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 :Ooh, I would not say that. For any reasonable meaning of "can be
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.
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.
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 :Ooh, I would not say that. For any reasonable meaning of "can be
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.
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.
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 :Ooh, I would not say that. For any reasonable meaning of "can be
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.
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.
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 :Ooh, I would not say that. For any reasonable meaning of "can be >>>>>> defined" the set is countable, isn't it?
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.
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.
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 :Ooh, I would not say that. For any reasonable meaning of "can be
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.
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.
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 :Ooh, I would not say that. For any reasonable meaning of "can be >>>>>>> defined" the set is countable, isn't it?
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.
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.
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 :Ooh, I would not say that. For any reasonable meaning of "can be >>>>>>>> defined" the set is countable, isn't it?
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.
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.
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 :Ooh, I would not say that. For any reasonable meaning of "can be
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.
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 7:27 PM, olcott wrote:
On 1/26/2023 5:37 PM, Richard Damon wrote:
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 :Ooh, I would not say that. For any reasonable meaning of "can be >>>>>>> defined" the set is countable, isn't it?
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.
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 set of all finite subsets of the natural numbers is countable
https://en.wikipedia.org/wiki/Countable_set
Side note, read on this page about total order:
In both examples of well orders here, any subset has a least element;
and in both examples of non-well orders, some subsets do not have a
least element. This is the key definition that determines whether a
total order is also a well order.
Note, the rationals "in usual order" are a non-well ordered set, and
thus some subsets (like the open interval) do not have a least member.
This is why the intervale (0, 1] doesn't have a "lowest" value.
https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable
https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable
Thus the set of all finite subsets of finite strings is countable.
Thus the set all finite string pairs is countable because all of these
pairs are subsets having two elements thus finite length,
Note, MY claim was the number of FUNCTIONS was uncountable, not the
number of subsets, thus not all Functions F(N) -> N are computable, as
there are more functions than computations.
On 1/26/2023 5:37 PM, Richard Damon wrote:
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 :Ooh, I would not say that. For any reasonable meaning of "can be >>>>>> defined" the set is countable, isn't it?
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.
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 set of all finite subsets of the natural numbers is countable https://en.wikipedia.org/wiki/Countable_set
In both examples of well orders here, any subset has a least element; and in both examples of non-well orders, some subsets do not have a least element. This is the key definition that determines whether a total order is also a well order.
https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable
https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable
Thus the set of all finite subsets of finite strings is countable.
Thus the set all finite string pairs is countable because all of these
pairs are subsets having two elements thus finite length,
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 1/26/2023 9:25 PM, Richard Damon wrote:
On 1/26/23 7:27 PM, olcott wrote:
On 1/26/2023 5:37 PM, Richard Damon wrote:
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 :Ooh, I would not say that. For any reasonable meaning of "can be >>>>>>>> defined" the set is countable, isn't it?
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.
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 set of all finite subsets of the natural numbers is countable
https://en.wikipedia.org/wiki/Countable_set
Side note, read on this page about total order:
In both examples of well orders here, any subset has a least element;
and in both examples of non-well orders, some subsets do not have a
least element. This is the key definition that determines whether a
total order is also a well order.
Note, the rationals "in usual order" are a non-well ordered set, and
thus some subsets (like the open interval) do not have a least member.
This is why the intervale (0, 1] doesn't have a "lowest" value.
https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable
https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable
Thus the set of all finite subsets of finite strings is countable.
Thus the set all finite string pairs is countable because all of
these pairs are subsets having two elements thus finite length,
Note, MY claim was the number of FUNCTIONS was uncountable, not the
number of subsets, thus not all Functions F(N) -> N are computable, as
there are more functions than computations.
There is a single TM can you not count to one?
A single TM can derive the sum of any arbitrary finite set
of finite strings of ASCII digits that are space delimited.
Proving that this cannot be done requires a specific
counter-example finite set of finite strings of ASCII digits
where the sum cannot be computed.
To be correct the TM need not calculate the sum of every finite
set of finite strings of ASCII digits, it merely has to always
compute this sum correctly for any arbitrary element of the
finite set of finite strings.
On 1/26/23 10:46 PM, olcott wrote:
On 1/26/2023 9:25 PM, Richard Damon wrote:
On 1/26/23 7:27 PM, olcott wrote:
On 1/26/2023 5:37 PM, Richard Damon wrote:
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 :Ooh, I would not say that. For any reasonable meaning of "can be >>>>>>>>> defined" the set is countable, isn't it?
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.
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 set of all finite subsets of the natural numbers is countable
https://en.wikipedia.org/wiki/Countable_set
Side note, read on this page about total order:
In both examples of well orders here, any subset has a least
element; and in both examples of non-well orders, some subsets do
not have a least element. This is the key definition that determines
whether a total order is also a well order.
Note, the rationals "in usual order" are a non-well ordered set, and
thus some subsets (like the open interval) do not have a least
member. This is why the intervale (0, 1] doesn't have a "lowest" value.
https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable
https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable
Thus the set of all finite subsets of finite strings is countable.
Thus the set all finite string pairs is countable because all of
these pairs are subsets having two elements thus finite length,
Note, MY claim was the number of FUNCTIONS was uncountable, not the
number of subsets, thus not all Functions F(N) -> N are computable,
as there are more functions than computations.
There is a single TM can you not count to one?
?????
Where do you get that claim?
A single TM can derive the sum of any arbitrary finite set
of finite strings of ASCII digits that are space delimited.
So?
Proving that this cannot be done requires a specific
counter-example finite set of finite strings of ASCII digits
where the sum cannot be computed.
Neer said you couldn't compute SOME functions.
To be correct the TM need not calculate the sum of every finite
set of finite strings of ASCII digits, it merely has to always
compute this sum correctly for any arbitrary element of the
finite set of finite strings.
NBot talking about "Sums"
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite
set of finite strings of ASCII digits, it merely has to always
compute this sum correctly for any arbitrary element of the
finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the
computability of halting on the basis that there are no countability
issues with the computation of sums.
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite
set of finite strings of ASCII digits, it merely has to always
compute this sum correctly for any arbitrary element of the
finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the
computability of halting on the basis that there are no countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite
set of finite strings of ASCII digits, it merely has to always
compute this sum correctly for any arbitrary element of the
finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the
computability of halting on the basis that there are no countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite set of
finite strings then H cannot have any countability issue with any
finite pair of finite strings.
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite
set of finite strings of ASCII digits, it merely has to always
compute this sum correctly for any arbitrary element of the
finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the
computability of halting on the basis that there are no countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite set of
finite strings then H cannot have any countability issue with any
finite pair of finite strings.
How do you make a countable infinite number of machine make an
uncountable number of maps.
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite
set of finite strings of ASCII digits, it merely has to always
compute this sum correctly for any arbitrary element of the
finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the
computability of halting on the basis that there are no countability >>>>> issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite set of
finite strings then H cannot have any countability issue with any
finite pair of finite strings.
How do you make a countable infinite number of machine make an
uncountable number of maps.
I ask you again can you count to one?
There is only one machine that always has a finite number of inputs.
We don't need any maps we only need one set of inputs deriving a single output for any arbitrary set of inputs. When we think of this as sums
then it is obvious that countability is not an issue.
Is there any finite set of finite strings of ASCII digits that cannot be summed by a TM? No, therefore computability is proven.
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite >>>>>>>>> set of finite strings of ASCII digits, it merely has to always >>>>>>>>> compute this sum correctly for any arbitrary element of the
finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the
computability of halting on the basis that there are no countability >>>>>>> issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite set of >>>>> finite strings then H cannot have any countability issue with any
finite pair of finite strings.
How do you make a countable infinite number of machine make an
uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of inputs.
We don't need any maps we only need one set of inputs deriving a single
output for any arbitrary set of inputs. When we think of this as sums
then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime factors of
its input it the same machine that computes its inputs factorial?
You DO understand that a Turing Machine can be treated as a
computation that computes a specific mapping of inputs to outputs?
A TM must only compute the mapping from any arbitrary finite set of
inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits that cannot be >>> summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing mapping of
one arbitrary finite set of inputs to its corresponding finite set of
outputs with any computation at all proves that there is no such issue
for summing or halt status detection.
