On 1/25/23 9:19 PM, olcott wrote:
> On 1/25/2023 8:06 PM, Richard Damon wrote:
>> How?
>>
>> it specificially talks about problems based on starting at a specified
>> initial state and getting to a specified final state, using a sequence
>> of transfomation from a defined set.
>>
>
> That very obviously includes decision problems
> https://en.wikipedia.org/wiki/Decision_problem
>

Nope, see other post.

In a very real sense they are OPPOSITE problems.

On 1/25/2023 8:16 PM, Richard Damon wrote:
> On 1/25/23 9:07 PM, olcott wrote:
>> On 1/25/2023 7:42 PM, Richard Damon wrote:
>>> On 1/25/23 8:35 PM, olcott wrote:
>>>> On 1/25/2023 7:11 PM, Richard Damon wrote:
>>>>> On 1/25/23 7:48 PM, olcott wrote:
>>>>>> On 1/25/2023 6:33 PM, Richard Damon wrote:
>>>>>
>>>>>>> One simple comment that comes to mind that points out the error
>>>>>>> in your thinking:
>>>>>>>
>>>>>>> The number of possible computing machines is a countable
>>>>>>> infinite, because we can express every such machine as a finite
>>>>>>> string of a finite symbol set.
>>>>>>>
>>>>>>> The number of possible deciders that can be defined is an
>>>>>>> UNCOUNTABLE infinite.
>>>>>>>
>>>>>>
>>>>>> I had a very very long debate about this with a PhD computer
>>>>>> scientist.
>>>>>> When we make one single universal halt decider that correctly
>>>>>> determines
>>>>>> the halt status of any arbitrary input pair then the countability
>>>>>> issue
>>>>>> ceases to exist.
>>>>>>
>>>>>
>>>>> So, by assuming an impossible machine exists, you can prove a false
>>>>> statement.
>>>>>
>>>>
>>>> By hypothesizing a single halt decider for the infinite set of
>>>> arbitrary
>>>> finite string pairs countability ceases to be an issue.
>>>>
>>>
>>> So your answer to trying to do the impossible is to just assume you
>>> can do something impossible.
>>>
>>
>> *Not at all*
>> Under the conditions that I specified countability definitely does
>> become moot, thus conclusively proving that countability
>> is not always an issue.
>>
>>
>
> So in a land of fairy dust power magical mythical unicorns, you think
> you can count them.
>

Anyone can count to one. Do you disagree?

One TM and an arbitrary finite string pair also applies to deriving the
sum of natural numbers as pairs of finite strings. We need not count
these strings we only need to compute the sum of any arbitrary pair.

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

On 1/25/23 9:26 PM, olcott wrote:
> On 1/25/2023 8:16 PM, Richard Damon wrote:
>> On 1/25/23 9:07 PM, olcott wrote:
>>> On 1/25/2023 7:42 PM, Richard Damon wrote:
>>>> On 1/25/23 8:35 PM, olcott wrote:
>>>>> On 1/25/2023 7:11 PM, Richard Damon wrote:
>>>>>> On 1/25/23 7:48 PM, olcott wrote:
>>>>>>> On 1/25/2023 6:33 PM, Richard Damon wrote:
>>>>>>
>>>>>>>> One simple comment that comes to mind that points out the error
>>>>>>>> in your thinking:
>>>>>>>>
>>>>>>>> The number of possible computing machines is a countable
>>>>>>>> infinite, because we can express every such machine as a finite
>>>>>>>> string of a finite symbol set.
>>>>>>>>
>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>> UNCOUNTABLE infinite.
>>>>>>>>
>>>>>>>
>>>>>>> I had a very very long debate about this with a PhD computer
>>>>>>> scientist.
>>>>>>> When we make one single universal halt decider that correctly
>>>>>>> determines
>>>>>>> the halt status of any arbitrary input pair then the countability
>>>>>>> issue
>>>>>>> ceases to exist.
>>>>>>>
>>>>>>
>>>>>> So, by assuming an impossible machine exists, you can prove a
>>>>>> false statement.
>>>>>>
>>>>>
>>>>> By hypothesizing a single halt decider for the infinite set of
>>>>> arbitrary
>>>>> finite string pairs countability ceases to be an issue.
>>>>>
>>>>
>>>> So your answer to trying to do the impossible is to just assume you
>>>> can do something impossible.
>>>>
>>>
>>> *Not at all*
>>> Under the conditions that I specified countability definitely does
>>> become moot, thus conclusively proving that countability
>>> is not always an issue.
>>>
>>>
>>
>> So in a land of fairy dust power magical mythical unicorns, you think
>> you can count them.
>>
>
> Anyone can count to one. Do you disagree?
>
> One TM and an arbitrary finite string pair also applies to deriving the
> sum of natural numbers as pairs of finite strings. We need not count
> these strings we only need to compute the sum of any arbitrary pair.
>

Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.

The number of possible functions to compute goes exponentially to the
number of input values.

8 inputs to decide, 256 functions
9 inputs to decide, 512 functions
10 inputs to decide, 1024 functions

since the input can be a countable infinite number of values, there are
2 to a countable infinite number of functions to make.

It turns out that no method to try to map that number of functions to a
countable number exists, as it grows too fast.

Please post YOUR mapping that shows it is countable.

Note, you can't assume an actual Halt Decider exists, as you haven't
actually shown it does. Only that the common proof doesn't work for your
"alternate" definition of a Halt Decider that doesn't actually map to
the Halting Function, but it still won't work, as other proofs will
still break it.

On 1/25/2023 8:45 PM, Richard Damon wrote:
> On 1/25/23 9:26 PM, olcott wrote:
>> On 1/25/2023 8:16 PM, Richard Damon wrote:
>>> On 1/25/23 9:07 PM, olcott wrote:
>>>> On 1/25/2023 7:42 PM, Richard Damon wrote:
>>>>> On 1/25/23 8:35 PM, olcott wrote:
>>>>>> On 1/25/2023 7:11 PM, Richard Damon wrote:
>>>>>>> On 1/25/23 7:48 PM, olcott wrote:
>>>>>>>> On 1/25/2023 6:33 PM, Richard Damon wrote:
>>>>>>>
>>>>>>>>> One simple comment that comes to mind that points out the error
>>>>>>>>> in your thinking:
>>>>>>>>>
>>>>>>>>> The number of possible computing machines is a countable
>>>>>>>>> infinite, because we can express every such machine as a finite
>>>>>>>>> string of a finite symbol set.
>>>>>>>>>
>>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>>> UNCOUNTABLE infinite.
>>>>>>>>>
>>>>>>>>
>>>>>>>> I had a very very long debate about this with a PhD computer
>>>>>>>> scientist.
>>>>>>>> When we make one single universal halt decider that correctly
>>>>>>>> determines
>>>>>>>> the halt status of any arbitrary input pair then the
>>>>>>>> countability issue
>>>>>>>> ceases to exist.
>>>>>>>>
>>>>>>>
>>>>>>> So, by assuming an impossible machine exists, you can prove a
>>>>>>> false statement.
>>>>>>>
>>>>>>
>>>>>> By hypothesizing a single halt decider for the infinite set of
>>>>>> arbitrary
>>>>>> finite string pairs countability ceases to be an issue.
>>>>>>
>>>>>
>>>>> So your answer to trying to do the impossible is to just assume you
>>>>> can do something impossible.
>>>>>
>>>>
>>>> *Not at all*
>>>> Under the conditions that I specified countability definitely does
>>>> become moot, thus conclusively proving that countability
>>>> is not always an issue.
>>>>
>>>>
>>>
>>> So in a land of fairy dust power magical mythical unicorns, you think
>>> you can count them.
>>>
>>
>> Anyone can count to one. Do you disagree?
>>
>> One TM and an arbitrary finite string pair also applies to deriving the
>> sum of natural numbers as pairs of finite strings. We need not count
>> these strings we only need to compute the sum of any arbitrary pair.
>>
>
> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>
> The number of possible functions to compute goes exponentially to the
> number of input values.

