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

It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.

A simulating halt decider simulates N steps of its input until its input
halts on its own or it correctly determines that a pure simulation of
its input would never stop running.

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 1/13/22 3:33 PM, olcott wrote:
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
> the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.
>

So, that doesn't say anything about what the H^ built from an H that
does this will do. It just proves that an H^ built from an H that
doesn't abort will be non-halting, but that H won't answer, so is wrong.

> A simulating halt decider simulates N steps of its input until its input
> halts on its own or it correctly determines that a pure simulation of
> its input would never stop running.
>

WRONG. Already proven incorrect. 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)
>

On 1/13/2022 5:55 PM, Richard Damon wrote:
> On 1/13/22 3:33 PM, olcott wrote:
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop
>> running.
>>
>
> So, that doesn't say anything about what the H^ built from an H that
> does this will do. It just proves that an H^ built from an H that
> doesn't abort will be non-halting, but that H won't answer, so is wrong.
>

>> A simulating halt decider simulates N steps of its input until its
>> input halts on its own or it correctly determines that a pure
>> simulation of its input would never stop running.
>>
>
> WRONG. Already proven incorrect. FAIL.

You have never provided any proof all that you have ever had is dogmatic
assertions.

>
>
>> 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 1/13/22 7:11 PM, olcott wrote:
> On 1/13/2022 5:55 PM, Richard Damon wrote:
>> On 1/13/22 3:33 PM, olcott wrote:
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never
>>> stop running.
>>>
>>
>> So, that doesn't say anything about what the H^ built from an H that
>> does this will do. It just proves that an H^ built from an H that
>> doesn't abort will be non-halting, but that H won't answer, so is wrong.
>>
>
>>> A simulating halt decider simulates N steps of its input until its
>>> input halts on its own or it correctly determines that a pure
>>> simulation of its input would never stop running.
>>>
>>
>> WRONG. Already proven incorrect. FAIL.
>
> You have never provided any proof all that you have ever had is dogmatic
> assertions.

Subject: Re: Concise refutation of halting problem proofs V47, REFUTED,
FAIR WARNING.
Date: Sun, 9 Jan 2022 04:55:11 +0000
Message-ID: <j8sCJ.29853$8Q7.4996@fx10.iad>

You have NEVER provided a proper rebuttal to this, so you have thus
conceded that you have no rebuttal to it

And thus you just LIED.

FAIL.

On 1/13/2022 6:14 PM, Richard Damon wrote:
> On 1/13/22 7:11 PM, olcott wrote:
>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>> On 1/13/22 3:33 PM, olcott wrote:
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never
>>>> stop running.
>>>>
>>>
>>> So, that doesn't say anything about what the H^ built from an H that
>>> does this will do. It just proves that an H^ built from an H that
>>> doesn't abort will be non-halting, but that H won't answer, so is wrong.
>>>
>>
>>>> A simulating halt decider simulates N steps of its input until its
>>>> input halts on its own or it correctly determines that a pure
>>>> simulation of its input would never stop running.
>>>>
>>>
>>> WRONG. Already proven incorrect. FAIL.
>>
>> You have never provided any proof all that you have ever had is
>> dogmatic assertions.
>
> Subject: Re: Concise refutation of halting problem proofs V47, REFUTED,
> FAIR WARNING.
> Date: Sun, 9 Jan 2022 04:55:11 +0000
> Message-ID: <j8sCJ.29853$8Q7.4996@fx10.iad>
>
>
> You have NEVER provided a proper rebuttal to this, so you have thus
> conceded that you have no rebuttal to it
>
> And thus you just LIED.
>
> FAIL.

I am not going over every point of your dishonest dodge.
I state a necessary truth that cannot possisbly have any correct
rebuttal and you provide some incoherent double talk that superficially
appears to be a rebuttal to people paying less than 1% of their attention.

--
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 1/13/22 7:35 PM, olcott wrote:
> On 1/13/2022 6:14 PM, Richard Damon wrote:
>> On 1/13/22 7:11 PM, olcott wrote:
>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>
>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never
>>>>> stop running.
>>>>>
>>>>
>>>> So, that doesn't say anything about what the H^ built from an H that
>>>> does this will do. It just proves that an H^ built from an H that
>>>> doesn't abort will be non-halting, but that H won't answer, so is
>>>> wrong.
>>>>
>>>
>>>>> A simulating halt decider simulates N steps of its input until its
>>>>> input halts on its own or it correctly determines that a pure
>>>>> simulation of its input would never stop running.
>>>>>
>>>>
>>>> WRONG. Already proven incorrect. FAIL.
>>>
>>> You have never provided any proof all that you have ever had is
>>> dogmatic assertions.
>>
>> Subject: Re: Concise refutation of halting problem proofs V47,
>> REFUTED, FAIR WARNING.
>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>> Message-ID: <j8sCJ.29853$8Q7.4996@fx10.iad>
>>
>>
>> You have NEVER provided a proper rebuttal to this, so you have thus
>> conceded that you have no rebuttal to it
>>
>> And thus you just LIED.
>>
>> FAIL.
>
> I am not going over every point of your dishonest dodge.
> I state a necessary truth that cannot possisbly have any correct
> rebuttal and you provide some incoherent double talk that superficially
> appears to be a rebuttal to people paying less than 1% of their attention.
>

No, you state a LIE, as I have proven. If what I say is untrue, then you
can point out the error, like I have to what you have said.

Your FAILURE to do so just proves that you have no basis for your
statements, and YOUR comments are the 'dishonest dodge' because YOU are
the one who refuses to engage in an honest dialog.

PERIOD.

You DON'T use the right definitions, so you aren't talking about what
you claim you are making claim abouts.

All you have done is just PROVED that your logic must be unsound.

FAIL.

