On 2/28/2022 11:31 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 2/28/2022 8:50 AM, Ben Bacarisse wrote:
>>> olcott <polcott2@gmail.com> writes:
>>>
>>>> Even Linz was confused by this. embedded_H is not supposed to report
>>>> on itself or the computation that it is contained within.
>>> No one thinks it should. You don't know what Linz says even after all
>>> these years. If you want to know what Linz says, I am open to pertinent
>>> questions on the topic.
>>
>> You for one have insisted that it should as your primary rebuttal to
>> my work for six straight months.
>
> Quote please. You have a long track record of misunderstanding the points
> put to you.
>
> My "primary rebuttal" comes from your own claim that false is the
> correct answer despite the fact that computation represented halts,
> i.e. that you are not addressing the halting problem.
>

See there you go, you are asserting that the fact that Ĥ ⟨Ĥ⟩ halts
contradicts that fact that embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determines that
its input never halts.

Ĥ ⟨Ĥ⟩ is the computation that contains embedded_H and embedded_H is not
supposed to determine the halt status of itself or the computation that
contains it.

Because halt deciders are deciders they only compute the mapping from
their inputs to an accept or reject state. Because Ĥ ⟨Ĥ⟩ is not an
actual input to embedded_H it is out-of-scope for embedded_H.

THIS WILL BE MY ONLY REPLY UNTIL EVERY POINT ABOVE IS FULLY ADDRESSED

Do you understand that a decider computes the mapping ONLY from its
inputs to an accept or reject state and does not compute any other mapping?

--
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 3/22/2022 6:18 PM, Richard Damon wrote:
> On 3/22/22 10:53 AM, olcott wrote:
>> On 3/22/2022 9:39 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> Yet you cannot point out a single error in my halting problem
>>>> refutation.
>>>
>>> The mistakes have been pointed out so many times that it's reasonable to
>>> assume you can't see them or are simply ignoring them.  The latest
>>> monster error is that if (as you claim) this is true of your Ĥ:
>>>
>>>    "Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn"
>>>
>>> but
>>>
>>>    "H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
>>>
>>> then Ĥ is not even a Truing machine, let alone the specific Ĥ in Linz's
>>> proof.
>>>
>>
>> If the halt deciding criteria compares the finite strings of Turing
>> machine descriptions as its halt deciding basis then it will find that
>> H and the copy of H embedded at Ĥ.qx are not the identical (embedded_H
>> is a longer finite string) thus providing the basis for H to see that
>> embedded_H will transition to Ĥ.qn and halt.
>>
>>
>
> Except that inside embedded_H IS an exact copy of H, which should match
> the 'string compare' if you could possible built it.
>

Yes this is correct.

Ben's point was based on a lack of understanding the x86 machine
architecture where otherwise identical copies of machine code will have
different machines addresses that can be used as halt deciding criteria.

> Since H doesn't have a unique, or even a finite number of
> representations, it makes it hard to try to 'string compare' to find
> copies.

The Linz proof has identical copies, alternative proofs don't even
specify any copies.

--
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 3/22/2022 10:36 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 3/22/2022 9:39 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> Yet you cannot point out a single error in my halting problem
>>>> refutation.
>>> The mistakes have been pointed out so many times that it's reasonable to
>>> assume you can't see them or are simply ignoring them. The latest
>>> monster error is that if (as you claim) this is true of your Ĥ:
>>> "Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn"
>>> but
>>> "H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
>>> then Ĥ is not even a Truing machine, let alone the specific Ĥ in Linz's
>>> proof.
>>
>> If the halt deciding criteria compares the finite strings of Turing
>> machine descriptions...
>
> But Ĥ is not a Turing machine and ⟨Ĥ⟩ is not a Turing machine

Technically Linz refers to a whole class of computations. You can assume
that this class is empty on the basis of assuming that the Linz proof is
correct.

> description. You are not talking about Turing machines. Turing
> machines are not magic --

> identical state transition functions entail
> identical configuration sequences for the identical inputs.
>

Yes that is correct.

>> as its halt deciding basis then it will find that H and the copy of H
>> embedded at Ĥ.qx are not the identical (embedded_H is a longer finite
>> string) thus providing the basis for H to see that embedded_H will
>> transition to Ĥ.qn and halt.
>
> H and Ĥ have to be the Turing machines. When you write these symbols,

If you start with the foundational assumption that Linz is correct you
can simply assume that my rebuttal is incorrect without even looking at
it. This is not how correct reasoning works.

