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devel / comp.theory / Simplest Possible Halting Problem Proof Rebuttal

SubjectAuthor
* Simplest Possible Halting Problem Proof Rebuttalolcott
+- Simplest Possible Halting Problem Proof RebuttalRichard Damon
`* Simplest Possible Halting Problem Proof RebuttalPaul N
 +* Simplest Possible Halting Problem Proof Rebuttalolcott
 |`- Simplest Possible Halting Problem Proof RebuttalRichard Damon
 +- Simplest Possible Halting Problem Proof Rebuttalolcott
 `* Simplest Possible Halting Problem Proof Rebuttalolcott
  +* Simplest Possible Halting Problem Proof Rebuttalolcott
  |`- Simplest Possible Halting Problem Proof RebuttalRichard Damon
  +- Simplest Possible Halting Problem Proof RebuttalRichard Damon
  `- Simplest Possible Halting Problem Proof RebuttalRichard Damon

1
Simplest Possible Halting Problem Proof Rebuttal

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From: polcott2@gmail.com (olcott)
Newsgroups: sci.logic,comp.theory,comp.ai.philosophy
Subject: Simplest Possible Halting Problem Proof Rebuttal
Date: Wed, 18 Oct 2023 22:20:09 -0500
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 by: olcott - Thu, 19 Oct 2023 03:20 UTC

*As soon as this is understood to be correct then*
The inability to do the logically impossible never places any actual
limits on anyone or anything.

Then it is understood that the logical impossibility of solving the
halting problem the way it is currently defined places no actual limit
on computation.

It is equally logically impossible to define a CAD system that correctly
draws square circles.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: Simplest Possible Halting Problem Proof Rebuttal

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From: richard@damon-family.org (Richard Damon)
Newsgroups: sci.logic,comp.theory,comp.ai.philosophy
Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
Date: Wed, 18 Oct 2023 23:36:25 -0400
Organization: i2pn2 (i2pn.org)
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 by: Richard Damon - Thu, 19 Oct 2023 03:36 UTC

On 10/18/23 11:20 PM, olcott wrote:
> *As soon as this is understood to be correct then*

And how soon till you understand that isn't what you are showing?

> The inability to do the logically impossible never places any actual
> limits on anyone or anything.

So, you need to be clear what IS the "logically impossible" task.

Remember, that there IS a correct answer to the question about if a
given machine will halt, this is a simple fact.

What is impossible is to produce a program that always gives that
correct answer.

Thus, we have a limit to what can be computed.

>
> Then it is understood that the logical impossibility of solving the
> halting problem the way it is currently defined places no actual limit
> on computation.

So, you really don't understand the problem do you.

If there is no actual limit on computation, please write the program
that determines the answer to the Collatz conjecture, or the Twin Primes
problem?

How about find the formula to answer the Busy Beaver problem?

Or, are you confusing yourself with bad defintions, and trying to say
that if all possible programs are possible, that means there is no
limits to computation, as any actual limit is just ignored because it
was always a logical impossibility.

>
> It is equally logically impossible to define a CAD system that correctly
> draws square circles.
>

What?

Since it is logically impossible for a given program to give an answer
other than the one it was programmed to give for that input, it is thus
logically impossible for a given program to correctly decide for all
programs if they will halt or not.

It also seem impossible for you to come up with an actual correct
logical argument.

You are just showing you are talking double talk, and buring your head
in the sand.

Re: Simplest Possible Halting Problem Proof Rebuttal

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Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
From: gw7rib@aol.com (Paul N)
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 by: Paul N - Thu, 19 Oct 2023 11:21 UTC

On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
> *As soon as this is understood to be correct then*
> The inability to do the logically impossible never places any actual
> limits on anyone or anything.
>
> Then it is understood that the logical impossibility of solving the
> halting problem the way it is currently defined places no actual limit
> on computation.
>
> It is equally logically impossible to define a CAD system that correctly
> draws square circles.

Exactly. However, the equivalent to what you are saying is to say that that everyone's proof that it is impossible to define a CAD system that draws square circles is wrong, and that you do actually have a CAD system which does so. When other people try to rebut this by pointing out that square circles don't exist, you say that because they don't, your system's failure to draw them is not a problem and that therefore you are perfectly entitled to insist that your system can draw them.