On 1/13/2022 7:39 PM, Richard Damon wrote:
> On 1/13/22 7:35 PM, olcott wrote:
>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>> On 1/13/22 7:11 PM, olcott wrote:
>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never
>>>>>> stop running.
>>>>>>
>>>>>
>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>> that does this will do. It just proves that an H^ built from an H
>>>>> that doesn't abort will be non-halting, but that H won't answer, so
>>>>> is wrong.
>>>>>
>>>>
>>>>>> A simulating halt decider simulates N steps of its input until its
>>>>>> input halts on its own or it correctly determines that a pure
>>>>>> simulation of its input would never stop running.
>>>>>>
>>>>>
>>>>> WRONG. Already proven incorrect. FAIL.
>>>>
>>>> You have never provided any proof all that you have ever had is
>>>> dogmatic assertions.
>>>
>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>> REFUTED, FAIR WARNING.
>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>> Message-ID: <j8sCJ.29853$8Q7.4996@fx10.iad>
>>>
>>>
>>> You have NEVER provided a proper rebuttal to this, so you have thus
>>> conceded that you have no rebuttal to it
>>>
>>> And thus you just LIED.
>>>
>>> FAIL.
>>
>> I am not going over every point of your dishonest dodge.
>> I state a necessary truth that cannot possisbly have any correct
>> rebuttal and you provide some incoherent double talk that
>> superficially appears to be a rebuttal to people paying less than 1%
>> of their attention.
>>
>
> No, you state a LIE, as I have proven. If what I say is untrue, then you
> can point out the error, like I have to what you have said.
>

The error is that your scatter brain couldn't even stay focused on the
point. You couldn't even pay attention to the fact that I was not
talking about Ĥ.

void P(ptr x)
{ if (H(x, y))
HERE: goto HERE;
}

When H bases its halt status decision on the behavior of the simulation
of its input then P demonstrates an infinitely repeating pattern that
cannot possibly ever reach its final state.

--
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 Friday, 14 January 2022 at 09:52:56 UTC+8, olcott wrote:
> On 1/13/2022 7:39 PM, Richard Damon wrote:
> > On 1/13/22 7:35 PM, olcott wrote:
> >> On 1/13/2022 6:14 PM, Richard Damon wrote:
> >>> On 1/13/22 7:11 PM, olcott wrote:
> >>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
> >>>>> On 1/13/22 3:33 PM, olcott wrote:
> >>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> >>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> >>>>>>
> >>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
> >>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never
> >>>>>> stop running.
> >>>>>>
> >>>>>
> >>>>> So, that doesn't say anything about what the H^ built from an H
> >>>>> that does this will do. It just proves that an H^ built from an H
> >>>>> that doesn't abort will be non-halting, but that H won't answer, so
> >>>>> is wrong.
> >>>>>
> >>>>
> >>>>>> A simulating halt decider simulates N steps of its input until its
> >>>>>> input halts on its own or it correctly determines that a pure
> >>>>>> simulation of its input would never stop running.
> >>>>>>
> >>>>>
> >>>>> WRONG. Already proven incorrect. FAIL.
> >>>>
> >>>> You have never provided any proof all that you have ever had is
> >>>> dogmatic assertions.
> >>>
> >>> Subject: Re: Concise refutation of halting problem proofs V47,
> >>> REFUTED, FAIR WARNING.
> >>> Date: Sun, 9 Jan 2022 04:55:11 +0000
> >>> Message-ID: <j8sCJ.29853$8Q7....@fx10.iad>
> >>>
> >>>
> >>> You have NEVER provided a proper rebuttal to this, so you have thus
> >>> conceded that you have no rebuttal to it
> >>>
> >>> And thus you just LIED.
> >>>
> >>> FAIL.
> >>
> >> I am not going over every point of your dishonest dodge.
> >> I state a necessary truth that cannot possisbly have any correct
> >> rebuttal and you provide some incoherent double talk that
> >> superficially appears to be a rebuttal to people paying less than 1%
> >> of their attention.
> >>
> >
> > No, you state a LIE, as I have proven. If what I say is untrue, then you
> > can point out the error, like I have to what you have said.
> >
> The error is that your scatter brain couldn't even stay focused on the
> point. You couldn't even pay attention to the fact that I was not
> talking about Ĥ.
>
> void P(ptr x)
> {
> if (H(x, y))
> HERE: goto HERE;
> }
>
>
> When H bases its halt status decision on the behavior of the simulation
> of its input then P demonstrates an infinitely repeating pattern that
> cannot possibly ever reach its final state.

This is the logical and real result. The Halt Decider cannot possibly ever
reach its final state. This is exactly the ANSWER.
If you don't agree, you have to unsay these words of yours.

> --
> 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 1/13/22 8:52 PM, olcott wrote:
> On 1/13/2022 7:39 PM, Richard Damon wrote:
>> On 1/13/22 7:35 PM, olcott wrote:
>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>
>>>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx
>>>>>>> must abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would
>>>>>>> never stop running.
>>>>>>>
>>>>>>
>>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>>> that does this will do. It just proves that an H^ built from an H
>>>>>> that doesn't abort will be non-halting, but that H won't answer,
>>>>>> so is wrong.
>>>>>>
>>>>>
>>>>>>> A simulating halt decider simulates N steps of its input until
>>>>>>> its input halts on its own or it correctly determines that a pure
>>>>>>> simulation of its input would never stop running.
>>>>>>>
>>>>>>
>>>>>> WRONG. Already proven incorrect. FAIL.
>>>>>
>>>>> You have never provided any proof all that you have ever had is
>>>>> dogmatic assertions.
>>>>
>>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>>> REFUTED, FAIR WARNING.
>>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>>> Message-ID: <j8sCJ.29853$8Q7.4996@fx10.iad>
>>>>
>>>>
>>>> You have NEVER provided a proper rebuttal to this, so you have thus
>>>> conceded that you have no rebuttal to it
>>>>
>>>> And thus you just LIED.
>>>>
>>>> FAIL.
>>>
>>> I am not going over every point of your dishonest dodge.
>>> I state a necessary truth that cannot possisbly have any correct
>>> rebuttal and you provide some incoherent double talk that
>>> superficially appears to be a rebuttal to people paying less than 1%
>>> of their attention.
>>>
>>
>> No, you state a LIE, as I have proven. If what I say is untrue, then
>> you can point out the error, like I have to what you have said.
>>
>
> The error is that your scatter brain couldn't even stay focused on the
> point. You couldn't even pay attention to the fact that I was not
> talking about Ĥ.

Then you aren't talking about Linz or the real Halting Problem, and are
just spreading lies that you are.

FAIL.

(By the way, read what you wrote about previously, you WERE talking
about H^, you need to keep your lies straighter).

>
> void P(ptr x)
> {
>   if (H(x, y))
>     HERE: goto HERE;
> }
>

Note, this is the first reference to P in this thread, so it is just
your attempt to divert attention from your fundamental problems.

>
> When H bases its halt status decision on the behavior of the simulation
> of its input then P demonstrates an infinitely repeating pattern that
> cannot possibly ever reach its final state.
>

Excet that P CAN'T have an infinitely repeating pattern for any H that
does abort it.