If X always works then X always works with Y or Z.
You are just proving you are totally ignorant of what you are talking
about.
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite >>>>>>>>>> set of finite strings of ASCII digits, it merely has to always >>>>>>>>>> compute this sum correctly for any arbitrary element of the >>>>>>>>>> finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the
computability of halting on the basis that there are no
countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite set of >>>>>> finite strings then H cannot have any countability issue with any
finite pair of finite strings.
How do you make a countable infinite number of machine make an
uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of inputs.
We don't need any maps we only need one set of inputs deriving a single >>>> output for any arbitrary set of inputs. When we think of this as sums
then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime factors of
its input it the same machine that computes its inputs factorial?
You DO understand that a Turing Machine can be treated as a
computation that computes a specific mapping of inputs to outputs?
A TM must only compute the mapping from any arbitrary finite set of
inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits that
cannot be
summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing mapping of
one arbitrary finite set of inputs to its corresponding finite set of
outputs with any computation at all proves that there is no such issue
for summing or halt status detection.
So you are claiming that proof by example is true, that if one example
out of a set is true, the statement is true for all?
The fact that ONE machine can be counted, doesn't mean that all can be.
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical Pathological Lying Idiot.
That statement must be true, because it only makes an assurtion if it is true,
Thus you ARE a Hypocritical Pathological Lying Idiot.
You are just proving you like to make all the mistakes that history has
found in logic, because you refuse to learn from it.
You are just proving you are totally ignorant of what you are talking
about.
On 1/27/2023 10:54 PM, Richard Damon wrote:
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite >>>>>>>>>>> set of finite strings of ASCII digits, it merely has to always >>>>>>>>>>> compute this sum correctly for any arbitrary element of the >>>>>>>>>>> finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the >>>>>>>>> computability of halting on the basis that there are no
countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite
set of
finite strings then H cannot have any countability issue with any >>>>>>> finite pair of finite strings.
How do you make a countable infinite number of machine make an
uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of inputs. >>>>> We don't need any maps we only need one set of inputs deriving a
single
output for any arbitrary set of inputs. When we think of this as sums >>>>> then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime factors
of its input it the same machine that computes its inputs factorial?
You DO understand that a Turing Machine can be treated as a
computation that computes a specific mapping of inputs to outputs?
A TM must only compute the mapping from any arbitrary finite set of
inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits that
cannot be
summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing mapping of
one arbitrary finite set of inputs to its corresponding finite set of
outputs with any computation at all proves that there is no such issue
for summing or halt status detection.
So you are claiming that proof by example is true, that if one example
out of a set is true, the statement is true for all?
The fact that ONE machine can be counted, doesn't mean that all can be.
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical Pathological
Lying Idiot.
As always when you run out of reasoning you resort to ad hominem because gullible fools will take ad hominem as rebuttal. Ad hominem makes you
look quite foolish to anyone accustomed to academic decorum.
There can be no countability issue with computing the mapping
from a finite set of finite string inputs to their corresponding output
for any arbitrary finite string inputs or computation.
A counter-example is categorically impossible.
Try and show any finite set of finite strings that cannot be summed.
On 1/28/2023 7:00 AM, Richard Damon wrote:
On 1/28/23 12:19 AM, olcott wrote:
On 1/27/2023 10:54 PM, Richard Damon wrote:
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every >>>>>>>>>>>>> finite
set of finite strings of ASCII digits, it merely has to always >>>>>>>>>>>>> compute this sum correctly for any arbitrary element of the >>>>>>>>>>>>> finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the >>>>>>>>>>> computability of halting on the basis that there are no
countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite >>>>>>>>> set of
finite strings then H cannot have any countability issue with any >>>>>>>>> finite pair of finite strings.
How do you make a countable infinite number of machine make an >>>>>>>> uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of inputs. >>>>>>> We don't need any maps we only need one set of inputs deriving a >>>>>>> single
output for any arbitrary set of inputs. When we think of this as >>>>>>> sums
then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime factors >>>>>> of its input it the same machine that computes its inputs factorial? >>>>>>
You DO understand that a Turing Machine can be treated as a
computation that computes a specific mapping of inputs to outputs? >>>>>>
A TM must only compute the mapping from any arbitrary finite set of
inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits that
cannot be
summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing mapping of >>>>> one arbitrary finite set of inputs to its corresponding finite set of >>>>> outputs with any computation at all proves that there is no such issue >>>>> for summing or halt status detection.
So you are claiming that proof by example is true, that if one
example out of a set is true, the statement is true for all?
The fact that ONE machine can be counted, doesn't mean that all can be. >>>>
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical Pathological
Lying Idiot.
As always when you run out of reasoning you resort to ad hominem because >>> gullible fools will take ad hominem as rebuttal. Ad hominem makes you
look quite foolish to anyone accustomed to academic decorum.
No, I am showing you are using invalid logic.
Your statement is only true if you assume it to be true, and use it as
one of your truth makers that can make it true. This is not an
allowed operation, as if you do, then you get that exact same
arguement I was using.
So, *IF* you want to say your argument is valid, you *MUST& accept
that mine is too, and thus you are what I said.
The fact that you don't get it, might say my results are valid anyway
independent of validity of the statement.
In looking at your connection to truth makers view, X reaches back to
some base truth makers, but also links back to itself. This leads to
an infinite loop in the connection logic, where there is always a
piece of the connection not connected to things alreayd known to be
truth makers, so the statement has never been established as True,
which requires that ALL neccessary premises for the statement be
connected to know truth makers.
This is the flaw of you thinking you work backwards from the premise
to the truth makers, you can decieve yourself with such a loop,
thinking it has been established, when it is just a floating island of
unsupported logic. If you start at the Truth Makers, you can't run
into this problem, as you never had the presumptive statement
available to incorrectly use.
A counter-example to my reasoning does not exist.
“Analytic” sentences, such as “Pediatricians are doctors,” have historically been characterized as ones that are true by virtue of the meanings of their words alone and/or can be known to be so solely by
knowing those meanings. https://plato.stanford.edu/entries/analytic-synthetic/
This is correctly paraphrased as
Analytical truth is the connection from an expression X of formal or
natural language L using truth preserving operations to expressions of L
that have been stipulated to be true.
We know that cat are animals is true entirely on the basis of the
meaning of the words. The meaning of words are expressions of L that
have been stipulated to be true.
There can be no countability issue with computing the mapping
from a finite set of finite string inputs to their corresponding output
for any arbitrary finite string inputs or computation.
??? I don't think you know what you are saying here.
A counter-example is categorically impossible.
Try and show any finite set of finite strings that cannot be summed.
Maybe, because you are saying something meaningless, just a bunch of
word salad.
If there is a countability issue with calculating the sum of arbitrary
finite sets of finite strings of ASCII digits that prevented a correct
sum from being calculated then there would be an element of this set
that cannot be summed.
Not being able to count all of the sets of finite strings to be summed
in no way prevents any element of the sets of finite strings from being correctly summed, thus has no impact on computability.
As long as a TM can correctly determine the sum of any arbitrary finite
set of finite strings of ASCII digits then sum is computable.
On 1/28/23 12:19 AM, olcott wrote:
On 1/27/2023 10:54 PM, Richard Damon wrote:
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every finite >>>>>>>>>>>> set of finite strings of ASCII digits, it merely has to always >>>>>>>>>>>> compute this sum correctly for any arbitrary element of the >>>>>>>>>>>> finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the >>>>>>>>>> computability of halting on the basis that there are no
countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite >>>>>>>> set of
finite strings then H cannot have any countability issue with any >>>>>>>> finite pair of finite strings.
How do you make a countable infinite number of machine make an
uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of inputs. >>>>>> We don't need any maps we only need one set of inputs deriving a
single
output for any arbitrary set of inputs. When we think of this as sums >>>>>> then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime factors
of its input it the same machine that computes its inputs factorial? >>>>>
You DO understand that a Turing Machine can be treated as a
computation that computes a specific mapping of inputs to outputs?