Yes of course everyone knows that we must have a separate TM to compute
the sum of each distinct input pair. It is utterly impossible to define
a single machine that could compute the sum of 3+4 and also compute the
sum of 2+1.

Even if by some magic this would be possible everyone knows that it is
utterly impossible to chain these calls together to derive the sum of
more elements than a pair.

sum(sum(2,3), sum(4,5)) is utterly unimaginable total nonsense.
Recursive calls are a form of psychosis.

Did you notice that was sarcasm?

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

On 1/25/23 10:26 PM, olcott wrote:
> On 1/25/2023 8:45 PM, Richard Damon wrote:
>> On 1/25/23 9:26 PM, olcott wrote:
>>> On 1/25/2023 8:16 PM, Richard Damon wrote:
>>>> On 1/25/23 9:07 PM, olcott wrote:
>>>>> On 1/25/2023 7:42 PM, Richard Damon wrote:
>>>>>> On 1/25/23 8:35 PM, olcott wrote:
>>>>>>> On 1/25/2023 7:11 PM, Richard Damon wrote:
>>>>>>>> On 1/25/23 7:48 PM, olcott wrote:
>>>>>>>>> On 1/25/2023 6:33 PM, Richard Damon wrote:
>>>>>>>>
>>>>>>>>>> One simple comment that comes to mind that points out the
>>>>>>>>>> error in your thinking:
>>>>>>>>>>
>>>>>>>>>> The number of possible computing machines is a countable
>>>>>>>>>> infinite, because we can express every such machine as a
>>>>>>>>>> finite string of a finite symbol set.
>>>>>>>>>>
>>>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>>>> UNCOUNTABLE infinite.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> I had a very very long debate about this with a PhD computer
>>>>>>>>> scientist.
>>>>>>>>> When we make one single universal halt decider that correctly
>>>>>>>>> determines
>>>>>>>>> the halt status of any arbitrary input pair then the
>>>>>>>>> countability issue
>>>>>>>>> ceases to exist.
>>>>>>>>>
>>>>>>>>
>>>>>>>> So, by assuming an impossible machine exists, you can prove a
>>>>>>>> false statement.
>>>>>>>>
>>>>>>>
>>>>>>> By hypothesizing a single halt decider for the infinite set of
>>>>>>> arbitrary
>>>>>>> finite string pairs countability ceases to be an issue.
>>>>>>>
>>>>>>
>>>>>> So your answer to trying to do the impossible is to just assume
>>>>>> you can do something impossible.
>>>>>>
>>>>>
>>>>> *Not at all*
>>>>> Under the conditions that I specified countability definitely does
>>>>> become moot, thus conclusively proving that countability
>>>>> is not always an issue.
>>>>>
>>>>>
>>>>
>>>> So in a land of fairy dust power magical mythical unicorns, you
>>>> think you can count them.
>>>>
>>>
>>> Anyone can count to one. Do you disagree?
>>>
>>> One TM and an arbitrary finite string pair also applies to deriving the
>>> sum of natural numbers as pairs of finite strings. We need not count
>>> these strings we only need to compute the sum of any arbitrary pair.
>>>
>>
>> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>>
>> The number of possible functions to compute goes exponentially to the
>> number of input values.
>
> Yes of course everyone knows that we must have a separate TM to compute
> the sum of each distinct input pair. It is utterly impossible to define
> a single machine that could compute the sum of 3+4 and also compute the
> sum of 2+1.

????? That is a standard first year exercise.

>
> Even if by some magic this would be possible everyone knows that it is
> utterly impossible to chain these calls together to derive the sum of
> more elements than a pair.

So, "Sum" is a computable function.

You falling into the proof by example problem agian.

>
> sum(sum(2,3), sum(4,5)) is utterly unimaginable total nonsense.
> Recursive calls are a form of psychosis.
>
> Did you notice that was sarcasm?
>
>

So, rather than try to actually come up with an answewr, you deflect
with garbage.

Guess that shows how much basis you have to work from.

You still don't seem to understand that there are an order of magnatude
more functions than machines, (order as in degree of infinity) so you
can't compute all the functions, there aren't enough machines to do it.

Your brain is just too small to understand that, because it doesn't even
seem to understand the actual concept of infinity. Infinity isn't just a
very big number.

Python <python@invalid.org> writes:

> Le 26/01/2023 à 01:33, Richard Damon a écrit :
>> One simple comment that comes to mind that points out the error in your
>> thinking:
>> The number of possible computing machines is a countable infinite,
>> because we can express every such machine as a finite string of a
>> finite symbol set.
>> The number of possible deciders that can be defined is an UNCOUNTABLE
>> infinite.

Ooh, I would not say that. For any reasonable meaning of "can be
defined" the set is countable, isn't it?

The set of subsets of N is uncountable, so most of these sets are not
decidable, but only a countable subset can be defined. That's how I'd
put anyway.

>> Thus, there are deciders that can not be computed, in
>> fact, MOST deciders can't be computed. (Admittedly a randomly chosen
>> decider likely isn't useful).

(I'd phrase it in terms of sets that can't be decided.)

>> I suspect you aren't going to understand this, as you can't seem to
>> handle the actual concept of infinity.
>
> Exactly. This is something Ben Bacarisse pointed out several times
> here. Then the fact there is no program being an Halt Decider is
> not much a surprise. It is only the first of this kind of problem that
> has been found not to be able to be solve by a program.

As RD says, the counting argument does not guarantee that we care about
any of the undecidable sets. This together with the fact that only
countable many can be finitely defined means that we are doubly lucky
(or unlucky, depending on how you see it) to have examples of
well-defined sets that are both undecidable and interesting. All the
undecidable ones /could/ have been boring or ones we can't finitely
define.

The halting deniers studiously ignore all the other well-defined
non-computable functions because they have no basis for rejecting them
as pathological. And they always ignore the busy beaver function
because it's non-computability has in independent proof.

--
Ben.

On 1/25/2023 9:35 PM, Ben Bacarisse wrote:
> Richard Damon <Richard@Damon-Family.org> writes:
>
>> On 1/25/23 8:06 PM, Ben Bacarisse wrote:
>>> Python <python@invalid.org> writes:
>>>
>>>> Peter Olcott wrote:
>>>> ...
>>>>> No machine can possibly be defined that divides all pairs of finite
>>>>> strings into those that represent machines would halt on their input
>>>>> when directly executed and those that do not
>>>>
>>>> So at least you admit that there no program can be written that
>>>> is an halt decider! It took time for you to abandon your delusions!
>>>
>>> Not so fast! Remember that PO is often fractally wrong. Not only is he
>>> usually wrong about the big picture he's usually wrong about all the
>>> details too. In this case, he's having trouble expressing what he
>>> means. This is actually just another iteration of his "it's an invalid
>>> question" stance, badly phrased.
>>
>> Meaning some times he just admits he is wrong because he fails to
>> fashion a good enough lie and slips up and tells the truth that he
>> doesn't beleive.
>
> Not quite. Python cut off the end of the sentence. The "...because"
> text is what shows it's the same old "some instances have no correct
> answer" in new clothing. PO almost certainly does not stand by what
> Python quoted without the because clause.
>

It is the case that ZFC did eliminate Russell's Paradox by eliminating
its basis of a set containing itself. Under the original definition of
the problem within naive set theory Russell's Paradox still exists.