Re: Simplest Possible Halting Problem Proof Rebuttal

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
Date: Thu, 19 Oct 2023 11:24:23 -0500
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 by: olcott - Thu, 19 Oct 2023 16:24 UTC

On 10/19/2023 6:21 AM, Paul N wrote:
> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
>> *As soon as this is understood to be correct then*
>> The inability to do the logically impossible never places any actual
>> limits on anyone or anything.
>>
>> Then it is understood that the logical impossibility of solving the
>> halting problem the way it is currently defined places no actual limit
>> on computation.
>>
>> It is equally logically impossible to define a CAD system that correctly
>> draws square circles.
>
> Exactly. However, the equivalent to what you are saying is to say that that everyone's proof that it is impossible to define a CAD system that draws square circles is wrong, and that you do actually have a CAD system which does so. When other people try to rebut this by pointing out that square circles don't exist, you say that because they don't, your system's failure to draw them is not a problem and that therefore you are perfectly entitled to insist that your system can draw them.

Welcome back, you are my best reviewer because you assess my reasoning
and respond to it rather than spewing boiler-plate replies that ignore
what I say. You are referring to two different proofs that assume two
different halt status criteria.

(a) One is required to report on the direct execution of D(D).
(b) The other reports on D correctly simulated by H.

It is a *logical impossibility* for decider H to return the halt status
of input D that does the opposite of whatever Boolean value that H
returns when H is required to report on the behavior of the direct
execution of D(D).

*The inability to do the logically impossible never places*
*any actual limits on anyone or anything* That no CAD system can
possibly correctly draw a square circle places no actual limits on
computation.

That is the proof that the PhD computer science professor
totally agreed with and independently derived his own version
of this proof. I just spoke with him several times yesterday
and we are on the exact same page. The following is my
alternative proof.

This is a different halt status criteria that reports on the actual
behavior of the actual input on the basis of D correctly simulated by H
that does take into account that D calls H in recursive simulation. The
directly executed D(D) has an entirely different execution trace.

*MIT Professor Michael Sipser has agreed that*
*the following verbatim paragraph is correct*
(a) If simulating halt decider H correctly simulates its input D until H
correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: Simplest Possible Halting Problem Proof Rebuttal

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
Date: Thu, 19 Oct 2023 11:26:59 -0500
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 by: olcott - Thu, 19 Oct 2023 16:26 UTC

On 10/19/2023 6:21 AM, Paul N wrote:
> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
>> *As soon as this is understood to be correct then*
>> The inability to do the logically impossible never places any actual
>> limits on anyone or anything.
>>
>> Then it is understood that the logical impossibility of solving the
>> halting problem the way it is currently defined places no actual limit
>> on computation.
>>
>> It is equally logically impossible to define a CAD system that correctly
>> draws square circles.
>
> Exactly. However, the equivalent to what you are saying is to say that that everyone's proof that it is impossible to define a CAD system that draws square circles is wrong, and that you do actually have a CAD system which does so. When other people try to rebut this by pointing out that square circles don't exist, you say that because they don't, your system's failure to draw them is not a problem and that therefore you are perfectly entitled to insist that your system can draw them.

Welcome back, you are my best reviewer because you assess my reasoning
and respond to it rather than spewing boiler-plate replies that ignore
what I say. You are referring to two different proofs that assume two
different halt status criteria.

(a) One is required to report on the direct execution of D(D).
(b) The other reports on D correctly simulated by H.

It is a *logical impossibility* for decider H to return the halt status
of input D that does the opposite of whatever Boolean value that H
returns when H is required to report on the behavior of the direct
execution of D(D).

*The inability to do the logically impossible never places*
*any actual limits on anyone or anything* That no CAD system can
possibly correctly draw a square circle places no actual limits on
computation.

That is the proof that the PhD computer science professor
totally agreed with and independently derived his own version
of this proof. I just spoke with him several times yesterday
and we are on the exact same page. The following is my
alternative proof.

This is a different halt status criteria that reports on the actual
behavior of the actual input on the basis of D correctly simulated by H
that does take into account that D calls H in recursive simulation. The
directly executed D(D) has an entirely different execution trace.