A TM must only compute the mapping from any arbitrary finite set of
inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits that
cannot be
summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing mapping of >>>> one arbitrary finite set of inputs to its corresponding finite set of
outputs with any computation at all proves that there is no such issue >>>> for summing or halt status detection.
So you are claiming that proof by example is true, that if one
example out of a set is true, the statement is true for all?
The fact that ONE machine can be counted, doesn't mean that all can be.
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical Pathological
Lying Idiot.
As always when you run out of reasoning you resort to ad hominem because
gullible fools will take ad hominem as rebuttal. Ad hominem makes you
look quite foolish to anyone accustomed to academic decorum.
No, I am showing you are using invalid logic.
Your statement is only true if you assume it to be true, and use it as
one of your truth makers that can make it true. This is not an allowed operation, as if you do, then you get that exact same arguement I was
using.
So, *IF* you want to say your argument is valid, you *MUST& accept that
mine is too, and thus you are what I said.
The fact that you don't get it, might say my results are valid anyway independent of validity of the statement.
In looking at your connection to truth makers view, X reaches back to
some base truth makers, but also links back to itself. This leads to an infinite loop in the connection logic, where there is always a piece of
the connection not connected to things alreayd known to be truth makers,
so the statement has never been established as True, which requires that
ALL neccessary premises for the statement be connected to know truth
makers.
This is the flaw of you thinking you work backwards from the premise to
the truth makers, you can decieve yourself with such a loop, thinking it
has been established, when it is just a floating island of unsupported
logic. If you start at the Truth Makers, you can't run into this
problem, as you never had the presumptive statement available to
incorrectly use.
There can be no countability issue with computing the mapping
from a finite set of finite string inputs to their corresponding output
for any arbitrary finite string inputs or computation.
??? I don't think you know what you are saying here.
A counter-example is categorically impossible.
Try and show any finite set of finite strings that cannot be summed.
Maybe, because you are saying something meaningless, just a bunch of
word salad.
On 1/28/23 10:49 AM, olcott wrote:
On 1/28/2023 7:00 AM, Richard Damon wrote:
On 1/28/23 12:19 AM, olcott wrote:
On 1/27/2023 10:54 PM, Richard Damon wrote:
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every >>>>>>>>>>>>>> finite
set of finite strings of ASCII digits, it merely has to >>>>>>>>>>>>>> always
compute this sum correctly for any arbitrary element of the >>>>>>>>>>>>>> finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the >>>>>>>>>>>> computability of halting on the basis that there are no >>>>>>>>>>>> countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any finite >>>>>>>>>> set of
finite strings then H cannot have any countability issue with any >>>>>>>>>> finite pair of finite strings.
How do you make a countable infinite number of machine make an >>>>>>>>> uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of
inputs.
We don't need any maps we only need one set of inputs deriving a >>>>>>>> single
output for any arbitrary set of inputs. When we think of this as >>>>>>>> sums
then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime
factors of its input it the same machine that computes its inputs >>>>>>> factorial?
You DO understand that a Turing Machine can be treated as a
computation that computes a specific mapping of inputs to outputs? >>>>>>>
A TM must only compute the mapping from any arbitrary finite set of >>>>>> inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits that >>>>>>>> cannot be
summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing
mapping of
one arbitrary finite set of inputs to its corresponding finite set of >>>>>> outputs with any computation at all proves that there is no such
issue
for summing or halt status detection.
So you are claiming that proof by example is true, that if one
example out of a set is true, the statement is true for all?
The fact that ONE machine can be counted, doesn't mean that all can
be.
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical Pathological
Lying Idiot.
As always when you run out of reasoning you resort to ad hominem
because
gullible fools will take ad hominem as rebuttal. Ad hominem makes
you look quite foolish to anyone accustomed to academic decorum.
No, I am showing you are using invalid logic.
Your statement is only true if you assume it to be true, and use it
as one of your truth makers that can make it true. This is not an
allowed operation, as if you do, then you get that exact same
arguement I was using.
So, *IF* you want to say your argument is valid, you *MUST& accept
that mine is too, and thus you are what I said.
The fact that you don't get it, might say my results are valid anyway
independent of validity of the statement.
In looking at your connection to truth makers view, X reaches back to
some base truth makers, but also links back to itself. This leads to
an infinite loop in the connection logic, where there is always a
piece of the connection not connected to things alreayd known to be
truth makers, so the statement has never been established as True,
which requires that ALL neccessary premises for the statement be
connected to know truth makers.
This is the flaw of you thinking you work backwards from the premise
to the truth makers, you can decieve yourself with such a loop,
thinking it has been established, when it is just a floating island
of unsupported logic. If you start at the Truth Makers, you can't run
into this problem, as you never had the presumptive statement
available to incorrectly use.
A counter-example to my reasoning does not exist.
Nope, I have given it.
“Analytic” sentences, such as “Pediatricians are doctors,” have
historically been characterized as ones that are true by virtue of the
meanings of their words alone and/or can be known to be so solely by
knowing those meanings.
https://plato.stanford.edu/entries/analytic-synthetic/
Right, and a sentence that depends on its own truth value (like if H
returns the right value ...) are proved to be able to lead to
cotradictipons.
This is correctly paraphrased as
Analytical truth is the connection from an expression X of formal or
natural language L using truth preserving operations to expressions of L
that have been stipulated to be true.
Right, and you can not create such a chain to the statement you are making.
We know that cat are animals is true entirely on the basis of the
meaning of the words. The meaning of words are expressions of L that
have been stipulated to be true.
So. Red Herring. Means nothing about the error of using a statement that references its own truth.
Your statement that H can do xxx if H correctly decides yyy is a
statement of the class that has been proven to lead to logical paradoxes.
Such a statement can never be actually connected to previously accepted
truth makers unless you can prove INDEPENDENTLY of that statement that H
does give a correct decision.
The fact that we know, by definition, that H(D,D) must be 1 to be
correct if D(D) halts,
If there is a countability issue with calculating the sum of arbitrary
finite sets of finite strings of ASCII digits that prevented a correct
sum from being calculated then there would be an element of this set
that cannot be summed.
No, you don't understand countability, and are trying to apply it to a
Red Herring. You seem to think that "Computation" == "Sum", which is
just a proof of your ignorance.
Not being able to count all of the sets of finite strings to be summed
in no way prevents any element of the sets of finite strings from being
correctly summed, thus has no impact on computability.
It isn't the set of finite strings that is uncountable, it is the set of mappings every finite string individualy to 0 or 1 that is uncountable.
The number of Accept/Reject Mapping functions over Turing
Machines/Inputs is an order of infinity larger (it is uncountably
infinte) than the number of Turing Machines that could be built (a countable infinity) to try to decide them, thus not all of these
mappings can be created by a Turing Machine.
As long as a TM can correctly determine the sum of any arbitrary finite
set of finite strings of ASCII digits then sum is computable.
Which is a Red Herring as no one is talking about "summing" being non-computable, so the fact that you keep bringing it up just shows you
are too stupid to see your ignorance.
On 1/28/2023 10:08 AM, Richard Damon wrote:
On 1/28/23 10:49 AM, olcott wrote:
On 1/28/2023 7:00 AM, Richard Damon wrote:
On 1/28/23 12:19 AM, olcott wrote:
On 1/27/2023 10:54 PM, Richard Damon wrote:
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every >>>>>>>>>>>>>>> finite
set of finite strings of ASCII digits, it merely has to >>>>>>>>>>>>>>> always
compute this sum correctly for any arbitrary element of the >>>>>>>>>>>>>>> finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the >>>>>>>>>>>>> computability of halting on the basis that there are no >>>>>>>>>>>>> countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any
finite set of
finite strings then H cannot have any countability issue with >>>>>>>>>>> any
finite pair of finite strings.
How do you make a countable infinite number of machine make an >>>>>>>>>> uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of >>>>>>>>> inputs.
We don't need any maps we only need one set of inputs deriving >>>>>>>>> a single
output for any arbitrary set of inputs. When we think of this >>>>>>>>> as sums
then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime
factors of its input it the same machine that computes its
inputs factorial?
