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On 5/12/2024 5:40 PM, Richard Damon wrote:On 5/12/24 6:21 PM, olcott wrote:
Oh, it definitely can. It must decide for every machine, includingNo decider is ever allowed to report on its own behavior thus
embedded_H as a simulating partial halt decider is NOT ALLOWED to
report on the direct execution of Ĥ ⟨Ĥ⟩ because this IS REPORTING ON
ITS OWN BEHAVIOR.
WHO SAYS THIS?
A decider must compute the mapping from an input.
Its actual self cannot possibly be an input.
No decider takes an actual Turing machine as input thus no decider canWell, an encoding of one.
possibly take its actual self as input.
(a) The behavior of the directly executed Ĥ ⟨Ĥ⟩ is after embedded_H ⟨Ĥ⟩Then they are either not the same machine, or simulated incorrectly.
⟨Ĥ⟩ has already aborted its simulation.
(b) The behavior of the simulated input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
is before embedded_H has aborted its simulation.
(c) These two behaviors (a) and (b) ARE NOT THE SAME. (a) will stop
running on its own (b) will never stop running unless aborted.
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