So, either you have an H that doesn't abort, and thus never answers, or
you do have an H that does abort, and gets the wrong answer.

Which way are you wrong?

Remember, H is a DEFINITE decider, not some quantum mechanics uncertain
machine.

Your logic shows H getting the right answer, but for the wrong machine,
so it doesn't actually prove anything.

Note. your H/P system has a fundamental issue when you try to convert it
to the Turing Machine H/H^ system. The method you have defined for how H
works doesn't translate to Turing Machines, because your H is entangled
with its input, so the input is NOT isolated properly. This allows you
to make your recursion test.

Define your inputs properly so that H can take ANY program as an input,
not just programs with H loaded in the same location of memory as the
deciding H has it, and you will find that you can't do what you are
trying to do.

On 1/13/2022 8:24 PM, wij wrote:
> On Friday, 14 January 2022 at 09:52:56 UTC+8, olcott wrote:
>> On 1/13/2022 7:39 PM, Richard Damon wrote:
>>> On 1/13/22 7:35 PM, olcott wrote:
>>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>
>>>>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>>>>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never
>>>>>>>> stop running.
>>>>>>>>
>>>>>>>
>>>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>>>> that does this will do. It just proves that an H^ built from an H
>>>>>>> that doesn't abort will be non-halting, but that H won't answer, so
>>>>>>> is wrong.
>>>>>>>
>>>>>>
>>>>>>>> A simulating halt decider simulates N steps of its input until its
>>>>>>>> input halts on its own or it correctly determines that a pure
>>>>>>>> simulation of its input would never stop running.
>>>>>>>>
>>>>>>>
>>>>>>> WRONG. Already proven incorrect. FAIL.
>>>>>>
>>>>>> You have never provided any proof all that you have ever had is
>>>>>> dogmatic assertions.
>>>>>
>>>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>>>> REFUTED, FAIR WARNING.
>>>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>>>> Message-ID: <j8sCJ.29853$8Q7....@fx10.iad>
>>>>>
>>>>>
>>>>> You have NEVER provided a proper rebuttal to this, so you have thus
>>>>> conceded that you have no rebuttal to it
>>>>>
>>>>> And thus you just LIED.
>>>>>
>>>>> FAIL.
>>>>
>>>> I am not going over every point of your dishonest dodge.
>>>> I state a necessary truth that cannot possisbly have any correct
>>>> rebuttal and you provide some incoherent double talk that
>>>> superficially appears to be a rebuttal to people paying less than 1%
>>>> of their attention.
>>>>
>>>
>>> No, you state a LIE, as I have proven. If what I say is untrue, then you
>>> can point out the error, like I have to what you have said.
>>>
>> The error is that your scatter brain couldn't even stay focused on the
>> point. You couldn't even pay attention to the fact that I was not
>> talking about Ĥ.
>>
>> void P(ptr x)
>> {
>> if (H(x, y))
>> HERE: goto HERE;
>> }
>>
>>
>> When H bases its halt status decision on the behavior of the simulation
>> of its input then P demonstrates an infinitely repeating pattern that
>> cannot possibly ever reach its final state.
>
> This is the logical and real result. The Halt Decider cannot possibly ever
> reach its final state. This is exactly the ANSWER.
> If you don't agree, you have to unsay these words of yours.
>

I spent 2 full-time years creating the actual H that actually correctly
determines the halt status of the above P. H recognizes the essentially
infinitely recursive pattern of the behavior of P.

>> --
>> Copyright 2021 Pete Olcott
>>
>> Talent hits a target no one else can hit;
>> Genius hits a target no one else can see.
>> Arthur Schopenhauer

--
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 1/13/2022 8:27 PM, Richard Damon wrote:
> On 1/13/22 8:52 PM, olcott wrote:
>> On 1/13/2022 7:39 PM, Richard Damon wrote:
>>> On 1/13/22 7:35 PM, olcott wrote:
>>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>
>>>>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx
>>>>>>>> must abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would
>>>>>>>> never stop running.
>>>>>>>>
>>>>>>>
>>>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>>>> that does this will do. It just proves that an H^ built from an H
>>>>>>> that doesn't abort will be non-halting, but that H won't answer,
>>>>>>> so is wrong.
>>>>>>>
>>>>>>
>>>>>>>> A simulating halt decider simulates N steps of its input until
>>>>>>>> its input halts on its own or it correctly determines that a
>>>>>>>> pure simulation of its input would never stop running.
>>>>>>>>
>>>>>>>
>>>>>>> WRONG. Already proven incorrect. FAIL.
>>>>>>
>>>>>> You have never provided any proof all that you have ever had is
>>>>>> dogmatic assertions.
>>>>>
>>>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>>>> REFUTED, FAIR WARNING.
>>>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>>>> Message-ID: <j8sCJ.29853$8Q7.4996@fx10.iad>
>>>>>
>>>>>
>>>>> You have NEVER provided a proper rebuttal to this, so you have thus
>>>>> conceded that you have no rebuttal to it
>>>>>
>>>>> And thus you just LIED.
>>>>>
>>>>> FAIL.
>>>>
>>>> I am not going over every point of your dishonest dodge.
>>>> I state a necessary truth that cannot possisbly have any correct
>>>> rebuttal and you provide some incoherent double talk that
>>>> superficially appears to be a rebuttal to people paying less than 1%
>>>> of their attention.
>>>>
>>>
>>> No, you state a LIE, as I have proven. If what I say is untrue, then
>>> you can point out the error, like I have to what you have said.
>>>
>>
>> The error is that your scatter brain couldn't even stay focused on the
>> point. You couldn't even pay attention to the fact that I was not
>> talking about Ĥ.
>

When I start a whole new thread and only talk about H and P I am not
talking about Ĥ nitwit.