You DO understand that a Turing Machine can be treated as a
computation that computes a specific mapping of inputs to outputs? >>>>>>>>
A TM must only compute the mapping from any arbitrary finite set of >>>>>>> inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits that >>>>>>>>> cannot be
summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing
mapping of
one arbitrary finite set of inputs to its corresponding finite
set of
outputs with any computation at all proves that there is no such >>>>>>> issue
for summing or halt status detection.
So you are claiming that proof by example is true, that if one
example out of a set is true, the statement is true for all?
The fact that ONE machine can be counted, doesn't mean that all
can be.
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical Pathological
Lying Idiot.
As always when you run out of reasoning you resort to ad hominem
because
gullible fools will take ad hominem as rebuttal. Ad hominem makes
you look quite foolish to anyone accustomed to academic decorum.
No, I am showing you are using invalid logic.
Your statement is only true if you assume it to be true, and use it
as one of your truth makers that can make it true. This is not an
allowed operation, as if you do, then you get that exact same
arguement I was using.
So, *IF* you want to say your argument is valid, you *MUST& accept
that mine is too, and thus you are what I said.
The fact that you don't get it, might say my results are valid
anyway independent of validity of the statement.
In looking at your connection to truth makers view, X reaches back
to some base truth makers, but also links back to itself. This leads
to an infinite loop in the connection logic, where there is always a
piece of the connection not connected to things alreayd known to be
truth makers, so the statement has never been established as True,
which requires that ALL neccessary premises for the statement be
connected to know truth makers.
This is the flaw of you thinking you work backwards from the premise
to the truth makers, you can decieve yourself with such a loop,
thinking it has been established, when it is just a floating island
of unsupported logic. If you start at the Truth Makers, you can't
run into this problem, as you never had the presumptive statement
available to incorrectly use.
A counter-example to my reasoning does not exist.
Nope, I have given it.
“Analytic” sentences, such as “Pediatricians are doctors,” have
historically been characterized as ones that are true by virtue of the
meanings of their words alone and/or can be known to be so solely by
knowing those meanings.
https://plato.stanford.edu/entries/analytic-synthetic/
Right, and a sentence that depends on its own truth value (like if H
returns the right value ...) are proved to be able to lead to
cotradictipons.
This is correctly paraphrased as
Analytical truth is the connection from an expression X of formal or
natural language L using truth preserving operations to expressions of L >>> that have been stipulated to be true.
Right, and you can not create such a chain to the statement you are
making.
We know that cat are animals is true entirely on the basis of the
meaning of the words. The meaning of words are expressions of L that
have been stipulated to be true.
So. Red Herring. Means nothing about the error of using a statement
that references its own truth.
Your statement that H can do xxx if H correctly decides yyy is a
statement of the class that has been proven to lead to logical paradoxes.
Such a statement can never be actually connected to previously
accepted truth makers unless you can prove INDEPENDENTLY of that
statement that H does give a correct decision.
The fact that we know, by definition, that H(D,D) must be 1 to be
correct if D(D) halts,
*Changing the subject away from the following is a dishonest dodge*
When the first seven instructions of D are correctly simulated by H it
can be seen that the simulated D would never stop running unless aborted
by H.
H: Begin Simulation Execution Trace Stored at:112ae5
Address_of_H:1383
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= ============= [000019b3][00112ad1][00112ad5] 55 push ebp // begin D
[000019b4][00112ad1][00112ad5] 8bec mov ebp,esp [000019b6][00112acd][00102aa1] 51 push ecx [000019b7][00112acd][00102aa1] 8b4508 mov eax,[ebp+08] [000019ba][00112ac9][000019b3] 50 push eax // push D
[000019bb][00112ac9][000019b3] 8b4d08 mov ecx,[ebp+08] [000019be][00112ac5][000019b3] 51 push ecx // push D
[000019bf][00112ac1][000019c4] e8bff9ffff call 00001383 // call H
H: Infinitely Recursive Simulation Detected Simulation Stopped
*D only stops running when H aborts its simulation of D*
If there is a countability issue with calculating the sum of arbitrary
finite sets of finite strings of ASCII digits that prevented a correct
sum from being calculated then there would be an element of this set
that cannot be summed.
No, you don't understand countability, and are trying to apply it to a
Red Herring. You seem to think that "Computation" == "Sum", which is
just a proof of your ignorance.
Not being able to count all of the sets of finite strings to be summed
in no way prevents any element of the sets of finite strings from being
correctly summed, thus has no impact on computability.
It isn't the set of finite strings that is uncountable, it is the set
of mappings every finite string individualy to 0 or 1 that is
uncountable.
You are restricting the set of natural number sums to 1 and 0?
Sum does not map every set of finite strings of ASCII digits to the
sum of these ASCII digit finite strings. It need not do that.
Sum maps *ARBITRARY ELEMENTS* of the *FINITE* sets of finite strings of
ASCII digits to their corresponding sum.
The number of Accept/Reject Mapping functions over Turing
Machines/Inputs is an order of infinity larger (it is uncountably
infinte) than the number of Turing Machines that could be built (a
countable infinity) to try to decide them, thus not all of these
mappings can be created by a Turing Machine.
As long as a TM can correctly determine the sum of any arbitrary finite
set of finite strings of ASCII digits then sum is computable.
Which is a Red Herring as no one is talking about "summing" being
non-computable, so the fact that you keep bringing it up just shows
you are too stupid to see your ignorance.
The same reasoning applies to halt deciders.
As long as a halt decider H correctly maps an arbitrary finite string
pair input to 1 or 0 on the basis of the behavior of its correct
simulation of this input then H is correct.
*D only stops running when H aborts its simulation of D*
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
You just don't understand the basics of category theory, and just fall
into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property doesn't
mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the wrong
answer.
If there are no counter-examples that prove that IS_SUM gets the wrong
answer then this logically entails that IS_SUM always gets the correct answer.
On 1/28/23 11:45 AM, olcott wrote:
On 1/28/2023 10:08 AM, Richard Damon wrote:
On 1/28/23 10:49 AM, olcott wrote:
On 1/28/2023 7:00 AM, Richard Damon wrote:
On 1/28/23 12:19 AM, olcott wrote:
On 1/27/2023 10:54 PM, Richard Damon wrote:
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of every >>>>>>>>>>>>>>>> finite
set of finite strings of ASCII digits, it merely has to >>>>>>>>>>>>>>>> always
compute this sum correctly for any arbitrary element of the >>>>>>>>>>>>>>>> finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the >>>>>>>>>>>>>> computability of halting on the basis that there are no >>>>>>>>>>>>>> countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any >>>>>>>>>>>> finite set of
finite strings then H cannot have any countability issue >>>>>>>>>>>> with any
finite pair of finite strings.
How do you make a countable infinite number of machine make >>>>>>>>>>> an uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of >>>>>>>>>> inputs.
We don't need any maps we only need one set of inputs deriving >>>>>>>>>> a single
output for any arbitrary set of inputs. When we think of this >>>>>>>>>> as sums
then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime
factors of its input it the same machine that computes its
inputs factorial?
You DO understand that a Turing Machine can be treated as a
computation that computes a specific mapping of inputs to outputs? >>>>>>>>>
A TM must only compute the mapping from any arbitrary finite set of >>>>>>>> inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits that >>>>>>>>>> cannot be
summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing
mapping of
one arbitrary finite set of inputs to its corresponding finite >>>>>>>> set of
outputs with any computation at all proves that there is no such >>>>>>>> issue
for summing or halt status detection.
So you are claiming that proof by example is true, that if one
example out of a set is true, the statement is true for all?
The fact that ONE machine can be counted, doesn't mean that all
can be.
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical Pathological >>>>>>> Lying Idiot.
As always when you run out of reasoning you resort to ad hominem
because
gullible fools will take ad hominem as rebuttal. Ad hominem makes
you look quite foolish to anyone accustomed to academic decorum.
No, I am showing you are using invalid logic.
Your statement is only true if you assume it to be true, and use it
as one of your truth makers that can make it true. This is not an
allowed operation, as if you do, then you get that exact same
arguement I was using.
So, *IF* you want to say your argument is valid, you *MUST& accept
that mine is too, and thus you are what I said.
