*The following shows how Linz H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the right*
*answer because Linz Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*

*Criterion Measure*
H is assumed to be a simulating termination analyzer that
aborts the simulation of any input that would cause its
own non-termination and returns NO. Otherwise H always
returns YES.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt

*It is true that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must transition to Ĥ.Hqn to prevent*
*its own infinite execution. It is true that this makes Ĥ ⟨Ĥ⟩ halt*
*This entails that H ⟨Ĥ⟩ ⟨Ĥ⟩ would be correct to transition to H.qy*

When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly apply the above criteria
then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to Ĥ.Hqn and H ⟨Ĥ⟩ ⟨Ĥ⟩ would
transition to H.qy

No one has refuted the above reasoning the best that anyone has
done is baselessly claim that either Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ or H ⟨Ĥ⟩ ⟨Ĥ⟩ is not
smart enough to apply their identical criterion measure to their
own input.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 5/03/24 21:24, olcott wrote:
> *The following shows how Linz H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the right*
> *answer because Linz Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>
> *Criterion Measure*
> H is assumed to be a simulating termination analyzer that
> aborts the simulation of any input that would cause its
> own non-termination and returns NO. Otherwise H always
> returns YES.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>
> *It is true that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must transition to Ĥ.Hqn to prevent*
> *its own infinite execution. It is true that this makes Ĥ ⟨Ĥ⟩ halt*
> *This entails that H ⟨Ĥ⟩ ⟨Ĥ⟩ would be correct to transition to H.qy*
>
> When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly apply the above criteria
> then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to Ĥ.Hqn and H ⟨Ĥ⟩ ⟨Ĥ⟩ would
> transition to H.qy
>
> No one has refuted the above reasoning the best that anyone has
> done is baselessly claim that either Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ or H ⟨Ĥ⟩ ⟨Ĥ⟩ is not
> smart enough to apply their identical criterion measure to their
> own input.
>

No one has refuted the claim that a copy of a Turing machine cannot get
a different result from the actual Turing machine.

If two Turing machines give different results for the same input, they
give different execution sequences for the same input. If two sequences
are different there is a first element which is different. Show the
first element which is different.

On 3/5/2024 7:52 PM, immibis wrote:
> On 5/03/24 21:24, olcott wrote:
>> *The following shows how Linz H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the right*
>> *answer because Linz Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>
>> *Criterion Measure*
>> H is assumed to be a simulating termination analyzer that
>> aborts the simulation of any input that would cause its
>> own non-termination and returns NO. Otherwise H always
>> returns YES.
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>
>> *It is true that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must transition to Ĥ.Hqn to prevent*
>> *its own infinite execution. It is true that this makes Ĥ ⟨Ĥ⟩ halt*
>> *This entails that H ⟨Ĥ⟩ ⟨Ĥ⟩ would be correct to transition to H.qy*
>>
>> When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly apply the above criteria
>> then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to Ĥ.Hqn and H ⟨Ĥ⟩ ⟨Ĥ⟩ would
>> transition to H.qy
>>
>> No one has refuted the above reasoning the best that anyone has
>> done is baselessly claim that either Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ or H ⟨Ĥ⟩ ⟨Ĥ⟩ is not
>> smart enough to apply their identical criterion measure to their
>> own input.
>>
>
> No one has refuted the claim that a copy of a Turing machine cannot get
> a different result from the actual Turing machine.
>

Richard has continued to baselessly disagree with this.
*You are correct that no one has refuted this*
It is good to have you here to balance Richard.

> If two Turing machines give different results for the same input, they
> give different execution sequences for the same input. If two sequences
> are different there is a first element which is different. Show the
> first element which is different.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt

Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process

The point where Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts its simulation does not
exist in H ⟨Ĥ⟩ ⟨Ĥ⟩.

We get the exact same result even if we eliminate the
infinite loop after Ĥ.Hqy.

