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Simplest Possible Halting Problem Proof Rebuttal
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olcott
Oct 18, 2023, 11:20:15 PM10/18/23
to
*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
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
Richard Damon
Oct 18, 2023, 11:36:29 PM10/18/23
to
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?
> *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.
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.
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.
>
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.
olcott
Oct 19, 2023, 12:37:33 PM10/19/23
to
On 10/19/2023 6:21 AM, Paul N wrote:
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.
> 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.
>> *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.
>
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.
olcott
Oct 20, 2023, 3:00:05 PM10/20/23
to
On 10/19/2023 6:21 AM, Paul N wrote:
> On Thursday, October 19, 2023 at 4:20:18 AM UTC+1, olcott 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.
>
>> 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*
olcott
Oct 20, 2023, 4:26:43 PM10/20/23
to
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.
Richard Damon
Oct 20, 2023, 11:08:03 PM10/20/23
to
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.
> 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.
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.
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.
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*
>
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.
Richard Damon
Oct 20, 2023, 11:08:13 PM10/20/23
to
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.
Richard Damon
Oct 20, 2023, 11:52:40 PM10/20/23
to
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
>> 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.
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.
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.
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*
>
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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