The fact that you don't get it, might say my results are valid
anyway independent of validity of the statement.
In looking at your connection to truth makers view, X reaches back
to some base truth makers, but also links back to itself. This
leads to an infinite loop in the connection logic, where there is
always a piece of the connection not connected to things alreayd
known to be truth makers, so the statement has never been
established as True, which requires that ALL neccessary premises
for the statement be connected to know truth makers.
This is the flaw of you thinking you work backwards from the
premise to the truth makers, you can decieve yourself with such a
loop, thinking it has been established, when it is just a floating
island of unsupported logic. If you start at the Truth Makers, you
can't run into this problem, as you never had the presumptive
statement available to incorrectly use.
A counter-example to my reasoning does not exist.
Nope, I have given it.
“Analytic” sentences, such as “Pediatricians are doctors,” have >>>> historically been characterized as ones that are true by virtue of the >>>> meanings of their words alone and/or can be known to be so solely by
knowing those meanings.
https://plato.stanford.edu/entries/analytic-synthetic/
Right, and a sentence that depends on its own truth value (like if H
returns the right value ...) are proved to be able to lead to
cotradictipons.
This is correctly paraphrased as
Analytical truth is the connection from an expression X of formal or
natural language L using truth preserving operations to expressions
of L
that have been stipulated to be true.
Right, and you can not create such a chain to the statement you are
making.
We know that cat are animals is true entirely on the basis of the
meaning of the words. The meaning of words are expressions of L that
have been stipulated to be true.
So. Red Herring. Means nothing about the error of using a statement
that references its own truth.
Your statement that H can do xxx if H correctly decides yyy is a
statement of the class that has been proven to lead to logical
paradoxes.
Such a statement can never be actually connected to previously
accepted truth makers unless you can prove INDEPENDENTLY of that
statement that H does give a correct decision.
The fact that we know, by definition, that H(D,D) must be 1 to be
correct if D(D) halts,
*Changing the subject away from the following is a dishonest dodge*
When the first seven instructions of D are correctly simulated by H it
can be seen that the simulated D would never stop running unless
aborted by H.
H: Begin Simulation Execution Trace Stored at:112ae5
Address_of_H:1383
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[000019b3][00112ad1][00112ad5] 55 push ebp // begin D
[000019b4][00112ad1][00112ad5] 8bec mov ebp,esp
[000019b6][00112acd][00102aa1] 51 push ecx
[000019b7][00112acd][00102aa1] 8b4508 mov eax,[ebp+08]
[000019ba][00112ac9][000019b3] 50 push eax // push D
[000019bb][00112ac9][000019b3] 8b4d08 mov ecx,[ebp+08]
[000019be][00112ac5][000019b3] 51 push ecx // push D
[000019bf][00112ac1][000019c4] e8bff9ffff call 00001383 // call H
H: Infinitely Recursive Simulation Detected Simulation Stopped
*D only stops running when H aborts its simulation of D*
If there is a countability issue with calculating the sum of arbitrary >>>> finite sets of finite strings of ASCII digits that prevented a correct >>>> sum from being calculated then there would be an element of this set
that cannot be summed.
No, you don't understand countability, and are trying to apply it to
a Red Herring. You seem to think that "Computation" == "Sum", which
is just a proof of your ignorance.
Not being able to count all of the sets of finite strings to be summed >>>> in no way prevents any element of the sets of finite strings from being >>>> correctly summed, thus has no impact on computability.
It isn't the set of finite strings that is uncountable, it is the set
of mappings every finite string individualy to 0 or 1 that is
uncountable.
You are restricting the set of natural number sums to 1 and 0?
No, I am restricting the output of a decider to accept or reject.
The set of mappings (functions) of N -> N is also an uncountable set,
but that just shows that not all functions on the Natural Numbers are computable. Showing that not a decision functions are computable is a stronger statement.
Sum does not map every set of finite strings of ASCII digits to the
sum of these ASCII digit finite strings. It need not do that.
Why do you keep talking about Sum" That is just Herring with Red sauce.
On 1/28/2023 11:35 AM, Richard Damon wrote:
On 1/28/23 11:45 AM, olcott wrote:
On 1/28/2023 10:08 AM, Richard Damon wrote:
On 1/28/23 10:49 AM, olcott wrote:
On 1/28/2023 7:00 AM, Richard Damon wrote:
On 1/28/23 12:19 AM, olcott wrote:
On 1/27/2023 10:54 PM, Richard Damon wrote:
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of >>>>>>>>>>>>>>>>> every finite
set of finite strings of ASCII digits, it merely has to >>>>>>>>>>>>>>>>> always
compute this sum correctly for any arbitrary element of >>>>>>>>>>>>>>>>> the
finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with the >>>>>>>>>>>>>>> computability of halting on the basis that there are no >>>>>>>>>>>>>>> countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any >>>>>>>>>>>>> finite set of
finite strings then H cannot have any countability issue >>>>>>>>>>>>> with any
finite pair of finite strings.
How do you make a countable infinite number of machine make >>>>>>>>>>>> an uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of >>>>>>>>>>> inputs.
We don't need any maps we only need one set of inputs
deriving a single
output for any arbitrary set of inputs. When we think of this >>>>>>>>>>> as sums
then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime >>>>>>>>>> factors of its input it the same machine that computes its >>>>>>>>>> inputs factorial?
You DO understand that a Turing Machine can be treated as a >>>>>>>>>> computation that computes a specific mapping of inputs to
outputs?
A TM must only compute the mapping from any arbitrary finite >>>>>>>>> set of
inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits >>>>>>>>>>> that cannot be
summed by a TM? No, therefore computability is proven.
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing
mapping of
one arbitrary finite set of inputs to its corresponding finite >>>>>>>>> set of
outputs with any computation at all proves that there is no
such issue
for summing or halt status detection.
So you are claiming that proof by example is true, that if one >>>>>>>> example out of a set is true, the statement is true for all?
The fact that ONE machine can be counted, doesn't mean that all >>>>>>>> can be.
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical
Pathological Lying Idiot.
As always when you run out of reasoning you resort to ad hominem >>>>>>> because
gullible fools will take ad hominem as rebuttal. Ad hominem makes >>>>>>> you look quite foolish to anyone accustomed to academic decorum.
No, I am showing you are using invalid logic.
Your statement is only true if you assume it to be true, and use
it as one of your truth makers that can make it true. This is not >>>>>> an allowed operation, as if you do, then you get that exact same
arguement I was using.
So, *IF* you want to say your argument is valid, you *MUST& accept >>>>>> that mine is too, and thus you are what I said.
The fact that you don't get it, might say my results are valid
anyway independent of validity of the statement.
In looking at your connection to truth makers view, X reaches back >>>>>> to some base truth makers, but also links back to itself. This
leads to an infinite loop in the connection logic, where there is
always a piece of the connection not connected to things alreayd
known to be truth makers, so the statement has never been
established as True, which requires that ALL neccessary premises
for the statement be connected to know truth makers.
This is the flaw of you thinking you work backwards from the
premise to the truth makers, you can decieve yourself with such a
loop, thinking it has been established, when it is just a floating >>>>>> island of unsupported logic. If you start at the Truth Makers, you >>>>>> can't run into this problem, as you never had the presumptive
statement available to incorrectly use.
A counter-example to my reasoning does not exist.
Nope, I have given it.
“Analytic” sentences, such as “Pediatricians are doctors,” have >>>>> historically been characterized as ones that are true by virtue of the >>>>> meanings of their words alone and/or can be known to be so solely by >>>>> knowing those meanings.
https://plato.stanford.edu/entries/analytic-synthetic/
Right, and a sentence that depends on its own truth value (like if H
returns the right value ...) are proved to be able to lead to
cotradictipons.
This is correctly paraphrased as
Analytical truth is the connection from an expression X of formal or >>>>> natural language L using truth preserving operations to expressions
of L
that have been stipulated to be true.
Right, and you can not create such a chain to the statement you are
making.
We know that cat are animals is true entirely on the basis of the
meaning of the words. The meaning of words are expressions of L that >>>>> have been stipulated to be true.