The halting problem proof begins with the correct basis that arbitrary
pairs of finite strings either represent a computation that halts on its
input or not.

The proof that no machine can correctly determine this set is analogous
to the Liar Paradox in that the input is specifically defined to do the
opposite of whatever the halt decider determines. This transforms the
halting problem into an ill-formed problem.

In the study of problem solving, any problem in which the initial
state or starting position, the allowable operations, and the goal
state are clearly specified, and a unique solution can be shown to
exist.

*a unique solution can be shown to exist*
or the problem itself is ill-defined

https://www.oxfordreference.com/display/10.1093/oi/authority.20110803121717729;jsessionid=DAD4AA6FA046509B2E4564A52201A947

*I have been saying this for decades*
Every undecidable decision problem is necessarily erroneous.

*Finally www.oxfordreference.com agrees*

When we redefine the problem from arbitrary pairs of finite strings to
arbitrary pairs of finite string inputs to a decider then the liar
paradox issue goes away. H(D,D) does correctly determine that its
correct simulation of its input would never stop running unless aborted.

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

On 1/25/2023 9:39 PM, Richard Damon wrote:
> On 1/25/23 10:26 PM, olcott wrote:
>> On 1/25/2023 8:45 PM, Richard Damon wrote:
>>> On 1/25/23 9:26 PM, olcott wrote:
>>>> On 1/25/2023 8:16 PM, Richard Damon wrote:
>>>>> On 1/25/23 9:07 PM, olcott wrote:
>>>>>> On 1/25/2023 7:42 PM, Richard Damon wrote:
>>>>>>> On 1/25/23 8:35 PM, olcott wrote:
>>>>>>>> On 1/25/2023 7:11 PM, Richard Damon wrote:
>>>>>>>>> On 1/25/23 7:48 PM, olcott wrote:
>>>>>>>>>> On 1/25/2023 6:33 PM, Richard Damon wrote:
>>>>>>>>>
>>>>>>>>>>> One simple comment that comes to mind that points out the
>>>>>>>>>>> error in your thinking:
>>>>>>>>>>>
>>>>>>>>>>> The number of possible computing machines is a countable
>>>>>>>>>>> infinite, because we can express every such machine as a
>>>>>>>>>>> finite string of a finite symbol set.
>>>>>>>>>>>
>>>>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>>>>> UNCOUNTABLE infinite.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> I had a very very long debate about this with a PhD computer
>>>>>>>>>> scientist.
>>>>>>>>>> When we make one single universal halt decider that correctly
>>>>>>>>>> determines
>>>>>>>>>> the halt status of any arbitrary input pair then the
>>>>>>>>>> countability issue
>>>>>>>>>> ceases to exist.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> So, by assuming an impossible machine exists, you can prove a
>>>>>>>>> false statement.
>>>>>>>>>
>>>>>>>>
>>>>>>>> By hypothesizing a single halt decider for the infinite set of
>>>>>>>> arbitrary
>>>>>>>> finite string pairs countability ceases to be an issue.
>>>>>>>>
>>>>>>>
>>>>>>> So your answer to trying to do the impossible is to just assume
>>>>>>> you can do something impossible.
>>>>>>>
>>>>>>
>>>>>> *Not at all*
>>>>>> Under the conditions that I specified countability definitely does
>>>>>> become moot, thus conclusively proving that countability
>>>>>> is not always an issue.
>>>>>>
>>>>>>
>>>>>
>>>>> So in a land of fairy dust power magical mythical unicorns, you
>>>>> think you can count them.
>>>>>
>>>>
>>>> Anyone can count to one. Do you disagree?
>>>>
>>>> One TM and an arbitrary finite string pair also applies to deriving the
>>>> sum of natural numbers as pairs of finite strings. We need not count
>>>> these strings we only need to compute the sum of any arbitrary pair.
>>>>
>>>
>>> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>>>
>>> The number of possible functions to compute goes exponentially to the
>>> number of input values.
>>
>> Yes of course everyone knows that we must have a separate TM to compute
>> the sum of each distinct input pair. It is utterly impossible to define
>> a single machine that could compute the sum of 3+4 and also compute the
>> sum of 2+1.
>
> ????? That is a standard first year exercise.
>
>>
>> Even if by some magic this would be possible everyone knows that it is
>> utterly impossible to chain these calls together to derive the sum of
>> more elements than a pair.
>
> So, "Sum" is a computable function.
>
> You falling into the proof by example problem agian.
>
>>
>> sum(sum(2,3), sum(4,5)) is utterly unimaginable total nonsense.
>> Recursive calls are a form of psychosis.
>>
>> Did you notice that was sarcasm?
>>
>>
>
> So, rather than try to actually come up with an answewr, you deflect
> with garbage.
>

I have proven that countability is not an issue.
That you fail to understand that is not my problem.

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

On 1/25/23 11:16 PM, olcott wrote:
> On 1/25/2023 9:39 PM, Richard Damon wrote:
>> On 1/25/23 10:26 PM, olcott wrote:
>>> On 1/25/2023 8:45 PM, Richard Damon wrote:
>>>> On 1/25/23 9:26 PM, olcott wrote:
>>>>> On 1/25/2023 8:16 PM, Richard Damon wrote:
>>>>>> On 1/25/23 9:07 PM, olcott wrote:
>>>>>>> On 1/25/2023 7:42 PM, Richard Damon wrote:
>>>>>>>> On 1/25/23 8:35 PM, olcott wrote:
>>>>>>>>> On 1/25/2023 7:11 PM, Richard Damon wrote:
>>>>>>>>>> On 1/25/23 7:48 PM, olcott wrote:
>>>>>>>>>>> On 1/25/2023 6:33 PM, Richard Damon wrote:
>>>>>>>>>>
>>>>>>>>>>>> One simple comment that comes to mind that points out the
>>>>>>>>>>>> error in your thinking:
>>>>>>>>>>>>
>>>>>>>>>>>> The number of possible computing machines is a countable
>>>>>>>>>>>> infinite, because we can express every such machine as a
>>>>>>>>>>>> finite string of a finite symbol set.
>>>>>>>>>>>>
>>>>>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>>>>>> UNCOUNTABLE infinite.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> I had a very very long debate about this with a PhD computer
>>>>>>>>>>> scientist.
>>>>>>>>>>> When we make one single universal halt decider that correctly
>>>>>>>>>>> determines
>>>>>>>>>>> the halt status of any arbitrary input pair then the
>>>>>>>>>>> countability issue
>>>>>>>>>>> ceases to exist.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> So, by assuming an impossible machine exists, you can prove a
>>>>>>>>>> false statement.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> By hypothesizing a single halt decider for the infinite set of
>>>>>>>>> arbitrary
>>>>>>>>> finite string pairs countability ceases to be an issue.
>>>>>>>>>
>>>>>>>>
>>>>>>>> So your answer to trying to do the impossible is to just assume
>>>>>>>> you can do something impossible.
>>>>>>>>
>>>>>>>
>>>>>>> *Not at all*
>>>>>>> Under the conditions that I specified countability definitely
>>>>>>> does become moot, thus conclusively proving that countability
>>>>>>> is not always an issue.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> So in a land of fairy dust power magical mythical unicorns, you
>>>>>> think you can count them.
>>>>>>
>>>>>
>>>>> Anyone can count to one. Do you disagree?
>>>>>
>>>>> One TM and an arbitrary finite string pair also applies to deriving
>>>>> the
>>>>> sum of natural numbers as pairs of finite strings. We need not count
>>>>> these strings we only need to compute the sum of any arbitrary pair.
>>>>>
>>>>
>>>> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>>>>
>>>> The number of possible functions to compute goes exponentially to
>>>> the number of input values.
>>>
>>> Yes of course everyone knows that we must have a separate TM to compute
>>> the sum of each distinct input pair. It is utterly impossible to define
>>> a single machine that could compute the sum of 3+4 and also compute the
>>> sum of 2+1.
>>
>> ????? That is a standard first year exercise.
>>
>>>
>>> Even if by some magic this would be possible everyone knows that it is
>>> utterly impossible to chain these calls together to derive the sum of
>>> more elements than a pair.
>>
>> So, "Sum" is a computable function.
>>
>> You falling into the proof by example problem agian.
>>
>>>
>>> sum(sum(2,3), sum(4,5)) is utterly unimaginable total nonsense.
>>> Recursive calls are a form of psychosis.
>>>
>>> Did you notice that was sarcasm?
>>>
>>>
>>
>> So, rather than try to actually come up with an answewr, you deflect
>> with garbage.
>>
>
> I have proven that countability is not an issue.
> That you fail to understand that is not my problem.
>