*MIT Professor Michael Sipser has agreed that*
*the following verbatim paragraph is correct*
(a) If simulating halt decider H correctly simulates its input D until H
correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: Simplest Possible Halting Problem Proof Rebuttal

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From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory
Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
Date: Thu, 19 Oct 2023 12:49:55 -0400
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 by: Richard Damon - Thu, 19 Oct 2023 16:49 UTC

On 10/19/23 12:24 PM, olcott wrote:
> On 10/19/2023 6:21 AM, Paul N wrote:
>> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
>>> *As soon as this is understood to be correct then*
>>> The inability to do the logically impossible never places any actual
>>> limits on anyone or anything.
>>>
>>> Then it is understood that the logical impossibility of solving the
>>> halting problem the way it is currently defined places no actual limit
>>> on computation.
>>>
>>> It is equally logically impossible to define a CAD system that correctly
>>> draws square circles.
>>
>> Exactly. However, the equivalent to what you are saying is to say that
>> that everyone's proof that it is impossible to define a CAD system
>> that draws square circles is wrong, and that you do actually have a
>> CAD system which does so. When other people try to rebut this by
>> pointing out that square circles don't exist, you say that because
>> they don't, your system's failure to draw them is not a problem and
>> that therefore you are perfectly entitled to insist that your system
>> can draw them.
>
> Welcome back, you are my best reviewer because you assess my reasoning
> and respond to it rather than spewing boiler-plate replies that ignore
> what I say. You are referring to two different proofs that assume two
> different halt status criteria.
>
> (a) One is required to report on the direct execution of D(D).
> (b) The other reports on D correctly simulated by H.

There is no "Halt Status Criteria" that is based on the input being
"Correctly Simulated by H".

Thus, any argument using that is just an invald Red Herring.

>
> It is a *logical impossibility* for decider H to return the halt status
> of input D that does the opposite of whatever Boolean value that H
> returns when H is required to report on the behavior of the direct
> execution of D(D).

That is an illogical statement.

It is PHYSICALLY (and logically) impossible for any given decider H to
return an answer other than the one that its programming give it.

A given H will only give the answer it is programmed to do, and thus any
logic that tries to presume otherwise is just invalid.

>
> *The inability to do the logically impossible never places*
> *any actual limits on anyone or anything* That no CAD system can
> possibly correctly draw a square circle places no actual limits on
> computation.

Right,

The answer to the Halting Quesion, your case (a), always exists, and
thus the question is correct.

The logical impossibility to make a decider that always answers that
question correctly (because we can always make the described
pathological program from it) means that you can't possible make such a
decider, and thus the Halting Problem is non-computable (not illogical).

This is just like it is impossible to construct a circle with the same
area as a given square with compass and straight edge. The circle
exists, it just isn't constructable by that method. In the same way, the
Halting Problem question has an answer, it just isn't computable (in
general).

>
> That is the proof that the PhD computer science professor
> totally agreed with and independently derived his own version
> of this proof. I just spoke with him several times yesterday
> and we are on the exact same page. The following is my
> alternative proof.

If you are refering to the statement that you quote below, it has been
pointed out that you just don't understand what he agreed to, because
you don't understand that to an actual Computer Scientist, the only
meaning of a "Correct Simulation" is one that shows the actual behavior
of the program described by the input.

>
> This is a different halt status criteria that reports on the actual
> behavior of the actual input on the basis of D correctly simulated by H
> that does take into account that D calls H in recursive simulation. The
> directly executed D(D) has an entirely different execution trace.
>

Which just admits that what you call a "correct simulation" is not
actually a correct simulation, but just another of your lies.

IF H can determine that D calls H, can try to take into account what H
does, but that taking into account must be what H actually does, not
what some other variant of H might have done, as D specifically used the
H that you are trying to claim is correct.

> *MIT Professor Michael Sipser has agreed that*
> *the following verbatim paragraph is correct*
> (a) If simulating halt decider H correctly simulates its input D until H
> correctly determines that its simulated D would never stop running
> unless aborted then
> (b) H can abort its simulation of D and correctly report that D
> specifies a non-halting sequence of configurations.
>
>
>

Right, so you need to show that H correctly simulates its input, by the
definition of an ACTUAL UTM that exactly reproduces the behavior of the
program described by the input.

Since H doesn't do that, your clause (a) isn't satisfied, and thus you
can't assert clause (b).

Remember, H is a SPECIFIC program. You can't pull the infinite class
trick, as calling it a "Halt Decider" says that it is a SPECIFIC
instance of that class of programs. Since you H does abort its
operation, it didn't correctly simulate the input, or even based its
answer on a "Correct Simulation" (possible done by some other program)
as such a correct simulation will show that, because H aborts and
returns 0, the input program is Halting, and thus can not be correctly
deteremined to not halt.