> you are referring to entities with magic properties. For actual Turing
> machines, If Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn then H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qn.
>

Not when Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn on the basis that Ĥ is invoking an
identical machine description with identical inputs twice in sequence.

The finite string of ⟨H⟩ is not identical to the finite string of ⟨Ĥ⟩
and the comparison of these finite strings is the halt deciding basis.

If we eliminate the appended input loop from H then H and Ĥ.qx would
have to derive the same result.

--
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 3/23/2022 6:36 AM, Richard Damon wrote:
> On 3/22/22 11:51 PM, olcott wrote:
>> On 3/22/2022 10:36 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 3/22/2022 9:39 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> Yet you cannot point out a single error in my halting problem
>>>>>> refutation.
>>>>> The mistakes have been pointed out so many times that it's
>>>>> reasonable to
>>>>> assume you can't see them or are simply ignoring them.  The latest
>>>>> monster error is that if (as you claim) this is true of your Ĥ:
>>>>>     "Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn"
>>>>> but
>>>>>     "H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
>>>>> then Ĥ is not even a Truing machine, let alone the specific Ĥ in
>>>>> Linz's
>>>>> proof.
>>>>
>>>> If the halt deciding criteria compares the finite strings of Turing
>>>> machine descriptions...
>>>
>>> But Ĥ is not a Turing machine and ⟨Ĥ⟩ is not a Turing machine
>>
>> Technically Linz refers to a whole class of computations. You can
>> assume that this class is empty on the basis of assuming that the Linz
>> proof is correct.
>>
>>> description.  You are not talking about Turing machines.  Turing
>>> machines are not magic --
>>
>>> identical state transition functions entail
>>> identical configuration sequences for the identical inputs.
>>>
>>
>> Yes that is correct.
>>
>>>> as its halt deciding basis then it will find that H and the copy of H
>>>> embedded at Ĥ.qx are not the identical (embedded_H is a longer finite
>>>> string) thus providing the basis for H to see that embedded_H will
>>>> transition to Ĥ.qn and halt.
>>>
>>> H and Ĥ have to be the Turing machines.  When you write these symbols,
>>
>> If you start with the foundational assumption that Linz is correct you
>> can simply assume that my rebuttal is incorrect without even looking
>> at it. This is not how correct reasoning works.
>>
>>> you are referring to entities with magic properties.  For actual Turing
>>> machines, If Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn then H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qn.
>>>
>>
>> Not when Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn on the basis that Ĥ is invoking an
>> identical machine description with identical inputs twice in sequence.
>
> Except that this is not a CORRECT descriptor of an infinite behVIOR.
>

There are no cases where a function or identical copy of a function
continues to be called with identical input that are not infinite behavior.

Provide a counter-example.

--
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 3/23/2022 10:08 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 3/22/2022 10:36 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 3/22/2022 9:39 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> Yet you cannot point out a single error in my halting problem
>>>>>> refutation.
>>>>> The mistakes have been pointed out so many times that it's reasonable to
>>>>> assume you can't see them or are simply ignoring them. The latest
>>>>> monster error is that if (as you claim) this is true of your Ĥ:
>>>>> "Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn"
>>>>> but
>>>>> "H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
>>>>> then Ĥ is not even a Truing machine, let alone the specific Ĥ in Linz's
>>>>> proof.
>>>>
>>>> If the halt deciding criteria compares the finite strings of Turing
>>>> machine descriptions...
>>> But Ĥ is not a Turing machine and ⟨Ĥ⟩ is not a Turing machine
>>
>> Technically Linz refers to a whole class of computations.
>
> No. Linz refer to Turing machine computations. Your PO-machines are
> not Turing machines because identical state transition functions entail
> different configuration sequences for identical inputs.
>
>>> description. You are not talking about Turing machines. Turing
>>> machines are not magic --
>>
>>> identical state transition functions entail
>>> identical configuration sequences for the identical inputs.
>>
>> Yes that is correct.
>
> It is for Turing machines, but not your magic PO-machines. This:
>
> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn and H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn
>
> is magic, and can't happen if H and H^ are Turing machines. It's odd
> that you don't get this. I don't think you have every grasped what a
> Turing machine is.
>
>> If you start with the foundational assumption that Linz is correct you
>> can simply assume that my rebuttal is incorrect without even looking
>> at it.
>
> I start wit the assumption that H is a Turing machine. You start
> with the assumption that H is not a Turing machine so sort of magic
> thing where an exact copy can entail a different sequence of
> transitions.
>
> Note that I am not rebutting anything -- your magic machine may well
> have the properties you claim. That's a another question, and one I
> don't care about.
>
>>> you are referring to entities with magic properties. For actual Turing
>>> machines, If Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn then H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qn.
>>
>> Not when Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn on the basis that Ĥ is invoking an
>> identical machine description with identical inputs twice in sequence.
>
> If H and Ĥ magic (which is probably what you intend the vague words "on the
> bases that" to imply) I agree. You machines can do anything you like,
> but it says nothing about the proof you think is wrong.
>
>> The finite string of ⟨H⟩ is not identical to the finite string of ⟨Ĥ⟩
>> and the comparison of these finite strings is the halt deciding basis.
>
> If Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn then so does H ⟨Ĥ⟩ ⟨Ĥ⟩ unless H and Ĥ are
> not Turing machines.
>