WHERE? You have CLAIMED, not proven.

You claim is based on Fairy Dust, not reality,

You are just proving you are insane or stupid.

>

On 1/25/23 10:58 PM, olcott wrote:
> On 1/25/2023 9:35 PM, Ben Bacarisse wrote:
>> Richard Damon <Richard@Damon-Family.org> writes:
>>
>>> On 1/25/23 8:06 PM, Ben Bacarisse wrote:
>>>> Python <python@invalid.org> writes:
>>>>
>>>>> Peter Olcott wrote:
>>>>> ...
>>>>>> No machine can possibly be defined that divides all pairs of finite
>>>>>> strings into those that represent machines would halt on their input
>>>>>> when directly executed and those that do not
>>>>>
>>>>> So at least you admit that there no program can be written that
>>>>> is an halt decider! It took time for you to abandon your delusions!
>>>>
>>>> Not so fast!  Remember that PO is often fractally wrong.  Not only
>>>> is he
>>>> usually wrong about the big picture he's usually wrong about all the
>>>> details too.  In this case, he's having trouble expressing what he
>>>> means.  This is actually just another iteration of his "it's an invalid
>>>> question" stance, badly phrased.
>>>
>>> Meaning some times he just admits he is wrong because he fails to
>>> fashion a good enough lie and slips up and tells the truth that he
>>> doesn't beleive.
>>
>> Not quite.  Python cut off the end of the sentence.  The "...because"
>> text is what shows it's the same old "some instances have no correct
>> answer" in new clothing.  PO almost certainly does not stand by what
>> Python quoted without the because clause.
>>
>
> It is the case that ZFC did eliminate Russell's Paradox by eliminating
> its basis of a set containing itself. Under the original definition of
> the problem within naive set theory Russell's Paradox still exists.

Not by just defining that sets can't contain themselves, but by limiting
the kind of things that sets can contain.

>
> The halting problem proof begins with the correct basis that arbitrary
> pairs of finite strings either represent a computation that halts on its
> input or not.
>
> The proof that no machine can correctly determine this set is analogous
> to the Liar Paradox in that the input is specifically defined to do the
> opposite of whatever the halt decider determines. This transforms the
> halting problem into an ill-formed problem.

Nope. Since the input is part of the domain of machine/inputs, the
logical semntics of what that machine is doing is irrelevent. Basd on
the rules the prospective Halt Decider is using, it will either Halt or
be Non-Halting, the the Problem has a correct answer, so the Halting
Funciton exists as a well defined function, it just is a fact that the
decider can't give the right answer.

The the problem that is ill-defined, is the problem to try to create the
Halt Decider, because it doesn't exist.

You just can't seem to distinguish between the actual problem asked for,
and the problem of trying to create a working Halt Decider.

>
>    In the study of problem solving, any problem in which the initial
>    state or starting position, the allowable operations, and the goal
>    state are clearly specified, and a unique solution can be shown to
>    exist.
>

You keep repeating the Strawman, This proves you are STUPID

> *a unique solution can be shown to exist*
> or the problem itself is ill-defined
>
> https://www.oxfordreference.com/display/10.1093/oi/authority.20110803121717729;jsessionid=DAD4AA6FA046509B2E4564A52201A947

Which is about PUZZLE Problems.

>
> *I have been saying this for decades*
> Every undecidable decision problem is necessarily erroneous.

Nope, just uncomputable.

>
> *Finally www.oxfordreference.com agrees*
>
> When we redefine the problem from arbitrary pairs of finite strings to
> arbitrary pairs of finite string inputs to a decider then the liar
> paradox issue goes away. H(D,D) does correctly determine that its
> correct simulation of its input would never stop running unless aborted.
>

So, you are admitting that H doesn't answer the actual Halting problem,
but think it is ok to give a wrong answer because it is impossible for
it to give the right answer?

This just proves that you are totally ignorant.

On 1/25/2023 8:45 PM, Richard Damon wrote:
> On 1/25/23 9:26 PM, olcott wrote:
>> On 1/25/2023 8:16 PM, Richard Damon wrote:
>>> On 1/25/23 9:07 PM, olcott wrote:
>>>> On 1/25/2023 7:42 PM, Richard Damon wrote:
>>>>> On 1/25/23 8:35 PM, olcott wrote:
>>>>>> On 1/25/2023 7:11 PM, Richard Damon wrote:
>>>>>>> On 1/25/23 7:48 PM, olcott wrote:
>>>>>>>> On 1/25/2023 6:33 PM, Richard Damon wrote:
>>>>>>>
>>>>>>>>> One simple comment that comes to mind that points out the error
>>>>>>>>> in your thinking:
>>>>>>>>>
>>>>>>>>> The number of possible computing machines is a countable
>>>>>>>>> infinite, because we can express every such machine as a finite
>>>>>>>>> string of a finite symbol set.
>>>>>>>>>
>>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>>> UNCOUNTABLE infinite.
>>>>>>>>>
>>>>>>>>
>>>>>>>> I had a very very long debate about this with a PhD computer
>>>>>>>> scientist.
>>>>>>>> When we make one single universal halt decider that correctly
>>>>>>>> determines
>>>>>>>> the halt status of any arbitrary input pair then the
>>>>>>>> countability issue
>>>>>>>> ceases to exist.
>>>>>>>>
>>>>>>>
>>>>>>> So, by assuming an impossible machine exists, you can prove a
>>>>>>> false statement.
>>>>>>>
>>>>>>
>>>>>> By hypothesizing a single halt decider for the infinite set of
>>>>>> arbitrary
>>>>>> finite string pairs countability ceases to be an issue.
>>>>>>
>>>>>
>>>>> So your answer to trying to do the impossible is to just assume you
>>>>> can do something impossible.
>>>>>
>>>>
>>>> *Not at all*
>>>> Under the conditions that I specified countability definitely does
>>>> become moot, thus conclusively proving that countability
>>>> is not always an issue.
>>>>
>>>>
>>>
>>> So in a land of fairy dust power magical mythical unicorns, you think
>>> you can count them.
>>>
>>
>> Anyone can count to one. Do you disagree?
>>
>> One TM and an arbitrary finite string pair also applies to deriving the
>> sum of natural numbers as pairs of finite strings. We need not count
>> these strings we only need to compute the sum of any arbitrary pair.
>>
>
> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>
> The number of possible functions to compute goes exponentially to the
> number of input values.
>
> 8 inputs to decide, 256 functions
> 9 inputs to decide, 512 functions
> 10 inputs to decide, 1024 functions
>

The above is pure Cockamamie bullshit.
One TM has an arbitrary number of space delimited finite strings of
ASCII digits that it computes the sum of.