> Then you aren't talking about Linz or the real Halting Problem, and are
> just spreading lies that you are.
>
> FAIL.
>
> (By the way, read what you wrote about previously, you WERE talking
> about H^, you need to keep your lies straighter).
>
>>
>> void P(ptr x)
>> {
>>    if (H(x, y))
>>      HERE: goto HERE;
>> }
>>
>
> Note, this is the first reference to P in this thread, so it is just
> your attempt to divert attention from your fundamental problems.
>
>>
>> When H bases its halt status decision on the behavior of the
>> simulation of its input then P demonstrates an infinitely repeating
>> pattern that cannot possibly ever reach its final state.
>>
>
> Excet that P CAN'T have an infinitely repeating pattern for any H that
> does abort it.
>
> So, either you have an H that doesn't abort, and thus never answers, or
> you do have an H that does abort, and gets the wrong answer.
>
> Which way are you wrong?
>
> Remember, H is a DEFINITE decider, not some quantum mechanics uncertain
> machine.
>
> Your logic shows H getting the right answer, but for the wrong machine,
> so it doesn't actually prove anything.
>
> Note. your H/P system has a fundamental issue when you try to convert it
> to the Turing Machine H/H^ system. The method you have defined for how H
> works doesn't translate to Turing Machines, because your H is entangled
> with its input, so the input is NOT isolated properly. This allows you
> to make your recursion test.
>
> Define your inputs properly so that H can take ANY program as an input,
> not just programs with H loaded in the same location of memory as the
> deciding H has it, and you will find that you can't do what you are
> trying to do.

--
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 1/13/22 9:40 PM, olcott wrote:
> On 1/13/2022 8:27 PM, Richard Damon wrote:
>> On 1/13/22 8:52 PM, olcott wrote:
>>> On 1/13/2022 7:39 PM, Richard Damon wrote:
>>>> On 1/13/22 7:35 PM, olcott wrote:
>>>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>
>>>>>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx
>>>>>>>>> must abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩
>>>>>>>>> would never stop running.
>>>>>>>>>

> When I start a whole new thread and only talk about H and P I am not
> talking about Ĥ nitwit.
>

Please read how you started THIS thread.

Care to retract your LIE?

This is your typical error.

FAIL.

Just shows how much 'truth' means to you.

On 1/13/22 9:39 PM, olcott wrote:
> On 1/13/2022 8:24 PM, wij wrote:
>> On Friday, 14 January 2022 at 09:52:56 UTC+8, olcott wrote:
>>> On 1/13/2022 7:39 PM, Richard Damon wrote:
>>>> On 1/13/22 7:35 PM, olcott wrote:
>>>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>
>>>>>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>>>>>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never
>>>>>>>>> stop running.
>>>>>>>>>
>>>>>>>>
>>>>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>>>>> that does this will do. It just proves that an H^ built from an H
>>>>>>>> that doesn't abort will be non-halting, but that H won't answer, so
>>>>>>>> is wrong.
>>>>>>>>
>>>>>>>
>>>>>>>>> A simulating halt decider simulates N steps of its input until its
>>>>>>>>> input halts on its own or it correctly determines that a pure
>>>>>>>>> simulation of its input would never stop running.
>>>>>>>>>
>>>>>>>>
>>>>>>>> WRONG. Already proven incorrect. FAIL.
>>>>>>>
>>>>>>> You have never provided any proof all that you have ever had is
>>>>>>> dogmatic assertions.
>>>>>>
>>>>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>>>>> REFUTED, FAIR WARNING.
>>>>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>>>>> Message-ID: <j8sCJ.29853$8Q7....@fx10.iad>
>>>>>>
>>>>>>
>>>>>> You have NEVER provided a proper rebuttal to this, so you have thus
>>>>>> conceded that you have no rebuttal to it
>>>>>>
>>>>>> And thus you just LIED.
>>>>>>
>>>>>> FAIL.
>>>>>
>>>>> I am not going over every point of your dishonest dodge.
>>>>> I state a necessary truth that cannot possisbly have any correct
>>>>> rebuttal and you provide some incoherent double talk that
>>>>> superficially appears to be a rebuttal to people paying less than 1%
>>>>> of their attention.
>>>>>
>>>>
>>>> No, you state a LIE, as I have proven. If what I say is untrue, then
>>>> you
>>>> can point out the error, like I have to what you have said.
>>>>
>>> The error is that your scatter brain couldn't even stay focused on the
>>> point. You couldn't even pay attention to the fact that I was not
>>> talking about Ĥ.
>>>
>>> void P(ptr x)
>>> {
>>> if (H(x, y))
>>> HERE: goto HERE;
>>> }
>>>
>>>
>>> When H bases its halt status decision on the behavior of the simulation
>>> of its input then P demonstrates an infinitely repeating pattern that
>>> cannot possibly ever reach its final state.
>>
>> This is the logical and real result. The Halt Decider cannot possibly
>> ever
>> reach its final state. This is exactly the ANSWER.
>> If you don't agree, you have to unsay these words of yours.
>>
>
> I spent 2 full-time years creating the actual H that actually correctly
> determines the halt status of the above P. H recognizes the essentially
> infinitely recursive pattern of the behavior of P.

You mean you WASTED 2 full-time years, as P doesn't have an infinitely
recursive pattern for an H that aborts its simulation as this H does.

All it does is either get the right answer for the wrong problem or the
wrong answer for the proper problem.

>
>>> --
>>> 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 1/13/2022 8:49 PM, Richard Damon wrote:
> On 1/13/22 9:39 PM, olcott wrote:
>> On 1/13/2022 8:24 PM, wij wrote:
>>> On Friday, 14 January 2022 at 09:52:56 UTC+8, olcott wrote:
>>>> On 1/13/2022 7:39 PM, Richard Damon wrote:
>>>>> On 1/13/22 7:35 PM, olcott wrote:
>>>>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>
>>>>>>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx
>>>>>>>>>> must
>>>>>>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never
>>>>>>>>>> stop running.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>>>>>> that does this will do. It just proves that an H^ built from an H
>>>>>>>>> that doesn't abort will be non-halting, but that H won't
>>>>>>>>> answer, so
>>>>>>>>> is wrong.
>>>>>>>>>
>>>>>>>>
>>>>>>>>>> A simulating halt decider simulates N steps of its input until
>>>>>>>>>> its
>>>>>>>>>> input halts on its own or it correctly determines that a pure
>>>>>>>>>> simulation of its input would never stop running.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> WRONG. Already proven incorrect. FAIL.
>>>>>>>>
>>>>>>>> You have never provided any proof all that you have ever had is
>>>>>>>> dogmatic assertions.
>>>>>>>
>>>>>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>>>>>> REFUTED, FAIR WARNING.
>>>>>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>>>>>> Message-ID: <j8sCJ.29853$8Q7....@fx10.iad>
>>>>>>>
>>>>>>>
>>>>>>> You have NEVER provided a proper rebuttal to this, so you have thus
>>>>>>> conceded that you have no rebuttal to it
>>>>>>>
>>>>>>> And thus you just LIED.
>>>>>>>
>>>>>>> FAIL.
>>>>>>
>>>>>> I am not going over every point of your dishonest dodge.
>>>>>> I state a necessary truth that cannot possisbly have any correct
>>>>>> rebuttal and you provide some incoherent double talk that
>>>>>> superficially appears to be a rebuttal to people paying less than 1%
>>>>>> of their attention.
>>>>>>
>>>>>
>>>>> No, you state a LIE, as I have proven. If what I say is untrue,
>>>>> then you
>>>>> can point out the error, like I have to what you have said.
>>>>>
>>>> The error is that your scatter brain couldn't even stay focused on the
>>>> point. You couldn't even pay attention to the fact that I was not
>>>> talking about Ĥ.
>>>>
>>>> void P(ptr x)
>>>> {
>>>> if (H(x, y))
>>>> HERE: goto HERE;
>>>> }
>>>>
>>>>
>>>> When H bases its halt status decision on the behavior of the simulation
>>>> of its input then P demonstrates an infinitely repeating pattern that
>>>> cannot possibly ever reach its final state.
>>>
>>> This is the logical and real result. The Halt Decider cannot possibly
>>> ever
>>> reach its final state. This is exactly the ANSWER.
>>> If you don't agree, you have to unsay these words of yours.
>>>
>>
>> I spent 2 full-time years creating the actual H that actually
>> correctly determines the halt status of the above P. H recognizes the
>> essentially infinitely recursive pattern of the behavior of P.
>
> You mean you WASTED 2 full-time years, as P doesn't have an infinitely
> recursive pattern for an H that aborts its simulation as this H does.
>

