Proving that embedded_H at Ĥ.qx correctly maps its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to
Ĥ.qn on the basis of the behavior of the UTM simulation of these inputs.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt

We know that H would correctly decide the halt status of its input on
the basis of correctly deciding the halt status of the UTM simulation of
its input.

We know this because a UTM simulation of the Turing machine description
is computationally equivalent to the direct execution of this same
Turing machine.

We know that the copy of H is at Ĥ.qx (AKA embedded_H) applies this
same criterion measure to every instance of embedded_H to any recursive
depth.

We know that this means that embedded_H is examining the behavior of Ĥ
applied to ⟨Ĥ⟩ as if embedded_H was (hypothetically) replaced by a UTM.

We know that the behavior of this hypothetical machine is the criterion
measure for embedded_H.

We know that Ĥ applied to ⟨Ĥ⟩ would never stop running if embedded_H was
replaced by a UTM because Ĥ applied to ⟨Ĥ⟩ copies its input then UTM ⟨Ĥ⟩
⟨Ĥ⟩ would repeat this cycle ...

We know that this means that when embedded_H computes the mapping from
its input ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qy or Ĥ.qn on the basis of the UTM simulation of
this input that its transition to Ĥ.qn is correct.

https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf

Linz, Peter 1990. An Introduction to Formal Languages and Automata.
Lexington/Toronto: D. C. Heath and Company. (315-320)

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

On 12/28/21 6:38 PM, olcott wrote:
> Proving that embedded_H at Ĥ.qx correctly maps its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to
> Ĥ.qn on the basis of the behavior of the UTM simulation of these inputs.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>

First line iff H -> H.qy, second line iff H -> H.qn

> *My criterion measure with Ben's notational conventions*
> H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
> H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt
>
> We know that H would correctly decide the halt status of its input on
> the basis of correctly deciding the halt status of the UTM simulation of
> its input.

Right H is correct iff its answer MATCHES what UTM does.

H does NOT actually use a REAL UTM, as that would fail to get to H.qn

>
> We know this because a UTM simulation of the Turing machine description
> is computationally equivalent to the direct execution of this same
> Turing machine.
>
> We know that the copy of H is at  Ĥ.qx (AKA embedded_H) applies this
> same criterion measure to every instance of embedded_H to any recursive
> depth.
>
> We know that this means that embedded_H is examining the behavior of Ĥ
> applied to ⟨Ĥ⟩ as if embedded_H was (hypothetically) replaced by a UTM.
>
> We know that the behavior of this hypothetical machine is the criterion
> measure for embedded_H.
>
> We know that Ĥ applied to ⟨Ĥ⟩ would never stop running if embedded_H was
> replaced by a UTM because Ĥ applied to ⟨Ĥ⟩ copies its input then UTM ⟨Ĥ⟩
> ⟨Ĥ⟩ would repeat this cycle ...

But you don't get to replace H with a UTM, because H isn't a UTM.

H goes to H.qn if the UTM would be Non-Halting, so H does NOT act like a
UTM (unless H does exactly act like a UTM, in which case you have shown
that H will never answer an thus fail to be a decider)

>
> We know that this means that when embedded_H computes the mapping from
> its input ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qy or Ĥ.qn on the basis of the UTM simulation of
> this input that its transition to Ĥ.qn is correct.

FALSE.

We KNOW that if H -> H.qn (your claimed 'Correct' Answer) then we know
that H^ ALSO goes to H^.qn and HALTS.

Thus we KNOW that UTM(<H^>,<H^>) Halts and that H should have gone to
H.qy, and was wrong to go to H.qn

The error H made was to assume that the embedded H would act like a UTN,
which it doesn't.

FAIL.
>
> https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
>
> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
> Lexington/Toronto: D. C. Heath and Company. (315-320)
>

Summary:

Your H -> H.qn is saying that its input <H^> <H^> would never Halt.

By construction this means that H^ <H^> -> H^.qn and Halts, and thus
UTM(<H^>,<H^>) which represents that computation will also Halt, and
thus H gave the wrong answer.

FAIL.

You 'claims' about the behavior are incorrect, due to making false
assumptions.

