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On 5/30/24 9:31 AM, olcott wrote:On 5/30/2024 2:40 AM, Mikko wrote:On 2024-05-30 01:15:21 +0000, olcott said:
The string directly SPECIFIES behavior to a UTM or toNote, the string doesn't DIRECTLY specify behavior, but only indirectly as a description/representation of the Turing Mach>x <is> a finite string Turing machine description that SPECIFIES behavior. The term: "representing" is inaccurate.>
No, x is a description of the Turing machine that specifies the behaviour
that H is required to report.
That is what I said.
When embedded_H is a real UTM then Ĥ ⟨Ĥ⟩ never stops and embedded_H is>The maning of x is that there is a universal>
Turing machine that, when given x and y, simulates what the described
Turing machine does when given y.
Yes that is also correct.>But it isn't unless H is also a UTM, and then H never returns.
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
When embedded_H is a UTM then it never halts.
You like to keep returning to that deception.
>Except that isn't what the question is, the question is what the actual behavior of the machine described, or equivalently, the simulation by a REAL UTM (one that never stops till done).
When embedded_H is a simulating halt decider then its correctly
simulated input never reaches its own simulated final state of
⟨Ĥ.qn⟩ and halts. H itself does halt and correctly rejects its
input as non-halting.
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