I would rather talk to no one than talk to you.

> All it does is either get the right answer for the wrong problem or the
> wrong answer for the proper problem.
>
>>
>>>> --
>>>> Copyright 2021 Pete Olcott
>>>>
>>>> Talent hits a target no one else can hit;
>>>> Genius hits a target no one else can see.
>>>> Arthur Schopenhauer
>>
>>
>

--
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 1/13/22 9:59 PM, olcott wrote:
> On 1/13/2022 8:49 PM, Richard Damon wrote:
>> On 1/13/22 9:39 PM, olcott wrote:
>>> On 1/13/2022 8:24 PM, wij wrote:
>>>> On Friday, 14 January 2022 at 09:52:56 UTC+8, olcott wrote:
>>>>> On 1/13/2022 7:39 PM, Richard Damon wrote:
>>>>>> On 1/13/22 7:35 PM, olcott wrote:
>>>>>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>>>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>
>>>>>>>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx
>>>>>>>>>>> must
>>>>>>>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would
>>>>>>>>>>> never
>>>>>>>>>>> stop running.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>>>>>>> that does this will do. It just proves that an H^ built from an H
>>>>>>>>>> that doesn't abort will be non-halting, but that H won't
>>>>>>>>>> answer, so
>>>>>>>>>> is wrong.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>>>> A simulating halt decider simulates N steps of its input
>>>>>>>>>>> until its
>>>>>>>>>>> input halts on its own or it correctly determines that a pure
>>>>>>>>>>> simulation of its input would never stop running.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> WRONG. Already proven incorrect. FAIL.
>>>>>>>>>
>>>>>>>>> You have never provided any proof all that you have ever had is
>>>>>>>>> dogmatic assertions.
>>>>>>>>
>>>>>>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>>>>>>> REFUTED, FAIR WARNING.
>>>>>>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>>>>>>> Message-ID: <j8sCJ.29853$8Q7....@fx10.iad>
>>>>>>>>
>>>>>>>>
>>>>>>>> You have NEVER provided a proper rebuttal to this, so you have thus
>>>>>>>> conceded that you have no rebuttal to it
>>>>>>>>
>>>>>>>> And thus you just LIED.
>>>>>>>>
>>>>>>>> FAIL.
>>>>>>>
>>>>>>> I am not going over every point of your dishonest dodge.
>>>>>>> I state a necessary truth that cannot possisbly have any correct
>>>>>>> rebuttal and you provide some incoherent double talk that
>>>>>>> superficially appears to be a rebuttal to people paying less than 1%
>>>>>>> of their attention.
>>>>>>>
>>>>>>
>>>>>> No, you state a LIE, as I have proven. If what I say is untrue,
>>>>>> then you
>>>>>> can point out the error, like I have to what you have said.
>>>>>>
>>>>> The error is that your scatter brain couldn't even stay focused on the
>>>>> point. You couldn't even pay attention to the fact that I was not
>>>>> talking about Ĥ.
>>>>>
>>>>> void P(ptr x)
>>>>> {
>>>>> if (H(x, y))
>>>>> HERE: goto HERE;
>>>>> }
>>>>>
>>>>>
>>>>> When H bases its halt status decision on the behavior of the
>>>>> simulation
>>>>> of its input then P demonstrates an infinitely repeating pattern that
>>>>> cannot possibly ever reach its final state.
>>>>
>>>> This is the logical and real result. The Halt Decider cannot
>>>> possibly ever
>>>> reach its final state. This is exactly the ANSWER.
>>>> If you don't agree, you have to unsay these words of yours.
>>>>
>>>
>>> I spent 2 full-time years creating the actual H that actually
>>> correctly determines the halt status of the above P. H recognizes the
>>> essentially infinitely recursive pattern of the behavior of P.
>>
>> You mean you WASTED 2 full-time years, as P doesn't have an infinitely
>> recursive pattern for an H that aborts its simulation as this H does.
>>
>
> I would rather talk to no one than talk to you.

Because I tell you the truth you don't want to hear?

I HAVE proven your logic is broken, and pointed out exactly why.

You can ignore me if you want, but my response will just confirm that
you are avoiding dealing with the truth and burying yourself in your lies.

You HAVE wasted many years on your fruitless search for something that
just is not true because you can't face the prospects that you are wrong.

Sorry, but you are wrong and have shown that you really don't understand
what truth actually means.