The H you describe either gives wrong answers or does not decide H^ in
finite time, either of which makes it NOT a correct Halt Decider.

olcott <NoOne@NoWhere.com> writes:

> *My criterion measure with Ben's notational conventions*
> H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
> H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt

You think the means something different to this:

H.q0 wM w ⊢* H.qy iff M applied to w halts
H.q0 wM w ⊢* H.qn iff M applied to w does not halt

No useful discussion can come from using a notion you don't understand.

And until you accept that two identical TMs, given identical tapes, will
generate identical configuration sequences, any you say about a TM built
a copy of another will be bogus.

--
Ben.

On 12/28/2021 6:32 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> *My criterion measure with Ben's notational conventions*
>> H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
>> H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
>
> You think the means something different to this:
>
> H.q0 wM w ⊢* H.qy iff M applied to w halts
> H.q0 wM w ⊢* H.qn iff M applied to w does not halt
>
> No useful discussion can come from using a notion you don't understand.
>

We have finally come to mutual agreement on this.
Please go back an review the rest of that post.

> And until you accept that two identical TMs, given identical tapes, will
> generate identical configuration sequences, any you say about a TM built
> a copy of another will be bogus.
>

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

On 12/28/2021 6:32 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> *My criterion measure with Ben's notational conventions*
>> H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
>> H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
>
> You think the means something different to this:
>
> H.q0 wM w ⊢* H.qy iff M applied to w halts
> H.q0 wM w ⊢* H.qn iff M applied to w does not halt
>

I agree that the above two are the same.

Proving that embedded_H at Ĥ.qx correctly maps its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to
Ĥ.qn on the basis of the behavior of the UTM simulation of these inputs.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt

We know that H would correctly decide the halt status of its input on
the basis of correctly deciding the halt status of the UTM simulation of
its input.

We know this because a UTM simulation of the Turing machine description
is computationally equivalent to the direct execution of this same
Turing machine.

We know that the copy of H is at Ĥ.qx (AKA embedded_H) applies this
same criterion measure to every instance of embedded_H to any recursive
depth.

We know that this means that embedded_H is examining the behavior of Ĥ
applied to ⟨Ĥ⟩ as if embedded_H was (hypothetically) replaced by a UTM.

We know that the behavior of this hypothetical machine is the criterion
measure for embedded_H.

We know that Ĥ applied to ⟨Ĥ⟩ would never stop running if embedded_H was
replaced by a UTM because Ĥ applied to ⟨Ĥ⟩ copies its input then UTM ⟨Ĥ⟩
⟨Ĥ⟩ would repeat this cycle ...

We know that this means that when embedded_H computes the mapping from
its input ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qy or Ĥ.qn on the basis of the UTM simulation of
this input that its transition to Ĥ.qn is correct.

https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf

Linz, Peter 1990. An Introduction to Formal Languages and Automata.
Lexington/Toronto: D. C. Heath and Company. (315-320)

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

olcott <NoOne@NoWhere.com> writes:

> On 12/28/2021 6:32 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> *My criterion measure with Ben's notational conventions*
>>>
>>> H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
>>> H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
>>
>> You think the means something different to this:
>>
>> H.q0 wM w ⊢* H.qy iff M applied to w halts
>> H.q0 wM w ⊢* H.qn iff M applied to w does not halt
>>
>> No useful discussion can come from using a notion you don't understand.
>
> We have finally come to mutual agreement on this.

What is "this"? You now agree that

H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt

and

H.q0 wM w ⊢* H.qy iff M applied to w halts
H.q0 wM w ⊢* H.qn iff M applied to w does not halt

mean the same thing? Or do you simply agree that no useful discussion
can come from using a notion you don't understand?

> Please go back an review the rest of that post.

Please go forward and address the elephant in the room:

>> And until you accept that two identical TMs, given identical tapes, will
>> generate identical configuration sequences, any you say about a TM built
>> a copy of another will be bogus.

--
Ben.