So. Red Herring. Means nothing about the error of using a statement
that references its own truth.
Your statement that H can do xxx if H correctly decides yyy is a
statement of the class that has been proven to lead to logical
paradoxes.
Such a statement can never be actually connected to previously
accepted truth makers unless you can prove INDEPENDENTLY of that
statement that H does give a correct decision.
The fact that we know, by definition, that H(D,D) must be 1 to be
correct if D(D) halts,
*Changing the subject away from the following is a dishonest dodge*
When the first seven instructions of D are correctly simulated by H
it can be seen that the simulated D would never stop running unless
aborted by H.
H: Begin Simulation Execution Trace Stored at:112ae5
Address_of_H:1383
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[000019b3][00112ad1][00112ad5] 55 push ebp // begin D
[000019b4][00112ad1][00112ad5] 8bec mov ebp,esp
[000019b6][00112acd][00102aa1] 51 push ecx
[000019b7][00112acd][00102aa1] 8b4508 mov eax,[ebp+08]
[000019ba][00112ac9][000019b3] 50 push eax // push D
[000019bb][00112ac9][000019b3] 8b4d08 mov ecx,[ebp+08]
[000019be][00112ac5][000019b3] 51 push ecx // push D
[000019bf][00112ac1][000019c4] e8bff9ffff call 00001383 // call H
H: Infinitely Recursive Simulation Detected Simulation Stopped
*D only stops running when H aborts its simulation of D*
If there is a countability issue with calculating the sum of arbitrary >>>>> finite sets of finite strings of ASCII digits that prevented a correct >>>>> sum from being calculated then there would be an element of this set >>>>> that cannot be summed.
No, you don't understand countability, and are trying to apply it to
a Red Herring. You seem to think that "Computation" == "Sum", which
is just a proof of your ignorance.
Not being able to count all of the sets of finite strings to be summed >>>>> in no way prevents any element of the sets of finite strings from
being
correctly summed, thus has no impact on computability.
It isn't the set of finite strings that is uncountable, it is the
set of mappings every finite string individualy to 0 or 1 that is
uncountable.
You are restricting the set of natural number sums to 1 and 0?
No, I am restricting the output of a decider to accept or reject.
The set of mappings (functions) of N -> N is also an uncountable set,
but that just shows that not all functions on the Natural Numbers are
computable. Showing that not a decision functions are computable is a
stronger statement.
Sum does not map every set of finite strings of ASCII digits to the
sum of these ASCII digit finite strings. It need not do that.
Why do you keep talking about Sum" That is just Herring with Red sauce.
We can transform sum into a simple decider.
A TM decider takes a space delimited finite list of finite strings of
ASCII digits: [0123456789]+ such that the first string is the sum of the remaining elements. Each element of the list is delimited by a
single space. The last element is marked by four trailing spaces.
IS_SUM computes the sum and compares this computed sum to the first
element on this list. IS_SUM accepts or rejects its input on the basis
that the computed == the first element of its finite list of finite
strings.
IS_SUM need not compute the mapping from every set of finite strings to
its accept or reject state, it only needs to compute the mapping from
any arbitrary element of the finite sets of finite strings of ASCII
digits.
On 1/28/23 3:16 PM, olcott wrote:
On 1/28/2023 11:35 AM, Richard Damon wrote:
On 1/28/23 11:45 AM, olcott wrote:
On 1/28/2023 10:08 AM, Richard Damon wrote:
On 1/28/23 10:49 AM, olcott wrote:
On 1/28/2023 7:00 AM, Richard Damon wrote:
On 1/28/23 12:19 AM, olcott wrote:
On 1/27/2023 10:54 PM, Richard Damon wrote:
On 1/27/23 11:47 PM, olcott wrote:
On 1/27/2023 10:05 PM, Richard Damon wrote:
On 1/27/23 10:59 PM, olcott wrote:
On 1/27/2023 9:47 PM, Richard Damon wrote:
On 1/27/23 10:34 PM, olcott wrote:
On 1/27/2023 9:19 PM, Richard Damon wrote:
On 1/27/23 8:56 PM, olcott wrote:
On 1/26/2023 10:16 PM, Richard Damon wrote:
On 1/26/23 10:46 PM, olcott wrote:
To be correct the TM need not calculate the sum of >>>>>>>>>>>>>>>>>> every finite
set of finite strings of ASCII digits, it merely has >>>>>>>>>>>>>>>>>> to always
compute this sum correctly for any arbitrary element >>>>>>>>>>>>>>>>>> of the
finite set of finite strings.
NBot talking about "Sums"
I just proved that there are no countability issues with >>>>>>>>>>>>>>>> the
computability of halting on the basis that there are no >>>>>>>>>>>>>>>> countability
issues with the computation of sums.
Nope. Falllicy of proof byt example.
None-the-less if sum has no countability issue with any >>>>>>>>>>>>>> finite set of
finite strings then H cannot have any countability issue >>>>>>>>>>>>>> with any
finite pair of finite strings.
How do you make a countable infinite number of machine make >>>>>>>>>>>>> an uncountable number of maps.
I ask you again can you count to one?
YTes, SO WHAT.
There is only one machine that always has a finite number of >>>>>>>>>>>> inputs.
We don't need any maps we only need one set of inputs
deriving a single
output for any arbitrary set of inputs. When we think of >>>>>>>>>>>> this as sums
then it is obvious that countability is not an issue.
So you are saying that the machine that computes the prime >>>>>>>>>>> factors of its input it the same machine that computes its >>>>>>>>>>> inputs factorial?
You DO understand that a Turing Machine can be treated as a >>>>>>>>>>> computation that computes a specific mapping of inputs to >>>>>>>>>>> outputs?
A TM must only compute the mapping from any arbitrary finite >>>>>>>>>> set of
inputs to its finite set of outputs.
Maybe not, as you are too stupid.
Is there any finite set of finite strings of ASCII digits >>>>>>>>>>>> that cannot be
summed by a TM? No, therefore computability is proven. >>>>>>>>>>>>
But summing isn't the only operation that a TM can do.
That there is never any countability issue in the computing >>>>>>>>>> mapping of
one arbitrary finite set of inputs to its corresponding finite >>>>>>>>>> set of
outputs with any computation at all proves that there is no >>>>>>>>>> such issue
for summing or halt status detection.
So you are claiming that proof by example is true, that if one >>>>>>>>> example out of a set is true, the statement is true for all? >>>>>>>>>
The fact that ONE machine can be counted, doesn't mean that all >>>>>>>>> can be.
If X always works then X always works with Y or Z.
So if this statement is true, you are a Hypocritical
Pathological Lying Idiot.
As always when you run out of reasoning you resort to ad hominem >>>>>>>> because
gullible fools will take ad hominem as rebuttal. Ad hominem
makes you look quite foolish to anyone accustomed to academic
decorum.
No, I am showing you are using invalid logic.
Your statement is only true if you assume it to be true, and use >>>>>>> it as one of your truth makers that can make it true. This is
not an allowed operation, as if you do, then you get that exact
same arguement I was using.
So, *IF* you want to say your argument is valid, you *MUST&
accept that mine is too, and thus you are what I said.
The fact that you don't get it, might say my results are valid
anyway independent of validity of the statement.
In looking at your connection to truth makers view, X reaches
back to some base truth makers, but also links back to itself.
This leads to an infinite loop in the connection logic, where
there is always a piece of the connection not connected to things >>>>>>> alreayd known to be truth makers, so the statement has never been >>>>>>> established as True, which requires that ALL neccessary premises >>>>>>> for the statement be connected to know truth makers.
This is the flaw of you thinking you work backwards from the
premise to the truth makers, you can decieve yourself with such a >>>>>>> loop, thinking it has been established, when it is just a
floating island of unsupported logic. If you start at the Truth
Makers, you can't run into this problem, as you never had the
presumptive statement available to incorrectly use.
A counter-example to my reasoning does not exist.
Nope, I have given it.
“Analytic” sentences, such as “Pediatricians are doctors,” have >>>>>> historically been characterized as ones that are true by virtue of >>>>>> the
meanings of their words alone and/or can be known to be so solely by >>>>>> knowing those meanings.
https://plato.stanford.edu/entries/analytic-synthetic/
Right, and a sentence that depends on its own truth value (like if
H returns the right value ...) are proved to be able to lead to
cotradictipons.