All you are doing is proving yourself to be a LIAR, and an idiot.

Re: Simplest Possible Halting Problem Proof Rebuttal

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
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 by: olcott - Fri, 20 Oct 2023 18:59 UTC

On 10/19/2023 6:21 AM, Paul N wrote:
> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
>> *As soon as this is understood to be correct then*
>> The inability to do the logically impossible never places any actual
>> limits on anyone or anything.
>>
>> Then it is understood that the logical impossibility of solving the
>> halting problem the way it is currently defined places no actual limit
>> on computation.
>>
>> It is equally logically impossible to define a CAD system that correctly
>> draws square circles.
>
> Exactly. However, the equivalent to what you are saying is to say that that everyone's proof that it is impossible to define a CAD system that draws square circles is wrong, and that you do actually have a CAD system which does so. When other people try to rebut this by pointing out that square circles don't exist, you say that because they don't, your system's failure to draw them is not a problem and that therefore you are perfectly entitled to insist that your system can draw them.

When a halt decider H is required to report on the behavior of the
direct execution of D that does the opposite of whatever H says that
it will do this is is merely a logically impossible requirement exactly
like requiring a CAD system to draw square circles.

When we change the requirement so that it is not logically
impossible then termination analyzer H is correct to report
that D correctly simulated by H will never terminate normally
because D specified recursive simulation to H.

The correct simulation of D by H must include the call from D to H that
specifies that D calls H in recursive simulation.

Consistently all of the reviewers of my work insist that H must ignore
this recursive simulation and report that D(D) halts because when H does
not ignore this recursive simulation and aborts its simulation D(D) does
halt. *They have no idea that their view is inconsistent*

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: Simplest Possible Halting Problem Proof Rebuttal

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
Date: Fri, 20 Oct 2023 15:26:33 -0500
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 by: olcott - Fri, 20 Oct 2023 20:26 UTC

On 10/20/2023 1:59 PM, olcott wrote:
> On 10/19/2023 6:21 AM, Paul N wrote:
>> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
>>> *As soon as this is understood to be correct then*
>>> The inability to do the logically impossible never places any actual
>>> limits on anyone or anything.
>>>
>>> Then it is understood that the logical impossibility of solving the
>>> halting problem the way it is currently defined places no actual limit
>>> on computation.
>>>
>>> It is equally logically impossible to define a CAD system that correctly
>>> draws square circles.
>>
>> Exactly. However, the equivalent to what you are saying is to say that
>> that everyone's proof that it is impossible to define a CAD system
>> that draws square circles is wrong, and that you do actually have a
>> CAD system which does so. When other people try to rebut this by
>> pointing out that square circles don't exist, you say that because
>> they don't, your system's failure to draw them is not a problem and
>> that therefore you are perfectly entitled to insist that your system
>> can draw them.
>
> When a halt decider H is required to report on the behavior of the
> direct execution of D that does the opposite of whatever H says that
> it will do this is is merely a logically impossible requirement exactly
> like requiring a CAD system to draw square circles.
>
> When we change the requirement so that it is not logically
> impossible then termination analyzer H is correct to report
> that D correctly simulated by H will never terminate normally
> because D specified recursive simulation to H.
>
> The correct simulation of D by H must include the call from D to H that
> specifies that D calls H in recursive simulation.
>
> Consistently all of the reviewers of my work insist that H must ignore
> this recursive simulation and report that D(D) halts because when H does
> not ignore this recursive simulation and aborts its simulation D(D) does
> halt.   *They have no idea that their view is inconsistent*

When the definition of the halting problem results in requirement that
cannot be met because this requirement is a logical impossibility it is
this problem definition that must be rejected. The inability to do the
logically impossible never derives any limitation on anyone or anything.

The logical impossibility of solving the halting problem (within its
current definition) is exactly the same as the logical impossibility of
creating a CAD system that correctly draws square circles.