On 12/28/2021 8:26 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 12/28/2021 6:32 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> *My criterion measure with Ben's notational conventions*
>>>>
>>>> H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
>>>> H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
>>>
>>> You think the means something different to this:
>>>
>>> H.q0 wM w ⊢* H.qy iff M applied to w halts
>>> H.q0 wM w ⊢* H.qn iff M applied to w does not halt
>>>
>>> No useful discussion can come from using a notion you don't understand.
>>
>> We have finally come to mutual agreement on this.
>
> What is "this"? You now agree that
>
> H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
> H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
>
> and
>
> H.q0 wM w ⊢* H.qy iff M applied to w halts
> H.q0 wM w ⊢* H.qn iff M applied to w does not halt
>
> mean the same thing? Or do you simply agree that no useful discussion
> can come from using a notion you don't understand?
>
>> Please go back an review the rest of that post.
>
> Please go forward and address the elephant in the room:

In this case they mean that same thing. There is no ambiguity about
whether the halt decider is deciding whether or not itself halts.

In this case the way that Linz says it
the halt decider is deciding about whether or not itself halts:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn if Ĥ applied to ⟨Ĥ⟩ does not halt.

The language that I use below disambiguates this:
-------------------------------------------------------------------
*Proving that embedded_H at Ĥ.qx correctly maps its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to
Ĥ.qn on the basis of the behavior of the UTM simulation of these inputs*

We know that H would correctly decide the halt status of its input on
the basis of correctly deciding the halt status of the UTM simulation of
its input.

We know this because a UTM simulation of the Turing machine description
is computationally equivalent to the direct execution of this same
Turing machine.

We know that the copy of H is at Ĥ.qx (AKA embedded_H) applies this
same criterion measure to every instance of embedded_H to any recursive
depth.

We know that this means that embedded_H is examining the behavior of Ĥ
applied to ⟨Ĥ⟩ as if embedded_H was (hypothetically) replaced by a UTM.

We know that the behavior of this hypothetical machine is the criterion
measure for embedded_H.

We know that Ĥ applied to ⟨Ĥ⟩ would never stop running if embedded_H was
replaced by a UTM because Ĥ applied to ⟨Ĥ⟩ copies its input then UTM ⟨Ĥ⟩
⟨Ĥ⟩ would repeat this cycle ...

We know that this means that when embedded_H computes the mapping from
its input ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qy or Ĥ.qn on the basis of the UTM simulation of
this input that its transition to Ĥ.qn is correct.

https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf

Linz, Peter 1990. An Introduction to Formal Languages and Automata.
Lexington/Toronto: D. C. Heath and Company. (315-320)

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

On 12/28/21 9:41 PM, olcott wrote:
> On 12/28/2021 8:26 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 12/28/2021 6:32 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> *My criterion measure with Ben's notational conventions*
>>>>>
>>>>> H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
>>>>> H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt
>>>>
>>>> You think the means something different to this:
>>>>
>>>>    H.q0 wM w ⊢* H.qy  iff M applied to w halts
>>>>    H.q0 wM w ⊢* H.qn  iff M applied to w does not halt
>>>>
>>>> No useful discussion can come from using a notion you don't understand.
>>>
>>> We have finally come to mutual agreement on this.
>>
>> What is "this"?  You now agree that
>>
>>    H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
>>    H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt
>>
>> and
>>
>>    H.q0 wM w ⊢* H.qy  iff M applied to w halts
>>    H.q0 wM w ⊢* H.qn  iff M applied to w does not halt
>>
>> mean the same thing?  Or do you simply agree that no useful discussion
>> can come from using a notion you don't understand?
>>
>>> Please go back an review the rest of that post.
>>
>> Please go forward and address the elephant in the room:
>
> In this case they mean that same thing. There is no ambiguity about
> whether the halt decider is deciding whether or not itself halts.
>
> In this case the way that Linz says it
> the halt decider is deciding about whether or not itself halts:
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ if Ĥ applied to ⟨Ĥ⟩ halts, and
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn   if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> The language that I use below disambiguates this:
> -------------------------------------------------------------------
> *Proving that embedded_H at Ĥ.qx correctly maps its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to
> Ĥ.qn on the basis of the behavior of the UTM simulation of these inputs*
>

But 'embedded_H' is the same thing as H. If you claim different, then H^
was built wrong.

> We know that H would correctly decide the halt status of its input on
> the basis of correctly deciding the halt status of the UTM simulation of
> its input.

