Re: Working out the details of the steps of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> ⊢* Ĥ.Hqn

On 3/8/24 7:06 PM, olcott wrote:

On 3/8/2024 8:47 PM, Richard Damon wrote:

On 3/8/24 6:25 PM, olcott wrote:

On 3/8/2024 8:16 PM, Richard Damon wrote:

On 3/8/24 5:54 PM, olcott wrote:

H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> must report on what happens in the pathological program.
When we specify how the pathological program will transition to Ĥ.Hqn
then we know what H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> will see and how it will report.
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The problem is that the direct application of the Linz Turing Machine template doesn't generate the Pathological Olcott-Machine template, because their execution semantics are diffferent, due to the modification caused by the master UTM.

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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> is smarter than Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly be.

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Irreverent.
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THe ^ templete has a DEFINITION of the semantics of what it is to do.
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Don't follow them, and you are just lying about what you are doing.

 H ⟨H⟩ ⟨H⟩ ⊢* H.qy // H applied to ⟨H⟩ halts
H ⟨H⟩ ⟨H⟩ ⊢* H.qn // H applied to ⟨H⟩ does not halt
 I was agreeing with you in that Olcott machines include sequences
specified by their ⊢* that Turing Machines do not have.
 

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The actual Pathological input for Olcot-Machines needs to get to the state H^.H with (H^) (H^) <H> on the tape,

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That is not how Olcott machines are stipulated to operate.
Ĥ ⟨Ĥ⟩ has <Ĥ> appended to its own tape.
H ⟨Ĥ⟩ ⟨Ĥ⟩ has <H> appended to its own tape.
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*You are just ignoring how Olcott machines are specified to work*
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And, after H^ starts, it can do whatever it wants to the tape.

That is off topic for now because:

But WHY?
If you have a problem with your definion of an Olcot machine, don't you want to know BEFORE you spend a lot of time on it

 *I cannot tolerate never getting closure on anything*
*I cannot tolerate never getting closure on anything*
*I cannot tolerate never getting closure on anything*

Then go ahead and spend a week figuring out what you what this H to do, then find out that your definition of Olcott machines doesn't do what you think it does, and have to throw that all away.
IS that what you REALLY want?

 Ĥ.H computes the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> to Ĥ.Hqn
therefore
H computes the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ <H> to H.qy
 I still don't know the detailed steps of how
H computes the mapping from ⟨H⟩ ⟨H⟩ <H> to H.qy
YET WE CAN SEE THAT IT IS CORRECT

No, you see that you have a set of results that makes
H (H^) (H^) <H> -> qy correct,
but doesn't make the other result
H^.H (H^) (H^) <H^> -> qn correct.
H^.H (H^) (H^) <H^> correct as by going to qn, it is rejecting the input and thus saying that H^ (H^) <H^> is non-halting when it is halting.
For H to be correct, ALL COPIES of it need to answer accorting to the mapping.
You just don't understand your requireents
You have just admitted that H isn't a computaton of just the description of the computation to be decided, as a Halt Decider MUST be, as the Halt Mapping is just a function of the Computation to be decided.

 You still don't understand how the indirect
criteria does compute halting as a side-effect.
 

Excpet it gets this case WRONG.
And WRONG is WRONG, and to say it is correct is just a LIE.