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: Simplest Possible Halting Problem Proof Rebuttal

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From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
Date: Fri, 20 Oct 2023 20:07:58 -0700
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 by: Richard Damon - Sat, 21 Oct 2023 03:07 UTC

On 10/20/23 11:59 AM, olcott wrote:
> On 10/19/2023 6:21 AM, Paul N wrote:
>> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
>>> *As soon as this is understood to be correct then*
>>> The inability to do the logically impossible never places any actual
>>> limits on anyone or anything.
>>>
>>> Then it is understood that the logical impossibility of solving the
>>> halting problem the way it is currently defined places no actual limit
>>> on computation.
>>>
>>> It is equally logically impossible to define a CAD system that correctly
>>> draws square circles.
>>
>> Exactly. However, the equivalent to what you are saying is to say that
>> that everyone's proof that it is impossible to define a CAD system
>> that draws square circles is wrong, and that you do actually have a
>> CAD system which does so. When other people try to rebut this by
>> pointing out that square circles don't exist, you say that because
>> they don't, your system's failure to draw them is not a problem and
>> that therefore you are perfectly entitled to insist that your system
>> can draw them.
>
> When a halt decider H is required to report on the behavior of the
> direct execution of D that does the opposite of whatever H says that
> it will do this is is merely a logically impossible requirement exactly
> like requiring a CAD system to draw square circles.

Nope. Category error.

Halt Decider H, if is IS a correct halt decider, needs to give the
correct answer for all possible machines given as input.

THAT is a definition.

There IS a correct answer for this input (since it is specific machine,
that was built from a specific decider), thus the ACTUAL question isn't
a "logical impossible requriement'.

That H, just happens to give the wrong answer.

Note, "Give the right answer" isn't a valid programming design, so
talking about it being impossiible to design this machine doesn't make
the problem illogical, just uncomputable.

Note, the inability to make a square circle, is a logical impossibility
by the defintion of the problem. The inability to make a Halt Decider is
not a fact by the defintion, but was something that needed to be proved
by the conditions.

The fact that we can't actually construct a Halt Decider doesn't make
the Halting Question invalid, just nin-computable. There are MANY valid
decision problems that are not computable (as can be proven by a simple
counting arguement).

You clearly don't understand these basic facts.

>
> When we change the requirement so that it is not logically
> impossible then termination analyzer H is correct to report
> that D correctly simulated by H will never terminate normally
> because D specified recursive simulation to H.

And the actual requirement has a correct answer, so "some" decider can
give it.

That H isn't that decider is just a matter of fact.

As has been pointed out many times, H doesn't do a correct simulation,
as a correct simulation means a simulation that exactly recreates the
behavior of the described machine. Since D(D) Halts, any simulation that
doesn't show that is INCORRECT, as are your claims.

>
> The correct simulation of D by H must include the call from D to H that
> specifies that D calls H in recursive simulation.

Right, and also that H WILL return 0 from that call, since that is the
actual behavior of H.

>
> Consistently all of the reviewers of my work insist that H must ignore
> this recursive simulation and report that D(D) halts because when H does
> not ignore this recursive simulation and aborts its simulation D(D) does
> halt.   *They have no idea that their view is inconsistent*
>

No, YOU are the one saying it must ignore the ACTUAL behavior of that
call. It seems you know nothing of the topic, as you seem to think that
a specific program can be just arbitrarily replaced another program,
which is just an incorrect statement.

Re: Simplest Possible Halting Problem Proof Rebuttal

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From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: Simplest Possible Halting Problem Proof Rebuttal
Date: Fri, 20 Oct 2023 20:08:07 -0700
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 by: Richard Damon - Sat, 21 Oct 2023 03:08 UTC