>
>> All it does is either get the right answer for the wrong problem or
>> the wrong answer for the proper problem.
>>
>>>
>>>>> --
>>>>> 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 1/13/2022 11:09 PM, Richard Damon wrote:
> On 1/13/22 9:59 PM, olcott wrote:
>> On 1/13/2022 8:49 PM, Richard Damon wrote:
>>> On 1/13/22 9:39 PM, olcott wrote:
>>>> On 1/13/2022 8:24 PM, wij wrote:
>>>>> On Friday, 14 January 2022 at 09:52:56 UTC+8, olcott wrote:
>>>>>> On 1/13/2022 7:39 PM, Richard Damon wrote:
>>>>>>> On 1/13/22 7:35 PM, olcott wrote:
>>>>>>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>>>>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>>>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>
>>>>>>>>>>>> It is the case that the copy of H (called embedded_H) at
>>>>>>>>>>>> Ĥ.qx must
>>>>>>>>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would
>>>>>>>>>>>> never
>>>>>>>>>>>> stop running.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>>>>>>>> that does this will do. It just proves that an H^ built from
>>>>>>>>>>> an H
>>>>>>>>>>> that doesn't abort will be non-halting, but that H won't
>>>>>>>>>>> answer, so
>>>>>>>>>>> is wrong.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>>> A simulating halt decider simulates N steps of its input
>>>>>>>>>>>> until its
>>>>>>>>>>>> input halts on its own or it correctly determines that a pure
>>>>>>>>>>>> simulation of its input would never stop running.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> WRONG. Already proven incorrect. FAIL.
>>>>>>>>>>
>>>>>>>>>> You have never provided any proof all that you have ever had is
>>>>>>>>>> dogmatic assertions.
>>>>>>>>>
>>>>>>>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>>>>>>>> REFUTED, FAIR WARNING.
>>>>>>>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>>>>>>>> Message-ID: <j8sCJ.29853$8Q7....@fx10.iad>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> You have NEVER provided a proper rebuttal to this, so you have
>>>>>>>>> thus
>>>>>>>>> conceded that you have no rebuttal to it
>>>>>>>>>
>>>>>>>>> And thus you just LIED.
>>>>>>>>>
>>>>>>>>> FAIL.
>>>>>>>>
>>>>>>>> I am not going over every point of your dishonest dodge.
>>>>>>>> I state a necessary truth that cannot possisbly have any correct
>>>>>>>> rebuttal and you provide some incoherent double talk that
>>>>>>>> superficially appears to be a rebuttal to people paying less
>>>>>>>> than 1%
>>>>>>>> of their attention.
>>>>>>>>
>>>>>>>
>>>>>>> No, you state a LIE, as I have proven. If what I say is untrue,
>>>>>>> then you
>>>>>>> can point out the error, like I have to what you have said.
>>>>>>>
>>>>>> The error is that your scatter brain couldn't even stay focused on
>>>>>> the
>>>>>> point. You couldn't even pay attention to the fact that I was not
>>>>>> talking about Ĥ.
>>>>>>
>>>>>> void P(ptr x)
>>>>>> {
>>>>>> if (H(x, y))
>>>>>> HERE: goto HERE;
>>>>>> }
>>>>>>
>>>>>>
>>>>>> When H bases its halt status decision on the behavior of the
>>>>>> simulation
>>>>>> of its input then P demonstrates an infinitely repeating pattern that
>>>>>> cannot possibly ever reach its final state.
>>>>>
>>>>> This is the logical and real result. The Halt Decider cannot
>>>>> possibly ever
>>>>> reach its final state. This is exactly the ANSWER.
>>>>> If you don't agree, you have to unsay these words of yours.
>>>>>
>>>>
>>>> I spent 2 full-time years creating the actual H that actually
>>>> correctly determines the halt status of the above P. H recognizes
>>>> the essentially infinitely recursive pattern of the behavior of P.
>>>
>>> You mean you WASTED 2 full-time years, as P doesn't have an
>>> infinitely recursive pattern for an H that aborts its simulation as
>>> this H does.
>>>
>>
>> I would rather talk to no one than talk to you.
>
> Because I tell you the truth you don't want to hear?
>
> I HAVE proven your logic is broken, and pointed out exactly why.
>

Because you can't even pay attention to what is actually being said.

> You can ignore me if you want, but my response will just confirm that
> you are avoiding dealing with the truth and burying yourself in your lies.
>
> You HAVE wasted many years on your fruitless search for something that
> just is not true because you can't face the prospects that you are wrong.
>
> Sorry, but you are wrong and have shown that you really don't understand
> what truth actually means.
>
>>
>>> All it does is either get the right answer for the wrong problem or
>>> the wrong answer for the proper problem.
>>>
>>>>
>>>>>> --
>>>>>> Copyright 2021 Pete Olcott
>>>>>>
>>>>>> Talent hits a target no one else can hit;
>>>>>> Genius hits a target no one else can see.
>>>>>> Arthur Schopenhauer
>>>>
>>>>
>>>
>>
>>
>