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

On 1/25/23 11:52 PM, olcott wrote:
> On 1/25/2023 8:45 PM, Richard Damon wrote:
>> On 1/25/23 9:26 PM, olcott wrote:

>> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>>
>> The number of possible functions to compute goes exponentially to the
>> number of input values.
>>
>> 8 inputs to decide, 256 functions
>> 9 inputs to decide, 512 functions
>> 10 inputs to decide, 1024 functions
>>
>
> The above is pure Cockamamie bullshit.
> One TM has an arbitrary number of space delimited finite strings of
> ASCII digits that it computes the sum of.
>

So, you have no idea of what you are talking, and think all computations
are mearly sumations.

You are just proving you are too stupid to be worth arguing about this.

Show the bijection, or the pair of injections, or STFU.

On 1/26/2023 6:22 AM, Ben Bacarisse wrote:
> Richard Damon <Richard@Damon-Family.org> writes:
>
>> On 1/25/23 10:54 PM, Ben Bacarisse wrote:
>>> Python <python@invalid.org> writes:
>>>
>>>> Le 26/01/2023 à 01:33, Richard Damon a écrit :
>>>>> One simple comment that comes to mind that points out the error in your
>>>>> thinking:
>>>>> The number of possible computing machines is a countable infinite,
>>>>> because we can express every such machine as a finite string of a
>>>>> finite symbol set.
>>>>> The number of possible deciders that can be defined is an UNCOUNTABLE
>>>>> infinite.
>>> Ooh, I would not say that. For any reasonable meaning of "can be
>>> defined" the set is countable, isn't it?
>>
>> Right, I meant FUNCTIONS is an uncountable set.
>
> Yes, it can also be framed in terms of functions. For any countably
> infinite set X, the set X->{0,1} is uncountable, so most of those
> functions are not TM computable.
>

The sum of every element of the set of all finite subsets of finite
strings of of ASCII digits can be computed because we can define a TM
that takes an arbitrary number of space delimited finite strings.

Infinite input to a TM is uncomputable because the TM would never halt,
thus the set of subsets of finite strings of ASCII digits must exclude
infinite subsets.

The same thing would apply to a halt decider that takes arbitrary pairs
of finite strings. We know this because we know that a TM that computes
the sum of arbitrary pairs of finite strings of ASCII digits can be
defined: This is merely a simpler case of the above.

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

On 1/26/2023 5:47 AM, Richard Damon wrote:
> On 1/25/23 11:52 PM, olcott wrote:
>> On 1/25/2023 8:45 PM, Richard Damon wrote:
>>> On 1/25/23 9:26 PM, olcott wrote:
>
>>> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>>>
>>> The number of possible functions to compute goes exponentially to the
>>> number of input values.
>>>
>>> 8 inputs to decide, 256 functions
>>> 9 inputs to decide, 512 functions
>>> 10 inputs to decide, 1024 functions
>>>
>>
>> The above is pure Cockamamie bullshit.
>> One TM has an arbitrary number of space delimited finite strings of
>> ASCII digits that it computes the sum of.
>>
>
> So, you have no idea of what you are talking, and think all computations
> are mearly sumations.
>

If a TM that computes the sum of a finite number of arbitrary finite
strings of ASCII digits is computable then a TM that computes anything
else on a finite number of arbitrary finite strings is not uncomputable
for any countability reason.

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

On 1/25/2023 10:37 PM, Richard Damon wrote:
> On 1/25/23 10:58 PM, olcott wrote:
>> On 1/25/2023 9:35 PM, Ben Bacarisse wrote:
>>> Richard Damon <Richard@Damon-Family.org> writes:
>>>
>>>> On 1/25/23 8:06 PM, Ben Bacarisse wrote:
>>>>> Python <python@invalid.org> writes:
>>>>>
>>>>>> Peter Olcott wrote:
>>>>>> ...
>>>>>>> No machine can possibly be defined that divides all pairs of finite
>>>>>>> strings into those that represent machines would halt on their input
>>>>>>> when directly executed and those that do not
>>>>>>
>>>>>> So at least you admit that there no program can be written that
>>>>>> is an halt decider! It took time for you to abandon your delusions!
>>>>>
>>>>> Not so fast!  Remember that PO is often fractally wrong.  Not only
>>>>> is he
>>>>> usually wrong about the big picture he's usually wrong about all the
>>>>> details too.  In this case, he's having trouble expressing what he
>>>>> means.  This is actually just another iteration of his "it's an
>>>>> invalid
>>>>> question" stance, badly phrased.
>>>>
>>>> Meaning some times he just admits he is wrong because he fails to
>>>> fashion a good enough lie and slips up and tells the truth that he
>>>> doesn't beleive.
>>>
>>> Not quite.  Python cut off the end of the sentence.  The "...because"
>>> text is what shows it's the same old "some instances have no correct
>>> answer" in new clothing.  PO almost certainly does not stand by what
>>> Python quoted without the because clause.
>>>
>>
>> It is the case that ZFC did eliminate Russell's Paradox by eliminating
>> its basis of a set containing itself. Under the original definition of
>> the problem within naive set theory Russell's Paradox still exists.
>
> Not by just defining that sets can't contain themselves, but by limiting
> the kind of things that sets can contain.

That a set cannot contain itself is what eliminates Russell's Paradox.

>>
>> The halting problem proof begins with the correct basis that arbitrary
>> pairs of finite strings either represent a computation that halts on its
>> input or not.
>>
>> The proof that no machine can correctly determine this set is analogous
>> to the Liar Paradox in that the input is specifically defined to do the
>> opposite of whatever the halt decider determines. This transforms the
>> halting problem into an ill-formed problem.
>
> Nope. Since the input is part of the domain of machine/inputs, the

When we define a machine that correctly determines whether or not pairs
of arbitrary finite sting inputs would reach the final state of the
first element of this input then halting is computable.

Because the pathological input actually has different behavior when it
is correctly simulated by its corresponding halt decider and the halt
decider must base its halt status decision on the actual behavior of
this input then the halt decider is necessarily correct to reject its
pathological input as non-halting.

You still have not provided any counter example that shows that my
definition of correct simulation is incorrect:

Try and provide a 100% specific counter-example where you show a line
of machine code such as [mov eax, 1] and the simulator simulates another
different line instead such as [push ebx] and the simulator is correct.

If no such counter-example exists then it is proven that the ultimate
measure of correct simulation is that the simulator simulates line-by-
line exactly what the machine code specifies.

ALL OF THE ARGUMENTS AGAINST MY POSITION HAVE PROVEN TO BE COUNTER-FACTUAL

(a) The simulation of the input to H(D,D) by H is correct.

(b) D correctly simulated by H would never stop running unless aborted.

(c) Therefore H is necessarily correct to abort its simulation of D and
reject D as non-halting.

https://www.researchgate.net/publication/364657019_Simulating_Halt_Decider_Applied_to_the_Halting_Theorem

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

On 2023-01-26 08:43, olcott wrote:
> On 1/26/2023 6:22 AM, Ben Bacarisse wrote:
>> Richard Damon <Richard@Damon-Family.org> writes:
>>
>>> On 1/25/23 10:54 PM, Ben Bacarisse wrote:
>>>> Python <python@invalid.org> writes:
>>>>
>>>>> Le 26/01/2023 à 01:33, Richard Damon a écrit :
>>>>>> One simple comment that comes to mind that points out the error in
>>>>>> your
>>>>>> thinking:
>>>>>> The number of possible computing machines is a countable infinite,
>>>>>> because we can express every such machine as a finite string of a
>>>>>> finite symbol set.
>>>>>> The number of possible deciders that can be defined is an UNCOUNTABLE
>>>>>> infinite.
>>>> Ooh, I would not say that.  For any reasonable meaning of "can be
>>>> defined" the set is countable, isn't it?
>>>
>>> Right, I meant FUNCTIONS is an uncountable set.
>>
>> Yes, it can also be framed in terms of functions.  For any countably
>> infinite set X, the set X->{0,1} is uncountable, so most of those
>> functions are not TM computable.
>>
>
> The sum of every element of the set of all finite subsets of finite
> strings of of ASCII digits can be computed because we can define a TM
> that takes an arbitrary number of space delimited finite strings.

