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

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.

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

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

 

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

 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.
 *You are just ignoring how Olcott machines are specified to work*
 

And, after H^ starts, it can do whatever it wants to the tape.
Thus, it can get to H^.H with the values (H^) (H^) <H> on it.
If you think not, what is stopping it?
If you want ot change that, you need to change you definitions.
And watch out, that you don't make you system less than Turing Complete.