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On 3/9/2024 3:17 PM, Richard Damon wrote:So?On 3/9/24 10:33 AM, olcott wrote:Yes. (Notice that I am agreeing with you, yet never do that with me)*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,Yes. (Notice that I am agreeing with you, yet never do that with me)
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If it just fails to answer, then it has failed to be a correct Halt Decider.Yes. (Notice that I am agreeing with you, yet never do that with me)
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The fact that you reach this conflict in actions, is the reason Halt Deciding is uncomputable.*No. We know that Ĥ ⟨Ĥ⟩ either halts or fails to halt*
If it halts then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitioned to Ĥ.Hqn.
If it fails to halt then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ either transitioned
to Ĥ.Hqy or it loops.
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