And how 'bout them Mets?

André

> Infinite input to a TM is uncomputable because the TM would never halt,
> thus the set of subsets of finite strings of ASCII digits must exclude
> infinite subsets.
>
> The same thing would apply to a halt decider that takes arbitrary pairs
> of finite strings. We know this because we know that a TM that computes
> the sum of arbitrary pairs of finite strings of ASCII digits can be
> defined: This is merely a simpler case of the above.
>

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 1/26/2023 11:15 AM, André G. Isaak wrote:
> On 2023-01-26 08:43, olcott wrote:
>> On 1/26/2023 6:22 AM, Ben Bacarisse wrote:
>>> Richard Damon <Richard@Damon-Family.org> writes:
>>>
>>>> On 1/25/23 10:54 PM, Ben Bacarisse wrote:
>>>>> Python <python@invalid.org> writes:
>>>>>
>>>>>> Le 26/01/2023 à 01:33, Richard Damon a écrit :
>>>>>>> One simple comment that comes to mind that points out the error
>>>>>>> in your
>>>>>>> thinking:
>>>>>>> The number of possible computing machines is a countable infinite,
>>>>>>> because we can express every such machine as a finite string of a
>>>>>>> finite symbol set.
>>>>>>> The number of possible deciders that can be defined is an
>>>>>>> UNCOUNTABLE
>>>>>>> infinite.
>>>>> Ooh, I would not say that.  For any reasonable meaning of "can be
>>>>> defined" the set is countable, isn't it?
>>>>
>>>> Right, I meant FUNCTIONS is an uncountable set.
>>>
>>> Yes, it can also be framed in terms of functions.  For any countably
>>> infinite set X, the set X->{0,1} is uncountable, so most of those
>>> functions are not TM computable.
>>>
>>
>> The sum of every element of the set of all finite subsets of finite
>> strings of of ASCII digits can be computed because we can define a TM
>> that takes an arbitrary number of space delimited finite strings.
>
> And how 'bout them Mets?
>
> André

In other words my statement is irrefutably correct.

>
>> Infinite input to a TM is uncomputable because the TM would never halt,
>> thus the set of subsets of finite strings of ASCII digits must exclude
>> infinite subsets.
>>
>> The same thing would apply to a halt decider that takes arbitrary pairs
>> of finite strings. We know this because we know that a TM that computes
>> the sum of arbitrary pairs of finite strings of ASCII digits can be
>> defined: This is merely a simpler case of the above.
>>
>
>

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

On 1/26/23 11:46 AM, olcott wrote:
> On 1/26/2023 5:47 AM, Richard Damon wrote:
>> On 1/25/23 11:52 PM, olcott wrote:
>>> On 1/25/2023 8:45 PM, Richard Damon wrote:
>>>> On 1/25/23 9:26 PM, olcott wrote:
>>
>>>> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>>>>
>>>> The number of possible functions to compute goes exponentially to
>>>> the number of input values.
>>>>
>>>> 8 inputs to decide, 256 functions
>>>> 9 inputs to decide, 512 functions
>>>> 10 inputs to decide, 1024 functions
>>>>
>>>
>>> The above is pure Cockamamie bullshit.
>>> One TM has an arbitrary number of space delimited finite strings of
>>> ASCII digits that it computes the sum of.
>>>
>>
>> So, you have no idea of what you are talking, and think all
>> computations are mearly sumations.
>>
>
> If a TM that computes the sum of a finite number of arbitrary finite
> strings of ASCII digits is computable then a TM that computes anything
> else on a finite number of arbitrary finite strings is not uncomputable
> for any countability reason.
>

WHy? Summing is a very simple operation, that we can do that doesn't say
anthing about more complicated operations.

Please write a TM that computes the last prime.

How about one that computes the Busy Beaver number for a given input on
number of states.

You are just showing how STUPID you are.

On 1/26/23 10:43 AM, olcott wrote:
> On 1/26/2023 6:22 AM, Ben Bacarisse wrote:
>> Richard Damon <Richard@Damon-Family.org> writes:
>>
>>> On 1/25/23 10:54 PM, Ben Bacarisse wrote:
>>>> Python <python@invalid.org> writes:
>>>>
>>>>> Le 26/01/2023 à 01:33, Richard Damon a écrit :
>>>>>> One simple comment that comes to mind that points out the error in
>>>>>> your
>>>>>> thinking:
>>>>>> The number of possible computing machines is a countable infinite,
>>>>>> because we can express every such machine as a finite string of a
>>>>>> finite symbol set.
>>>>>> The number of possible deciders that can be defined is an UNCOUNTABLE
>>>>>> infinite.
>>>> Ooh, I would not say that.  For any reasonable meaning of "can be
>>>> defined" the set is countable, isn't it?
>>>
>>> Right, I meant FUNCTIONS is an uncountable set.
>>
>> Yes, it can also be framed in terms of functions.  For any countably
>> infinite set X, the set X->{0,1} is uncountable, so most of those
>> functions are not TM computable.
>>
>
> The sum of every element of the set of all finite subsets of finite
> strings of of ASCII digits can be computed because we can define a TM
> that takes an arbitrary number of space delimited finite strings.

Red herring as we are not talking about functions that are just a "sum".

>
> Infinite input to a TM is uncomputable because the TM would never halt,
> thus the set of subsets of finite strings of ASCII digits must exclude
> infinite subsets.

We are not talking about any subset that has an infinte number of
members, but the number of finite subsets of the Natural Numbers.

The count of this is an uncountable infinity, an order of infinity
bigger than the count of the Natural Numbers.

>
> The same thing would apply to a halt decider that takes arbitrary pairs
> of finite strings. We know this because we know that a TM that computes
> the sum of arbitrary pairs of finite strings of ASCII digits can be
> defined: This is merely a simpler case of the above.
>

Wrong, and using a Red Herring and the fallacy of proof by examole,

You are just showing how IGNORANT you are of what you are talking about.