No quite, the UTM simulation will MATCH the output in the sense that if
the UTM Halts, H can answer Qy, but H can NOT use the UTM to decide for
Qn, as the UTM will run forever, and H must stop in finite time.

>
> We know this because a UTM simulation of the Turing machine description
> is computationally equivalent to the direct execution of this same
> Turing machine.

Right, UTM(<H^>,<H^>) will halt/non-halt exactly the sames as H^(<H^>)

>
> We know that the copy of H is at  Ĥ.qx (AKA embedded_H) applies this
> same criterion measure to every instance of embedded_H to any recursive
> depth.

Ok.
>
> We know that this means that embedded_H is examining the behavior of Ĥ
> applied to ⟨Ĥ⟩ as if embedded_H was (hypothetically) replaced by a UTM.

Nope, same old mistake, H must decide in FINITE time, the UTM can take
infinite time.

An other thing to note is H(InfiniteLoop) is a finite computation, but
UTM(InfiniteLoop) is an infinite computation, so a computation that uses
H with an input that is potentially infinite, might well end up finite
BECAUSE H is finte and aborts its simulation.

This points out that it is IMPOSSIBLE for correct H to be part of an
infinite simulation loop. PERIOD.

>
> We know that the behavior of this hypothetical machine is the criterion
> measure for embedded_H.

LIE. It is the behavior of the ACTUAL machine that embedded_H must use.

You just admitted that you H isn't really a computation.
>
> We know that Ĥ applied to ⟨Ĥ⟩ would never stop running if embedded_H was
> replaced by a UTM because Ĥ applied to ⟨Ĥ⟩ copies its input then UTM ⟨Ĥ⟩
> ⟨Ĥ⟩ would repeat this cycle ...

Nope, wrong criteria, FAIL.

>
> We know that this means that when embedded_H computes the mapping from
> its input ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qy or Ĥ.qn on the basis of the UTM simulation of
> this input that its transition to Ĥ.qn is correct.

Since your embedded_H is NOT H, your have LIED that you built H^ correctly.

FAIL.

>
> https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
>
> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
> Lexington/Toronto: D. C. Heath and Company. (315-320)
>

Same old lies that you have done things right.

FAIL.

You are just talking about your POOP.

On 12/28/21 9:24 PM, olcott wrote:
> On 12/28/2021 6:32 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> *My criterion measure with Ben's notational conventions*
>>> H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
>>> H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt
>>
>> You think the means something different to this:
>>
>>   H.q0 wM w ⊢* H.qy  iff M applied to w halts
>>   H.q0 wM w ⊢* H.qn  iff M applied to w does not halt
>>
>
> I agree that the above two are the same.
>
> Proving that embedded_H at Ĥ.qx correctly maps its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to
> Ĥ.qn on the basis of the behavior of the UTM simulation of these inputs.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> *My criterion measure with Ben's notational conventions*
> H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
> H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt
>
> We know that H would correctly decide the halt status of its input on
> the basis of correctly deciding the halt status of the UTM simulation of
> its input.
>
> We know this because a UTM simulation of the Turing machine description
> is computationally equivalent to the direct execution of this same
> Turing machine.
>
> We know that the copy of H is at  Ĥ.qx (AKA embedded_H) applies this
> same criterion measure to every instance of embedded_H to any recursive
> depth.
>
> We know that this means that embedded_H is examining the behavior of Ĥ
> applied to ⟨Ĥ⟩ as if embedded_H was (hypothetically) replaced by a UTM.
>
> We know that the behavior of this hypothetical machine is the criterion
> measure for embedded_H.
>
> We know that Ĥ applied to ⟨Ĥ⟩ would never stop running if embedded_H was
> replaced by a UTM because Ĥ applied to ⟨Ĥ⟩ copies its input then UTM ⟨Ĥ⟩
> ⟨Ĥ⟩ would repeat this cycle ...
>
> We know that this means that when embedded_H computes the mapping from
> its input ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qy or Ĥ.qn on the basis of the UTM simulation of
> this input that its transition to Ĥ.qn is correct.
>
> https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
>
> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
> Lexington/Toronto: D. C. Heath and Company. (315-320)
>
>
>

Same old lies.

Embededded_H must do the same thing as H.

H/Embeded H is NOT allowed to replace H with a UTM.

FAIL.