Yes, but you can't just assume they do, you need to define on what condition they do that.On 9/03/24 22:34, olcott wrote:>On 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.
This must just be over your head. It is very very difficult.
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What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows what
wrong answer to provide?
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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.
Simulating halt deciders must make sure that they themselves
do not get stuck in infinite execution. This means that they
must abort every simulation that cannot possibly otherwise halt.
>Except that they are running the exact same code, so if the code says to abort, they will both abort.
This requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts
its simulation.
If the code doesn't say it will abort, it won't and get stuck not returning.
>But because it is doing the exact same code on the exact same data will make the same decisions.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H ⟨Ĥ⟩ ⟨Ĥ⟩
does not simulate itself in recursive simulation.
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If they COULD tell themselves apart, you might be able justify that, but H^ specifically was define to use an EXACT copy, and Turing Machine enable this to be done.
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