XPost: comp.theory, sci.logic, sci.math
On 3/31/2022 10:45 AM, Dennis Bush wrote:
On Thursday, March 31, 2022 at 11:35:11 AM UTC-4, olcott wrote:
On 3/31/2022 10:31 AM, Dennis Bush wrote:
On Thursday, March 31, 2022 at 10:48:09 AM UTC-4, olcott wrote:
On 3/31/2022 9:32 AM, Dennis Bush wrote:
On Thursday, March 31, 2022 at 10:11:43 AM UTC-4, olcott wrote:
Bottom line: a halt decider H given input <M><I> is required to report if M applied to <I> halts.
The correct simulation of <M><I> by the simulating halt decider at the >>>> point in the execution trace where the simulating halt decider actually >>>> is does demonstrate the actual behavior specified by <M><I> at this
point in the execution trace.
It is merely a very persistent false assumption that a correct
simulation of <M><I> at some other different point in the execution
trace must derive identical behavior.
Using an alternate definition doesn't work as doing so leads to the above. >>>>>
Q.E.D.
These three points are the basis of my correct analysis.
(1) Linz: computation that halts … the Turing machine will halt whenever >>>> it enters a final state. (Linz:1990:234)
The *turning machine*, not a partial simulation of a turing machine.
(2) That a correct simulation of a Turing machine description that would >>>> never reach its final state is computationally equivalent to the direct >>>> execution of this same Turing machine never reaching its final state.
The direct execution of the turing machine Ha^ applied to <Ha^> reaches a final state, therefore the correct simulation of that turing machine would reach a final state.
Hb applied to <Ha^><Ha^> reaches a final state of its input and is therefore a correct simulation. Ha applied to <Ha^><Ha^> does not reach a final state its input and is therefore not a correct simulation as demonstrated by Hb.
(3) That analyzing the behavior of a correct partial simulation of some >>>> of the steps of a Turing machine description can accurately predict that >>>> a full simulation would never reach its final state.
It *can*, but not in the case of Ha applied to <Ha^><Ha^> as demonstrated by Hb applied to <Ha^><Ha^>.
As I said before :
Any logic you use to show that Ha is correct to reject <Ha^><Ha^> can also be used to show that Ha3 is correct to reject <N><5>, and any logic you use to show that Ha3 is not correct to reject <N><5> can be used to show that that Ha is not correct to
reject <Ha^><Ha^>.
THE PART THAT YOU IGNORED
The correct simulation of <M><I> by the simulating halt decider at the
point in the execution trace where the simulating halt decider actually
is does demonstrate the actual behavior specified by <M><I> at this
point in the execution trace.
It is merely a very persistent false assumption that a correct
simulation of <M><I> at some other different point in the execution
trace must derive identical behavior.
A fundamental property of turning machines is that they always give the same output for a given input, otherwise it's not a turing machine.
The directly executed Ĥ applied to ⟨Ĥ⟩ is the first invocation of infinite recursion that only terminates normally because of its one-way dependency relationship on embedded_H aborting the second invocation of
this otherwise infinite recursion.
DIFFERENT SEQUENCES OF CONFIGURATIONS WILL HAVE DIFFERENT BEHAVIOR:
This makes the sequence of configurations of the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩
outside of Ĥ different than the the sequence of configurations of the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ inside of Ĥ. Different sequences of configurations
will have different behavior.
--
Copyright 2022 Pete Olcott
"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer
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