This is correctly paraphrased as
Analytical truth is the connection from an expression X of formal or >>>>>> natural language L using truth preserving operations to
expressions of L
that have been stipulated to be true.
Right, and you can not create such a chain to the statement you are
making.
We know that cat are animals is true entirely on the basis of the
meaning of the words. The meaning of words are expressions of L that >>>>>> have been stipulated to be true.
So. Red Herring. Means nothing about the error of using a statement
that references its own truth.
Your statement that H can do xxx if H correctly decides yyy is a
statement of the class that has been proven to lead to logical
paradoxes.
Such a statement can never be actually connected to previously
accepted truth makers unless you can prove INDEPENDENTLY of that
statement that H does give a correct decision.
The fact that we know, by definition, that H(D,D) must be 1 to be
correct if D(D) halts,
*Changing the subject away from the following is a dishonest dodge*
When the first seven instructions of D are correctly simulated by H
it can be seen that the simulated D would never stop running unless
aborted by H.
H: Begin Simulation Execution Trace Stored at:112ae5
Address_of_H:1383
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[000019b3][00112ad1][00112ad5] 55 push ebp // begin D
[000019b4][00112ad1][00112ad5] 8bec mov ebp,esp
[000019b6][00112acd][00102aa1] 51 push ecx
[000019b7][00112acd][00102aa1] 8b4508 mov eax,[ebp+08]
[000019ba][00112ac9][000019b3] 50 push eax // push D
[000019bb][00112ac9][000019b3] 8b4d08 mov ecx,[ebp+08]
[000019be][00112ac5][000019b3] 51 push ecx // push D
[000019bf][00112ac1][000019c4] e8bff9ffff call 00001383 // call H
H: Infinitely Recursive Simulation Detected Simulation Stopped
*D only stops running when H aborts its simulation of D*
If there is a countability issue with calculating the sum of
arbitrary
finite sets of finite strings of ASCII digits that prevented a
correct
sum from being calculated then there would be an element of this set >>>>>> that cannot be summed.
No, you don't understand countability, and are trying to apply it
to a Red Herring. You seem to think that "Computation" == "Sum",
which is just a proof of your ignorance.
Not being able to count all of the sets of finite strings to be
summed
in no way prevents any element of the sets of finite strings from
being
correctly summed, thus has no impact on computability.
It isn't the set of finite strings that is uncountable, it is the
set of mappings every finite string individualy to 0 or 1 that is
uncountable.
You are restricting the set of natural number sums to 1 and 0?
No, I am restricting the output of a decider to accept or reject.
The set of mappings (functions) of N -> N is also an uncountable set,
but that just shows that not all functions on the Natural Numbers are
computable. Showing that not a decision functions are computable is a
stronger statement.
Sum does not map every set of finite strings of ASCII digits to the
sum of these ASCII digit finite strings. It need not do that.
Why do you keep talking about Sum" That is just Herring with Red sauce.
We can transform sum into a simple decider.
A TM decider takes a space delimited finite list of finite strings of
ASCII digits: [0123456789]+ such that the first string is the sum of the
remaining elements. Each element of the list is delimited by a
single space. The last element is marked by four trailing spaces.
IS_SUM computes the sum and compares this computed sum to the first
element on this list. IS_SUM accepts or rejects its input on the basis
that the computed == the first element of its finite list of finite
strings.
IS_SUM need not compute the mapping from every set of finite strings to
its accept or reject state, it only needs to compute the mapping from
any arbitrary element of the finite sets of finite strings of ASCII
digits.
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
You just don't understand the basics of category theory, and just fall
into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property doesn't mean
that property applies to the whole class.
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a decider is?
You just don't understand the basics of category theory, and just
fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property doesn't
mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the wrong
answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets the wrong
answer then this logically entails that IS_SUM always gets the correct
answer.
But only to the one question it was built for.
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a decider is?
You just don't understand the basics of category theory, and just
fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property doesn't
mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the wrong
answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets the wrong
answer then this logically entails that IS_SUM always gets the correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with
DOES_HALT. DOES_HALT merely needs to compute the mapping from any
arbitrary input pair of finite strings to its accept or reject state.
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a decider is?
You just don't understand the basics of category theory, and just
fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property doesn't
mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the wrong
answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets the wrong >>>> answer then this logically entails that IS_SUM always gets the correct >>>> answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with
DOES_HALT. DOES_HALT merely needs to compute the mapping from any
arbitrary input pair of finite strings to its accept or reject state.
Can you PROVE that this is doable?
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a decider is?
You just don't understand the basics of category theory, and just
fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property
doesn't mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the
wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets the wrong >>>>> answer then this logically entails that IS_SUM always gets the correct >>>>> answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with
DOES_HALT. DOES_HALT merely needs to compute the mapping from any
arbitrary input pair of finite strings to its accept or reject state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of finite string inputs. If any of these finite strings contains a character that is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was
previously specified.
Because we can see from IS_SUM that there cannot be any countability
issues for any decider that takes a finite set of arbitrary finite
string inputs therefore there cannot be any countability issue for any decider that takes a pair of arbitrary finite sting inputs such as
DOES_HALT.
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a decider is? >>>>>>
You just don't understand the basics of category theory, and
just fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property
doesn't mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the
wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets the >>>>>>> wrong
answer then this logically entails that IS_SUM always gets the
correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with
DOES_HALT. DOES_HALT merely needs to compute the mapping from any
arbitrary input pair of finite strings to its accept or reject state. >>>>>
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of finite string
inputs. If any of these finite strings contains a character that is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was
previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_PRIME would have a single finite string of ASCII digits as its only
input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
and produces 1 (ONE), count them, map of inputs to outputs.
You still have a uncountable infinte number of mappings to go.
Maybe I need to always tell you things fifteen times every time that I
say them. You clearly have an attention deficit issue.
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a decider is? >>>>>
You just don't understand the basics of category theory, and just >>>>>>> fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property
doesn't mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the
wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets the
wrong
answer then this logically entails that IS_SUM always gets the
correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with
DOES_HALT. DOES_HALT merely needs to compute the mapping from any
arbitrary input pair of finite strings to its accept or reject state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of finite string
inputs. If any of these finite strings contains a character that is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was
previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_SUM takes the countably infinite number of inputs
and produces 1
(ONE), count them, map of inputs to outputs.
You still have a uncountable infinte number of mappings to go.
On 1/29/23 8:28 PM, olcott wrote:IS_SUM need not ever count any uncountable set. All that IS_SUM must do
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a decider is? >>>>>>>
You just don't understand the basics of category theory, and >>>>>>>>> just fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property
doesn't mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the >>>>>>>> wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets the >>>>>>>> wrong
answer then this logically entails that IS_SUM always gets the >>>>>>>> correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with
DOES_HALT. DOES_HALT merely needs to compute the mapping from any
arbitrary input pair of finite strings to its accept or reject state. >>>>>>
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of finite string >>>> inputs. If any of these finite strings contains a character that is not >>>> an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was
previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_PRIME would have a single finite string of ASCII digits as its only
input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
And computes ONE mapping, out of the uncountable infinte many mappings
of strings -> the set of answers for each input.
On 1/29/2023 7:41 PM, Richard Damon wrote:
On 1/29/23 8:28 PM, olcott wrote:IS_SUM need not ever count any uncountable set. All that IS_SUM must do
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a decider >>>>>>>> is?
You just don't understand the basics of category theory, and >>>>>>>>>> just fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property >>>>>>>>>> doesn't mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the >>>>>>>>> wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets >>>>>>>>> the wrong
answer then this logically entails that IS_SUM always gets the >>>>>>>>> correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with
DOES_HALT. DOES_HALT merely needs to compute the mapping from any >>>>>>> arbitrary input pair of finite strings to its accept or reject
state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of finite
string
inputs. If any of these finite strings contains a character that is
not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was
previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_PRIME would have a single finite string of ASCII digits as its
only input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
And computes ONE mapping, out of the uncountable infinte many mappings
of strings -> the set of answers for each input.
is compute each mapping that it is presented with, one finite set of
finite strings at a time.