When we remove the infinite loop from Ĥ then
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
and
Ĥ ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn

So in other words I am agreeing with a key element of what you are
saying. I don't have the time to waste on elements that are not a direct
part of my proof: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn

--
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 3/23/2022 10:08 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 3/22/2022 10:36 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 3/22/2022 9:39 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> Yet you cannot point out a single error in my halting problem
>>>>>> refutation.
>>>>> The mistakes have been pointed out so many times that it's reasonable to
>>>>> assume you can't see them or are simply ignoring them. The latest
>>>>> monster error is that if (as you claim) this is true of your Ĥ:
>>>>> "Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn"
>>>>> but
>>>>> "H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
>>>>> then Ĥ is not even a Truing machine, let alone the specific Ĥ in Linz's
>>>>> proof.
>>>>
>>>> If the halt deciding criteria compares the finite strings of Turing
>>>> machine descriptions...
>>> But Ĥ is not a Turing machine and ⟨Ĥ⟩ is not a Turing machine
>>
>> Technically Linz refers to a whole class of computations.
>
> No. Linz refer to Turing machine computations. Your PO-machines are
> not Turing machines because identical state transition functions entail
> different configuration sequences for identical inputs.
>
>>> description. You are not talking about Turing machines. Turing
>>> machines are not magic --
>>
>>> identical state transition functions entail
>>> identical configuration sequences for the identical inputs.
>>
>> Yes that is correct.
>
> It is for Turing machines, but not your magic PO-machines. This:
>
> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn and H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn
>
> is magic, and can't happen if H and H^ are Turing machines. It's odd
> that you don't get this. I don't think you have every grasped what a
> Turing machine is.
>
>> If you start with the foundational assumption that Linz is correct you
>> can simply assume that my rebuttal is incorrect without even looking
>> at it.
>
> I start wit the assumption that H is a Turing machine. You start
> with the assumption that H is not a Turing machine so sort of magic
> thing where an exact copy can entail a different sequence of
> transitions.
>
> Note that I am not rebutting anything -- your magic machine may well
> have the properties you claim. That's a another question, and one I
> don't care about.
>
>>> you are referring to entities with magic properties. For actual Turing
>>> machines, If Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn then H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qn.
>>
>> Not when Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn on the basis that Ĥ is invoking an
>> identical machine description with identical inputs twice in sequence.
>
> If H and Ĥ magic (which is probably what you intend the vague words "on the
> bases that" to imply) I agree. You machines can do anything you like,
> but it says nothing about the proof you think is wrong.
>
>> The finite string of ⟨H⟩ is not identical to the finite string of ⟨Ĥ⟩
>> and the comparison of these finite strings is the halt deciding basis.
>
> If Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn then so does H ⟨Ĥ⟩ ⟨Ĥ⟩ unless H and Ĥ are
> not Turing machines.
>

When we remove the infinite loop from Ĥ then
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
and
H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn

So in other words I am agreeing with a key element of what you are
saying. I don't have the time to waste on elements that are not a direct
part of my proof: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn

--
Copyright 2021 Pete Olcott

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