On 10/20/23 1:26 PM, olcott wrote:
> On 10/20/2023 1:59 PM, olcott wrote:
>> On 10/19/2023 6:21 AM, Paul N wrote:
>>> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
>>>> *As soon as this is understood to be correct then*
>>>> The inability to do the logically impossible never places any actual
>>>> limits on anyone or anything.
>>>>
>>>> Then it is understood that the logical impossibility of solving the
>>>> halting problem the way it is currently defined places no actual limit
>>>> on computation.
>>>>
>>>> It is equally logically impossible to define a CAD system that
>>>> correctly
>>>> draws square circles.
>>>
>>> Exactly. However, the equivalent to what you are saying is to say
>>> that that everyone's proof that it is impossible to define a CAD
>>> system that draws square circles is wrong, and that you do actually
>>> have a CAD system which does so. When other people try to rebut this
>>> by pointing out that square circles don't exist, you say that because
>>> they don't, your system's failure to draw them is not a problem and
>>> that therefore you are perfectly entitled to insist that your system
>>> can draw them.
>>
>> When a halt decider H is required to report on the behavior of the
>> direct execution of D that does the opposite of whatever H says that
>> it will do this is is merely a logically impossible requirement exactly
>> like requiring a CAD system to draw square circles.
>>
>> When we change the requirement so that it is not logically
>> impossible then termination analyzer H is correct to report
>> that D correctly simulated by H will never terminate normally
>> because D specified recursive simulation to H.
>>
>> The correct simulation of D by H must include the call from D to H that
>> specifies that D calls H in recursive simulation.
>>
>> Consistently all of the reviewers of my work insist that H must ignore
>> this recursive simulation and report that D(D) halts because when H does
>> not ignore this recursive simulation and aborts its simulation D(D) does
>> halt.   *They have no idea that their view is inconsistent*
>
> When the definition of the halting problem results in requirement that
> cannot be met because this requirement is a logical impossibility it is
> this problem definition that must be rejected. The inability to do the
> logically impossible never derives any limitation on anyone or anything.
>
> The logical impossibility of solving the halting problem (within its
> current definition) is exactly the same as the logical impossibility of
> creating a CAD system that correctly draws square circles.
>

Nope, you don't know the difference between an "invalid" decision
problem (one that doesn't have a correct answer for every input) from an
uncomputable problem.

Halting is computable, since every possible machine has a definite answer.

Your arguement isn't actually about a specific halting problem question,
since you don't have a specific input, since the input is defined based
on the decider it is to contradict, but you don't have a fixed machine,
as you say it doesn't have a specific defined behavior.

This shows that you don't understand what a "program" actually is.

Re: Simplest Possible Halting Problem Proof Rebuttal

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 by: Richard Damon - Sat, 21 Oct 2023 03:52 UTC

On 10/20/23 11:59 AM, olcott wrote:
> On 10/19/2023 6:21 AM, Paul N wrote:
>> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott wrote:
>>> *As soon as this is understood to be correct then*
>>> The inability to do the logically impossible never places any actual
>>> limits on anyone or anything.
>>>
>>> Then it is understood that the logical impossibility of solving the
>>> halting problem the way it is currently defined places no actual limit
>>> on computation.
>>>
>>> It is equally logically impossible to define a CAD system that correctly
>>> draws square circles.
>>
>> Exactly. However, the equivalent to what you are saying is to say that
>> that everyone's proof that it is impossible to define a CAD system
>> that draws square circles is wrong, and that you do actually have a
>> CAD system which does so. When other people try to rebut this by
>> pointing out that square circles don't exist, you say that because
>> they don't, your system's failure to draw them is not a problem and
>> that therefore you are perfectly entitled to insist that your system
>> can draw them.
>
> When a halt decider H is required to report on the behavior of the
> direct execution of D that does the opposite of whatever H says that
> it will do this is is merely a logically impossible requirement exactly
> like requiring a CAD system to draw square circles.

The fact that you are just repeating your claim and not answering the
errors pointed out just shows that you don't actually have an answer to
them. This holds for EVERY time you do that. Not answering an error
pointed out is effectively admitting you agree it is an error.

The level you do this just shows how little you understand anything you
are talking about, be it computations, or logic, or even what "Truth" is.

>
> When we change the requirement so that it is not logically
> impossible then termination analyzer H is correct to report
> that D correctly simulated by H will never terminate normally
> because D specified recursive simulation to H.

So, you ADMIT that you are working on a FALSE problem.

You can't claim to have an answer to a problem when you solved a
different problem.

This shows your total lack of understand of how logic works, and points
out how you work by deception.

>
> The correct simulation of D by H must include the call from D to H that
> specifies that D calls H in recursive simulation.

Yes, and it also must include the responce that H actually gives, which
is to return 0.

YOUR "simulation" doesn't show that, and thus is INCORRECT, and your
claims that it is correct is just a LIE.

>
> Consistently all of the reviewers of my work insist that H must ignore
> this recursive simulation and report that D(D) halts because when H does
> not ignore this recursive simulation and aborts its simulation D(D) does
> halt.   *They have no idea that their view is inconsistent*
>

THAT IS A LIE.

No one has said it must "ignore" the recursive simulation. It must
correctly predict the actual results of this base on the H that D was
built on.

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