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On 5/4/25 9:23 PM, olcott wrote:*Wrong*On 5/4/2025 8:04 PM, Ben Bacarisse wrote:Nope, because that isn't the input that it was given.Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:>
...As explained above, UTM(⟨Ĥ⟩ ⟨Ĥ⟩) simulates Ĥ run with input Ĥ (having the>
same halting behaviour) and Ĥ run with input Ĥ HALTS. So embedded_H does
not "gather enough information to deduce that UTM(⟨Ĥ⟩ ⟨Ĥ⟩) would never
halt". THAT IS JUST A FANTASY THAT YOU HAVE.
>
UTM(⟨Ĥ⟩ ⟨Ĥ⟩) DOES halt, so embedded_H can't possibly gather information
that genuinely implies it DOESN'T halt. The explanation is obvious:
embedded_H gathers information that *YOU* believe implies that UTM(⟨Ĥ⟩ ⟨Ĥ⟩)
would never halt, but *YOU ARE SIMPLY WRONG*.
He used to claim that false ("does not halt") was the correct answer,
/even though/ the computation in question halts! Those were simpler
days. Of course cranks will never admit to having been wrong about
anything other than a detail or two, so anyone who could be bothered
could try to get him to retract that old claim.
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D
until H correctly determines that its simulated D would never
stop running unless aborted then
>
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
In other words embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is correct to
reject its input if
>
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Would not halt.
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