On 1/26/23 12:04 PM, olcott wrote:
> On 1/25/2023 10:37 PM, Richard Damon wrote:
>> On 1/25/23 10:58 PM, olcott wrote:
>>> On 1/25/2023 9:35 PM, Ben Bacarisse wrote:
>>>> Richard Damon <Richard@Damon-Family.org> writes:
>>>>
>>>>> On 1/25/23 8:06 PM, Ben Bacarisse wrote:
>>>>>> Python <python@invalid.org> writes:
>>>>>>
>>>>>>> Peter Olcott wrote:
>>>>>>> ...
>>>>>>>> No machine can possibly be defined that divides all pairs of finite
>>>>>>>> strings into those that represent machines would halt on their
>>>>>>>> input
>>>>>>>> when directly executed and those that do not
>>>>>>>
>>>>>>> So at least you admit that there no program can be written that
>>>>>>> is an halt decider! It took time for you to abandon your delusions!
>>>>>>
>>>>>> Not so fast!  Remember that PO is often fractally wrong.  Not only
>>>>>> is he
>>>>>> usually wrong about the big picture he's usually wrong about all the
>>>>>> details too.  In this case, he's having trouble expressing what he
>>>>>> means.  This is actually just another iteration of his "it's an
>>>>>> invalid
>>>>>> question" stance, badly phrased.
>>>>>
>>>>> Meaning some times he just admits he is wrong because he fails to
>>>>> fashion a good enough lie and slips up and tells the truth that he
>>>>> doesn't beleive.
>>>>
>>>> Not quite.  Python cut off the end of the sentence.  The "...because"
>>>> text is what shows it's the same old "some instances have no correct
>>>> answer" in new clothing.  PO almost certainly does not stand by what
>>>> Python quoted without the because clause.
>>>>
>>>
>>> It is the case that ZFC did eliminate Russell's Paradox by eliminating
>>> its basis of a set containing itself. Under the original definition of
>>> the problem within naive set theory Russell's Paradox still exists.
>>
>> Not by just defining that sets can't contain themselves, but by
>> limiting the kind of things that sets can contain.
>
> That a set cannot contain itself is what eliminates Russell's Paradox.
>
>>>
>>> The halting problem proof begins with the correct basis that arbitrary
>>> pairs of finite strings either represent a computation that halts on its
>>> input or not.
>>>
>>> The proof that no machine can correctly determine this set is analogous
>>> to the Liar Paradox in that the input is specifically defined to do the
>>> opposite of whatever the halt decider determines. This transforms the
>>> halting problem into an ill-formed problem.
>>
>> Nope. Since the input is part of the domain of machine/inputs, the
>
> When we define a machine that correctly determines whether or not pairs
> of arbitrary finite sting inputs would reach the final state of the
> first element of this input then halting is computable.
>

But you need to show that you CAN do that. YOu haven't

H(P,P) says that P(P) will not halt when it does, so it fails.

> Because the pathological input actually has different behavior when it
> is correctly simulated by its corresponding halt decider and the halt
> decider must base its halt status decision on the actual behavior of
> this input then the halt decider is necessarily correct to reject its
> pathological input as non-halting.

No, it doesn't

You have been asked to point what point in the actual execution of
H(P,P) called by P(P) call by main differs from the execution of H(P,P)
calledd by main diverege.

You have failed to point that out, becaue it doesn't exist, because you
ar a LIAR and an IDOIT.

As stated before, your failure to even attempt to indicate this is taken
as your admission that your claim is a LIE, and you can not actually
prove your claim, also making you a damned hypocrite.

>
> You still have not provided any counter example that shows that my
> definition of correct simulation is incorrect:

LIE.

P(P) is the counter example.

>
> Try and provide a 100% specific counter-example where you show a line
> of machine code such as [mov eax, 1] and the simulator simulates another
> different line instead such as [push ebx] and the simulator is correct.

P(P), its call H will go through the exact same set of states when P(P)
call s as when main calls it, thus since H(P,P) returns 0 when called by
main, we know it does the same thing when called by P (or it just fails
to be the "pure function"/computation you have claimex it to be)

Since H simulats that call the H(P,P) as something that will never
returm, H has done an incorrect simulation.

Your inability to understand this just shows your STUPIDITY.

>
> If no such counter-example exists then it is proven that the ultimate
> measure of correct simulation is that the simulator simulates line-by-
> line exactly what the machine code specifies.
>
> ALL OF THE ARGUMENTS AGAINST MY POSITION HAVE PROVEN TO BE COUNTER-FACTUAL

Nope, YOUR statements, have been shown to be "counter-factual" and your
reasoning ab
>
> (a) The simulation of the input to H(D,D) by H is correct.

Nope, because it says H(P,P) will never return, when it doesn.

>
> (b) D correctly simulated by H would never stop running unless aborted.

Which is irrelvent and not a truth bearer, as H doen't do this, buyt
ALWAYS aborts its simulation.

H IN CORRECTLY presumes that the input is calling a DIFFERENT function
than it actually does, so gives the wrong asnwer.

>
> (c) Therefore H is necessarily correct to abort its simulation of D and
>     reject D as non-halting.

batting 0/3

YOU FAIL.

>
> https://www.researchgate.net/publication/364657019_Simulating_Halt_Decider_Applied_to_the_Halting_Theorem
>
>
>

Just shows you are a Hypocritical Pathological Lying Idiot.

On 2023-01-26 13:26, olcott wrote:
> On 1/26/2023 11:15 AM, André G. Isaak wrote:
>> On 2023-01-26 08:43, olcott wrote:
>>> On 1/26/2023 6:22 AM, Ben Bacarisse wrote:
>>>> Richard Damon <Richard@Damon-Family.org> writes:
>>>>
>>>>> On 1/25/23 10:54 PM, Ben Bacarisse wrote:
>>>>>> Python <python@invalid.org> writes:
>>>>>>
>>>>>>> Le 26/01/2023 à 01:33, Richard Damon a écrit :
>>>>>>>> One simple comment that comes to mind that points out the error
>>>>>>>> in your
>>>>>>>> thinking:
>>>>>>>> The number of possible computing machines is a countable infinite,
>>>>>>>> because we can express every such machine as a finite string of a
>>>>>>>> finite symbol set.
>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>> UNCOUNTABLE
>>>>>>>> infinite.
>>>>>> Ooh, I would not say that.  For any reasonable meaning of "can be
>>>>>> defined" the set is countable, isn't it?
>>>>>
>>>>> Right, I meant FUNCTIONS is an uncountable set.
>>>>
>>>> Yes, it can also be framed in terms of functions.  For any countably
>>>> infinite set X, the set X->{0,1} is uncountable, so most of those
>>>> functions are not TM computable.
>>>>
>>>
>>> The sum of every element of the set of all finite subsets of finite
>>> strings of of ASCII digits can be computed because we can define a TM
>>> that takes an arbitrary number of space delimited finite strings.
>>
>> And how 'bout them Mets?
>>
>> André
>
> In other words my statement is irrefutably correct.

Woooooosh!

André

>>
>>> Infinite input to a TM is uncomputable because the TM would never halt,
>>> thus the set of subsets of finite strings of ASCII digits must exclude
>>> infinite subsets.
>>>
>>> The same thing would apply to a halt decider that takes arbitrary pairs
>>> of finite strings. We know this because we know that a TM that computes
>>> the sum of arbitrary pairs of finite strings of ASCII digits can be
>>> defined: This is merely a simpler case of the above.
>>>
>>
>>
>

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 1/26/2023 5:48 PM, André G. Isaak wrote:
> On 2023-01-26 13:26, olcott wrote:
>> On 1/26/2023 11:15 AM, André G. Isaak wrote:
>>> On 2023-01-26 08:43, olcott wrote:
>>>> On 1/26/2023 6:22 AM, Ben Bacarisse wrote:
>>>>> Richard Damon <Richard@Damon-Family.org> writes:
>>>>>
>>>>>> On 1/25/23 10:54 PM, Ben Bacarisse wrote:
>>>>>>> Python <python@invalid.org> writes:
>>>>>>>
>>>>>>>> Le 26/01/2023 à 01:33, Richard Damon a écrit :
>>>>>>>>> One simple comment that comes to mind that points out the error
>>>>>>>>> in your
>>>>>>>>> thinking:
>>>>>>>>> The number of possible computing machines is a countable infinite,
>>>>>>>>> because we can express every such machine as a finite string of a
>>>>>>>>> finite symbol set.
>>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>>> UNCOUNTABLE
>>>>>>>>> infinite.
>>>>>>> Ooh, I would not say that.  For any reasonable meaning of "can be
>>>>>>> defined" the set is countable, isn't it?
>>>>>>
>>>>>> Right, I meant FUNCTIONS is an uncountable set.
>>>>>
>>>>> Yes, it can also be framed in terms of functions.  For any countably
>>>>> infinite set X, the set X->{0,1} is uncountable, so most of those
>>>>> functions are not TM computable.
>>>>>
>>>>
>>>> The sum of every element of the set of all finite subsets of finite
>>>> strings of of ASCII digits can be computed because we can define a TM
>>>> that takes an arbitrary number of space delimited finite strings.
>>>
>>> And how 'bout them Mets?
>>>
>>> André
>>
>> In other words my statement is irrefutably correct.
>
> Woooooosh!
>
> André
>

It is not possible that the relevance of your irrelevant response was
over my head.