--
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 1/14/22 12:28 AM, olcott wrote:
> On 1/13/2022 11:09 PM, Richard Damon wrote:
>> On 1/13/22 9:59 PM, olcott wrote:
>>> On 1/13/2022 8:49 PM, Richard Damon wrote:
>>>> On 1/13/22 9:39 PM, olcott wrote:
>>>>> On 1/13/2022 8:24 PM, wij wrote:
>>>>>> On Friday, 14 January 2022 at 09:52:56 UTC+8, olcott wrote:
>>>>>>> On 1/13/2022 7:39 PM, Richard Damon wrote:
>>>>>>>> On 1/13/22 7:35 PM, olcott wrote:
>>>>>>>>> On 1/13/2022 6:14 PM, Richard Damon wrote:
>>>>>>>>>> On 1/13/22 7:11 PM, olcott wrote:
>>>>>>>>>>> On 1/13/2022 5:55 PM, Richard Damon wrote:
>>>>>>>>>>>> On 1/13/22 3:33 PM, olcott wrote:
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>
>>>>>>>>>>>>> It is the case that the copy of H (called embedded_H) at
>>>>>>>>>>>>> Ĥ.qx must
>>>>>>>>>>>>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would
>>>>>>>>>>>>> never
>>>>>>>>>>>>> stop running.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> So, that doesn't say anything about what the H^ built from an H
>>>>>>>>>>>> that does this will do. It just proves that an H^ built from
>>>>>>>>>>>> an H
>>>>>>>>>>>> that doesn't abort will be non-halting, but that H won't
>>>>>>>>>>>> answer, so
>>>>>>>>>>>> is wrong.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>>> A simulating halt decider simulates N steps of its input
>>>>>>>>>>>>> until its
>>>>>>>>>>>>> input halts on its own or it correctly determines that a pure
>>>>>>>>>>>>> simulation of its input would never stop running.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> WRONG. Already proven incorrect. FAIL.
>>>>>>>>>>>
>>>>>>>>>>> You have never provided any proof all that you have ever had is
>>>>>>>>>>> dogmatic assertions.
>>>>>>>>>>
>>>>>>>>>> Subject: Re: Concise refutation of halting problem proofs V47,
>>>>>>>>>> REFUTED, FAIR WARNING.
>>>>>>>>>> Date: Sun, 9 Jan 2022 04:55:11 +0000
>>>>>>>>>> Message-ID: <j8sCJ.29853$8Q7....@fx10.iad>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You have NEVER provided a proper rebuttal to this, so you have
>>>>>>>>>> thus
>>>>>>>>>> conceded that you have no rebuttal to it
>>>>>>>>>>
>>>>>>>>>> And thus you just LIED.
>>>>>>>>>>
>>>>>>>>>> FAIL.
>>>>>>>>>
>>>>>>>>> I am not going over every point of your dishonest dodge.
>>>>>>>>> I state a necessary truth that cannot possisbly have any correct
>>>>>>>>> rebuttal and you provide some incoherent double talk that
>>>>>>>>> superficially appears to be a rebuttal to people paying less
>>>>>>>>> than 1%
>>>>>>>>> of their attention.
>>>>>>>>>
>>>>>>>>
>>>>>>>> No, you state a LIE, as I have proven. If what I say is untrue,
>>>>>>>> then you
>>>>>>>> can point out the error, like I have to what you have said.
>>>>>>>>
>>>>>>> The error is that your scatter brain couldn't even stay focused
>>>>>>> on the
>>>>>>> point. You couldn't even pay attention to the fact that I was not
>>>>>>> talking about Ĥ.
>>>>>>>
>>>>>>> void P(ptr x)
>>>>>>> {
>>>>>>> if (H(x, y))
>>>>>>> HERE: goto HERE;
>>>>>>> }
>>>>>>>
>>>>>>>
>>>>>>> When H bases its halt status decision on the behavior of the
>>>>>>> simulation
>>>>>>> of its input then P demonstrates an infinitely repeating pattern
>>>>>>> that
>>>>>>> cannot possibly ever reach its final state.
>>>>>>
>>>>>> This is the logical and real result. The Halt Decider cannot
>>>>>> possibly ever
>>>>>> reach its final state. This is exactly the ANSWER.
>>>>>> If you don't agree, you have to unsay these words of yours.
>>>>>>
>>>>>
>>>>> I spent 2 full-time years creating the actual H that actually
>>>>> correctly determines the halt status of the above P. H recognizes
>>>>> the essentially infinitely recursive pattern of the behavior of P.
>>>>
>>>> You mean you WASTED 2 full-time years, as P doesn't have an
>>>> infinitely recursive pattern for an H that aborts its simulation as
>>>> this H does.
>>>>
>>>
>>> I would rather talk to no one than talk to you.
>>
>> Because I tell you the truth you don't want to hear?
>>
>> I HAVE proven your logic is broken, and pointed out exactly why.
>>
>
> Because you can't even pay attention to what is actually being said.

That seems to be YOUR problem, isn't it.

On Thursday, January 13, 2022 at 9:34:08 PM UTC+1, olcott wrote:
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
> the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.
>
> A simulating halt decider simulates N steps of its input until its input
> halts on its own or it correctly determines that a pure simulation of
> its input would never stop running.
>
> 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

A Turing machine applied to some input reaches one final state or one infinite loop, not two different ones. This is a consequence of the definition of determinism.

On 1/14/2022 10:36 AM, Alex C wrote:
> On Thursday, January 13, 2022 at 9:34:08 PM UTC+1, olcott wrote:
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
>> the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.
>>
>> A simulating halt decider simulates N steps of its input until its input
>> halts on its own or it correctly determines that a pure simulation of
>> its input would never stop running.
>>
>> 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
>
> A Turing machine applied to some input reaches one final state or one infinite loop, not two different ones. This is a consequence of the definition of determinism.

In the case of Linz Ĥ applied to ⟨Ĥ⟩, embedded_H at Ĥ.qx correctly
transitions to its final state of Ĥ.qn on the basis that a mathematical
mapping between ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.qn is proven. It is proven on the basis
that the input ⟨Ĥ⟩ ⟨Ĥ⟩ specifies infinitely nested simulation to the
simulating halt decider of embedded_H.

--
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 1/13/2022 2:33 PM, olcott wrote:
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
> the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.
>
> A simulating halt decider simulates N steps of its input until its input
> halts on its own or it correctly determines that a pure simulation of
> its input would never stop running.
>
> 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)
>

dbush said:
@polcott There is a point of agreement, namely that you've successfully
shown that a simulating H cannot simulate the P built from it to
completion. Your confusion is that you think this is the same as
not-halting. By definition, it is not.

It must be a point of mutual agreement or it does not count as a point
of agreement.

I have successfully shown that sequences of configurations that never
reach the final state of this sequence can be recognized as meeting the
Linz definition of non-halting computations.

--
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 Friday, January 14, 2022 at 5:59:13 PM UTC+1, olcott wrote:
> On 1/14/2022 10:36 AM, Alex C wrote:
> > On Thursday, January 13, 2022 at 9:34:08 PM UTC+1, olcott wrote:
> >> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> >> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> >>
> >> It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
> >> the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.
> >>
> >> A simulating halt decider simulates N steps of its input until its input
> >> halts on its own or it correctly determines that a pure simulation of
> >> its input would never stop running.
> >>
> >> 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
> >
> > A Turing machine applied to some input reaches one final state or one infinite loop, not two different ones. This is a consequence of the definition of determinism.
> In the case of Linz Ĥ applied to ⟨Ĥ⟩, embedded_H at Ĥ.qx correctly
> transitions to its final state of Ĥ.qn on the basis that a mathematical
> mapping between ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.qn is proven. It is proven on the basis
> that the input ⟨Ĥ⟩ ⟨Ĥ⟩ specifies infinitely nested simulation to the
> simulating halt decider of embedded_H.
> --
> Copyright 2021 Pete Olcott
>
> Talent hits a target no one else can hit;
> Genius hits a target no one else can see.
> Arthur Schopenhauer

Ĥ.q0 ⟨Ĥ⟩ ⊢* ??
is a question. A question that has only one answer. We can find the answer by passing the arguments ⟨Ĥ.q0⟩ and ⟨Ĥ⟩ to a UTM. Otherwise it is like saying:

2+2 = 4
2+2 = 5

On 1/14/2022 12:11 PM, Alex C wrote:
> On Friday, January 14, 2022 at 5:59:13 PM UTC+1, olcott wrote:
>> On 1/14/2022 10:36 AM, Alex C wrote:
>>> On Thursday, January 13, 2022 at 9:34:08 PM UTC+1, olcott wrote:
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
>>>> the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.
>>>>
>>>> A simulating halt decider simulates N steps of its input until its input
>>>> halts on its own or it correctly determines that a pure simulation of
>>>> its input would never stop running.
>>>>
>>>> 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
>>>
>>> A Turing machine applied to some input reaches one final state or one infinite loop, not two different ones. This is a consequence of the definition of determinism.
>> In the case of Linz Ĥ applied to ⟨Ĥ⟩, embedded_H at Ĥ.qx correctly
>> transitions to its final state of Ĥ.qn on the basis that a mathematical
>> mapping between ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.qn is proven. It is proven on the basis
>> that the input ⟨Ĥ⟩ ⟨Ĥ⟩ specifies infinitely nested simulation to the
>> simulating halt decider of embedded_H.
>> --
>> Copyright 2021 Pete Olcott
>>
>> Talent hits a target no one else can hit;
>> Genius hits a target no one else can see.
>> Arthur Schopenhauer
>
> Ĥ.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ??
> is a question. A question that has only one answer. We can find the answer by passing the arguments ⟨Ĥ.q0⟩ and ⟨Ĥ⟩ to a UTM. Otherwise it is like saying:
>
> 2+2 = 4
> 2+2 = 5

That is not the question and you know it.
I can't really understand your motivation to lie.

This is the actual question: embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢*
and the answer is: Ĥ.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 1/14/2022 12:49 PM, olcott wrote:
> On 1/14/2022 12:11 PM, Alex C wrote:
>> On Friday, January 14, 2022 at 5:59:13 PM UTC+1, olcott wrote:
>>> On 1/14/2022 10:36 AM, Alex C wrote:
>>>> On Thursday, January 13, 2022 at 9:34:08 PM UTC+1, olcott wrote:
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>
>>>>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>>>>> abort
>>>>> the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop
>>>>> running.
>>>>>
>>>>> A simulating halt decider simulates N steps of its input until its
>>>>> input
>>>>> halts on its own or it correctly determines that a pure simulation of
>>>>> its input would never stop running.
>>>>>
>>>>> 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
>>>>
>>>> A Turing machine applied to some input reaches one final state or
>>>> one infinite loop, not two different ones. This is a consequence of
>>>> the definition of determinism.
>>> In the case of Linz Ĥ applied to ⟨Ĥ⟩, embedded_H at Ĥ.qx correctly
>>> transitions to its final state of Ĥ.qn on the basis that a mathematical
>>> mapping between ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.qn is proven. It is proven on the basis
>>> that the input ⟨Ĥ⟩ ⟨Ĥ⟩ specifies infinitely nested simulation to the
>>> simulating halt decider of embedded_H.
>>> --
>>> Copyright 2021 Pete Olcott
>>>
>>> Talent hits a target no one else can hit;
>>> Genius hits a target no one else can see.
>>> Arthur Schopenhauer
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ??
>> is a question. A question that has only one answer. We can find the
>> answer by passing the arguments ⟨Ĥ.q0⟩ and ⟨Ĥ⟩ to a UTM. Otherwise it
>> is like saying:
>>
>> 2+2 = 4
>> 2+2 = 5
>
> That is not the question and you know it.

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

There can only be a conditional branch at the point in the execution
trace where there is conditional code. The point in the execution trace
that you pointed to was an unconditional branch.

> I can't really understand your motivation to lie.
>
> This is the actual question: embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢*
> and the answer is: Ĥ.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 1/13/2022 2:33 PM, olcott wrote:
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> It is the case that the copy of H (called embedded_H) at Ĥ.qx must abort
> the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop running.
>
> A simulating halt decider simulates N steps of its input until its input
> halts on its own or it correctly determines that a pure simulation of
> its input would never stop running.
>
> 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)
>

@polcott There is a point of agreement, namely that you've successfully
shown that a simulating H cannot simulate the P built from it to
completion. Your confusion is that you think this is the same as
not-halting. By definition, it is not. – dbush

Proof that my reviewers are not interested in an honest dialogue:
(1) There are never any points of mutual agreement.
(2) Whenever I prove my point reviewers always change the subject rather
than acknowledge that I proved my point.

A simulating halt decider continues to simulate N steps of its input
input until (a) The simulation ends on its own after reaching the final
state specified by the input. (b) An infinite execution pattern has been
recognized.

It is ridiculously stupid to claim that the simulating halt decider
fails on the basis that it cannot simulate an input that specifies
infinite execution to completion.

H cannot simulate P to completion because P has no completion. All the
HP counter-example instance specify infinitely nested simulation to
every simulating halt decider.

--
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 1/14/2022 3:40 PM, olcott wrote:
> On 1/13/2022 2:33 PM, olcott wrote:
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> It is the case that the copy of H (called embedded_H) at Ĥ.qx must
>> abort the simulation of its input or Ĥ applied to ⟨Ĥ⟩ would never stop
>> running.
>>
>> A simulating halt decider simulates N steps of its input until its
>> input halts on its own or it correctly determines that a pure
>> simulation of its input would never stop running.
>>
>> 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)
>>
>
> @polcott There is a point of agreement, namely that you've successfully
> shown that a simulating H cannot simulate the P built from it to
> completion. Your confusion is that you think this is the same as
> not-halting. By definition, it is not. – dbush
>
> Proof that my reviewers are not interested in an honest dialogue:
> (1) There are never any points of mutual agreement.
> (2) Whenever I prove my point reviewers always change the subject rather
> than acknowledge that I proved my point.
>
> A simulating halt decider continues to simulate N steps of its input
> input until (a) The simulation ends on its own after reaching the final
> state specified by the input. (b) An infinite execution pattern has been
> recognized.
>
> It is ridiculously stupid to claim that the simulating halt decider
> fails on the basis that it cannot simulate an input that specifies
> infinite execution to completion.
>
> H cannot simulate P to completion because P has no completion. All the
> HP counter-example instance specify infinitely nested simulation to
> every simulating halt decider.
>

computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)

Because no finite number of N steps of the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by
embedded_H ever reach a final state of Ĥ, embedded_H correctly
determines that its input meets the Linz definition of a computation
that does not halt. At this point embedded_H aborts the simulation of
its input and correctly transitions to Ĥ.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