It is only the recursive simulation that forces Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
to abort its simulation. The infinite loop has no effect
on this.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/5/24 3:24 PM, olcott wrote:
> *The following shows how Linz H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the right*
> *answer because Linz Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>
> *Criterion Measure*
> H is assumed to be a simulating termination analyzer that
> aborts the simulation of any input that would cause its
> own non-termination and returns NO. Otherwise H always
> returns YES.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>
> *It is true that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must transition to Ĥ.Hqn to prevent*
> *its own infinite execution. It is true that this makes Ĥ ⟨Ĥ⟩ halt*
> *This entails that H ⟨Ĥ⟩ ⟨Ĥ⟩ would be correct to transition to H.qy*
>
> When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly apply the above criteria
> then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to Ĥ.Hqn and H ⟨Ĥ⟩ ⟨Ĥ⟩ would
> transition to H.qy
>
> No one has refuted the above reasoning the best that anyone has
> done is baselessly claim that either Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ or H ⟨Ĥ⟩ ⟨Ĥ⟩ is not
> smart enough to apply their identical criterion measure to their
> own input.
>

In other words, if you can LIE about H^.H being the exact same
computation as H, then you can get H to do something different than H^.H
and get the answer right.

THat means you are just admiting that to you, lies are just valid
statements, and nothing actually matters.

You are just proving that you live in a world of make-beleive and don't
care about what is actually true, only that you get to spout off your
POOP and claim it to be something important.

All you have done is destroyed your reputation, and made it so that
NOBODY will ever look at your ideas to see if you ideas of "correct
reasoning" have any basis, as obviously your "Correct Reasoning" is just
about wayt to lie.

This just shows that you really are just a self-made totally ignorant,
hypocritical, pathlogocial lying idiot, and you like it like that.

On 3/5/24 9:51 PM, olcott wrote:
> On 3/5/2024 7:52 PM, immibis wrote:
>> On 5/03/24 21:24, olcott wrote:
>>> *The following shows how Linz H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the right*
>>> *answer because Linz Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>
>>> *Criterion Measure*
>>> H is assumed to be a simulating termination analyzer that
>>> aborts the simulation of any input that would cause its
>>> own non-termination and returns NO. Otherwise H always
>>> returns YES.
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>
>>> *It is true that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must transition to Ĥ.Hqn to prevent*
>>> *its own infinite execution. It is true that this makes Ĥ ⟨Ĥ⟩ halt*
>>> *This entails that H ⟨Ĥ⟩ ⟨Ĥ⟩ would be correct to transition to H.qy*
>>>
>>> When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly apply the above criteria
>>> then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to Ĥ.Hqn and H ⟨Ĥ⟩ ⟨Ĥ⟩ would
>>> transition to H.qy
>>>
>>> No one has refuted the above reasoning the best that anyone has
>>> done is baselessly claim that either Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ or H ⟨Ĥ⟩ ⟨Ĥ⟩ is not
>>> smart enough to apply their identical criterion measure to their
>>> own input.
>>>
>>
>> No one has refuted the claim that a copy of a Turing machine cannot
>> get a different result from the actual Turing machine.
>>
>
> Richard has continued to baselessly disagree with this.
> *You are correct that no one has refuted this*
> It is good to have you here to balance Richard.
>
>> If two Turing machines give different results for the same input, they
>> give different execution sequences for the same input. If two
>> sequences are different there is a first element which is different.
>> Show the first element which is different.
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>
> Execution trace of Ĥ applied to ⟨Ĥ⟩
> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
> (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
> (c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
>
> The point where Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts its simulation does not
> exist in H ⟨Ĥ⟩ ⟨Ĥ⟩.

Why not? doesn't H (H^) (H^) see its input do all of those?

Note, the point where H^.H aborts its simulation is when it sees the
simulation reach that call to its simulate version of H,

H sees exactly the same thing.

What was diffferent in the traces?

LIART.

>
> We get the exact same result even if we eliminate the
> infinite loop after Ĥ.Hqy.
>
> It is only the recursive simulation that forces Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
> to abort its simulation. The infinite loop has no effect
> on this.
>

And the recursion is H (M) d simulating (M) and seeing it enter the
machine H (M) d in the simulation.

The point of H^ before that point is DEFINED how it works by
construction, so it doesn't abort ther.

You seem to have a fundamental misconseption of Programming that you
don't write code to try to acheive the desired results, but you seem to
think that you just put down the requirements you want to meet, wave
your magic programming wand and Poof, out comes the program that you
have no idea how it works.

DOesn't work that way.

There is no point in H^ that can abort the simulation it does except by
an act of H on its input.

To say otherwise is a LIE. if you disagree, show the machines that
meetthe actual construction requirements and where the differences happen.

Otherwise you are just ADMITTING you have been a total LIAR.