>>>
>>>> Infinite input to a TM is uncomputable because the TM would never halt,
>>>> thus the set of subsets of finite strings of ASCII digits must exclude
>>>> infinite subsets.
>>>>
>>>> The same thing would apply to a halt decider that takes arbitrary pairs
>>>> of finite strings. We know this because we know that a TM that computes
>>>> the sum of arbitrary pairs of finite strings of ASCII digits can be
>>>> defined: This is merely a simpler case of the above.
>>>>
>>>
>>>
>>
>
>

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

On 1/26/2023 5:37 PM, Richard Damon wrote:
> On 1/26/23 11:46 AM, olcott wrote:
>> On 1/26/2023 5:47 AM, Richard Damon wrote:
>>> On 1/25/23 11:52 PM, olcott wrote:
>>>> On 1/25/2023 8:45 PM, Richard Damon wrote:
>>>>> On 1/25/23 9:26 PM, olcott wrote:
>>>
>>>>> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
>>>>>
>>>>> The number of possible functions to compute goes exponentially to
>>>>> the number of input values.
>>>>>
>>>>> 8 inputs to decide, 256 functions
>>>>> 9 inputs to decide, 512 functions
>>>>> 10 inputs to decide, 1024 functions
>>>>>
>>>>
>>>> The above is pure Cockamamie bullshit.
>>>> One TM has an arbitrary number of space delimited finite strings of
>>>> ASCII digits that it computes the sum of.
>>>>
>>>
>>> So, you have no idea of what you are talking, and think all
>>> computations are mearly sumations.
>>>
>>
>> If a TM that computes the sum of a finite number of arbitrary finite
>> strings of ASCII digits is computable then a TM that computes anything
>> else on a finite number of arbitrary finite strings is not uncomputable
>> for any countability reason.
>>
>
> WHy? Summing is a very simple operation, that we can do that doesn't say
> anthing about more complicated operations.
>
It does completely prove that countability is not an issue for the
halting problem thus all proofs to the contrary have been refuted.

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

On 1/26/23 7:11 PM, olcott wrote:
> On 1/26/2023 5:48 PM, André G. Isaak wrote:
>> On 2023-01-26 13:26, olcott wrote:
>>> On 1/26/2023 11:15 AM, André G. Isaak wrote:
>>>> On 2023-01-26 08:43, olcott wrote:
>>>>> On 1/26/2023 6:22 AM, Ben Bacarisse wrote:
>>>>>> Richard Damon <Richard@Damon-Family.org> writes:
>>>>>>
>>>>>>> On 1/25/23 10:54 PM, Ben Bacarisse wrote:
>>>>>>>> Python <python@invalid.org> writes:
>>>>>>>>
>>>>>>>>> Le 26/01/2023 à 01:33, Richard Damon a écrit :
>>>>>>>>>> One simple comment that comes to mind that points out the
>>>>>>>>>> error in your
>>>>>>>>>> thinking:
>>>>>>>>>> The number of possible computing machines is a countable
>>>>>>>>>> infinite,
>>>>>>>>>> because we can express every such machine as a finite string of a
>>>>>>>>>> finite symbol set.
>>>>>>>>>> The number of possible deciders that can be defined is an
>>>>>>>>>> UNCOUNTABLE
>>>>>>>>>> infinite.
>>>>>>>> Ooh, I would not say that.  For any reasonable meaning of "can be
>>>>>>>> defined" the set is countable, isn't it?
>>>>>>>
>>>>>>> Right, I meant FUNCTIONS is an uncountable set.
>>>>>>
>>>>>> Yes, it can also be framed in terms of functions.  For any countably
>>>>>> infinite set X, the set X->{0,1} is uncountable, so most of those
>>>>>> functions are not TM computable.
>>>>>>
>>>>>
>>>>> The sum of every element of the set of all finite subsets of finite
>>>>> strings of of ASCII digits can be computed because we can define a TM
>>>>> that takes an arbitrary number of space delimited finite strings.
>>>>
>>>> And how 'bout them Mets?
>>>>
>>>> André
>>>
>>> In other words my statement is irrefutably correct.
>>
>> Woooooosh!
>>
>> André
>>
>
> It is not possible that the relevance of your irrelevant response was
> over my head.

Woooooosh!

>
>>>>
>>>>> Infinite input to a TM is uncomputable because the TM would never
>>>>> halt,
>>>>> thus the set of subsets of finite strings of ASCII digits must exclude
>>>>> infinite subsets.
>>>>>
>>>>> The same thing would apply to a halt decider that takes arbitrary
>>>>> pairs
>>>>> of finite strings. We know this because we know that a TM that
>>>>> computes
>>>>> the sum of arbitrary pairs of finite strings of ASCII digits can be
>>>>> defined: This is merely a simpler case of the above.
>>>>>
>>>>
>>>>
>>>
>>
>>
>

On 1/26/23 7:14 PM, olcott wrote:
> On 1/26/2023 5:37 PM, Richard Damon wrote:
>> On 1/26/23 11:46 AM, olcott wrote:
>>> On 1/26/2023 5:47 AM, Richard Damon wrote:
>>>> On 1/25/23 11:52 PM, olcott wrote:
>>>>> On 1/25/2023 8:45 PM, Richard Damon wrote:
>>>>>> On 1/25/23 9:26 PM, olcott wrote:
>>>>
>>>>>> Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is
>>>>>> computable.
>>>>>>
>>>>>> The number of possible functions to compute goes exponentially to
>>>>>> the number of input values.
>>>>>>
>>>>>> 8 inputs to decide, 256 functions
>>>>>> 9 inputs to decide, 512 functions
>>>>>> 10 inputs to decide, 1024 functions
>>>>>>
>>>>>
>>>>> The above is pure Cockamamie bullshit.
>>>>> One TM has an arbitrary number of space delimited finite strings of
>>>>> ASCII digits that it computes the sum of.
>>>>>
>>>>
>>>> So, you have no idea of what you are talking, and think all
>>>> computations are mearly sumations.
>>>>
>>>
>>> If a TM that computes the sum of a finite number of arbitrary finite
>>> strings of ASCII digits is computable then a TM that computes anything
>>> else on a finite number of arbitrary finite strings is not uncomputable
>>> for any countability reason.
>>>
>>
>> WHy? Summing is a very simple operation, that we can do that doesn't
>> say anthing about more complicated operations.
>>
> It does completely prove that countability is not an issue for the
> halting problem thus all proofs to the contrary have been refuted.
>

No, saying one functions is computable does not prove that another is.

You are just showing you don't understand what you are talking about.

As I remember, you failed at even trying to write a machine to do the
summing you are talking about, you are just going on the fact that
others says it can be done.

You ar showing you don't even understand the very nature of a proof, or
what it means to be true. You may be able to spout off words, but you
show ZERO undrstanding of the actual meaning.

You are just proving to the whole world how stupid and ignorant you are.