On 1/29/23 8:48 PM, olcott wrote:
On 1/29/2023 7:41 PM, Richard Damon wrote:
On 1/29/23 8:28 PM, olcott wrote:IS_SUM need not ever count any uncountable set. All that IS_SUM must
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a
decider is?
You just don't understand the basics of category theory, and >>>>>>>>>>> just fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property >>>>>>>>>>> doesn't mean that property applies to the whole class.
Try and provide a single counter-example where IS_SUM gets the >>>>>>>>>> wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets >>>>>>>>>> the wrong
answer then this logically entails that IS_SUM always gets the >>>>>>>>>> correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with >>>>>>>> DOES_HALT. DOES_HALT merely needs to compute the mapping from any >>>>>>>> arbitrary input pair of finite strings to its accept or reject >>>>>>>> state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of finite
string
inputs. If any of these finite strings contains a character that
is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was
previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_PRIME would have a single finite string of ASCII digits as its
only input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
And computes ONE mapping, out of the uncountable infinte many
mappings of strings -> the set of answers for each input.
do is compute each mapping that it is presented with, one finite set
of finite strings at a time.
So????
We aren't talking about IS_SUM being non-computable,
On 1/29/2023 8:06 PM, Richard Damon wrote:
On 1/29/23 8:48 PM, olcott wrote:
On 1/29/2023 7:41 PM, Richard Damon wrote:
On 1/29/23 8:28 PM, olcott wrote:IS_SUM need not ever count any uncountable set. All that IS_SUM must
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a
decider is?
You just don't understand the basics of category theory, and >>>>>>>>>>>> just fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property >>>>>>>>>>>> doesn't mean that property applies to the whole class. >>>>>>>>>>>>
Try and provide a single counter-example where IS_SUM gets >>>>>>>>>>> the wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets >>>>>>>>>>> the wrong
answer then this logically entails that IS_SUM always gets >>>>>>>>>>> the correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with >>>>>>>>> DOES_HALT. DOES_HALT merely needs to compute the mapping from any >>>>>>>>> arbitrary input pair of finite strings to its accept or reject >>>>>>>>> state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of finite >>>>>>> string
inputs. If any of these finite strings contains a character that >>>>>>> is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was >>>>>>> previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_PRIME would have a single finite string of ASCII digits as its
only input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
And computes ONE mapping, out of the uncountable infinte many
mappings of strings -> the set of answers for each input.
do is compute each mapping that it is presented with, one finite set
of finite strings at a time.
So????
We aren't talking about IS_SUM being non-computable,
Then the "you" of "we" is off topic. Countability was supposed to be an alternate proof that halting is undecidable.
On 1/29/23 9:27 PM, olcott wrote:
On 1/29/2023 8:06 PM, Richard Damon wrote:Right, countablility of mappings.
On 1/29/23 8:48 PM, olcott wrote:
On 1/29/2023 7:41 PM, Richard Damon wrote:
On 1/29/23 8:28 PM, olcott wrote:IS_SUM need not ever count any uncountable set. All that IS_SUM must
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a >>>>>>>>>>> decider is?
You just don't understand the basics of category theory, >>>>>>>>>>>>> and just fall into the fallacy of Proof by Example.
Just because one subset of a set happens to have a property >>>>>>>>>>>>> doesn't mean that property applies to the whole class. >>>>>>>>>>>>>
Try and provide a single counter-example where IS_SUM gets >>>>>>>>>>>> the wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM gets >>>>>>>>>>>> the wrong
answer then this logically entails that IS_SUM always gets >>>>>>>>>>>> the correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with >>>>>>>>>> DOES_HALT. DOES_HALT merely needs to compute the mapping from any >>>>>>>>>> arbitrary input pair of finite strings to its accept or reject >>>>>>>>>> state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of finite >>>>>>>> string
inputs. If any of these finite strings contains a character that >>>>>>>> is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was >>>>>>>> previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_PRIME would have a single finite string of ASCII digits as its
only input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
And computes ONE mapping, out of the uncountable infinte many
mappings of strings -> the set of answers for each input.
do is compute each mapping that it is presented with, one finite set
of finite strings at a time.
So????
We aren't talking about IS_SUM being non-computable,
Then the "you" of "we" is off topic. Countability was supposed to be
an alternate proof that halting is undecidable.
The computability of a single mapping show nothing about the
countability arguement.
No one says that NO mappings are computable, just that most are not.
You clearly don't understand categorical logic.
Again, your use of the fallacy of proof by example.
On 1/29/2023 8:52 PM, Richard Damon wrote:
On 1/29/23 9:27 PM, olcott wrote:
On 1/29/2023 8:06 PM, Richard Damon wrote:Right, countablility of mappings.
On 1/29/23 8:48 PM, olcott wrote:
On 1/29/2023 7:41 PM, Richard Damon wrote:
On 1/29/23 8:28 PM, olcott wrote:IS_SUM need not ever count any uncountable set. All that IS_SUM
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation.
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a >>>>>>>>>>>> decider is?
You just don't understand the basics of category theory, >>>>>>>>>>>>>> and just fall into the fallacy of Proof by Example. >>>>>>>>>>>>>>
Just because one subset of a set happens to have a >>>>>>>>>>>>>> property doesn't mean that property applies to the whole >>>>>>>>>>>>>> class.
Try and provide a single counter-example where IS_SUM gets >>>>>>>>>>>>> the wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM >>>>>>>>>>>>> gets the wrong
answer then this logically entails that IS_SUM always gets >>>>>>>>>>>>> the correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with >>>>>>>>>>> DOES_HALT. DOES_HALT merely needs to compute the mapping from >>>>>>>>>>> any
arbitrary input pair of finite strings to its accept or
reject state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of
finite string
inputs. If any of these finite strings contains a character
that is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was >>>>>>>>> previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_PRIME would have a single finite string of ASCII digits as its >>>>>>> only input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
And computes ONE mapping, out of the uncountable infinte many
mappings of strings -> the set of answers for each input.
must do is compute each mapping that it is presented with, one
finite set of finite strings at a time.
So????
We aren't talking about IS_SUM being non-computable,
Then the "you" of "we" is off topic. Countability was supposed to be
an alternate proof that halting is undecidable.
The computability of a single mapping show nothing about the
countability arguement.
No one says that NO mappings are computable, just that most are not.
No one can show even one mapping of a finite set of string inputs to
IS_SUM that is not computable, thus the conclusion that there are more
than one counter-example is totally refuted.
You clearly don't understand categorical logic.
Categorically exhaustive reasoning is my key innovation it utterly
eliminates the error of omission.
If it is impossible to show as many as a single counter-example then we
can correctly generalize this to say that more than one counter example
also does not exist.
Again, your use of the fallacy of proof by example.
On 1/30/23 12:08 AM, olcott wrote:
On 1/29/2023 8:52 PM, Richard Damon wrote:
On 1/29/23 9:27 PM, olcott wrote:
On 1/29/2023 8:06 PM, Richard Damon wrote:Right, countablility of mappings.
On 1/29/23 8:48 PM, olcott wrote:
On 1/29/2023 7:41 PM, Richard Damon wrote:
On 1/29/23 8:28 PM, olcott wrote:IS_SUM need not ever count any uncountable set. All that IS_SUM
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation. >>>>>>>>>>>>>>>
Not All Deciders can be built from a Sumation.
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a >>>>>>>>>>>>> decider is?
You just don't understand the basics of category theory, >>>>>>>>>>>>>>> and just fall into the fallacy of Proof by Example. >>>>>>>>>>>>>>>
Just because one subset of a set happens to have a >>>>>>>>>>>>>>> property doesn't mean that property applies to the whole >>>>>>>>>>>>>>> class.
Try and provide a single counter-example where IS_SUM gets >>>>>>>>>>>>>> the wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM >>>>>>>>>>>>>> gets the wrong
answer then this logically entails that IS_SUM always gets >>>>>>>>>>>>>> the correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues with >>>>>>>>>>>> DOES_HALT. DOES_HALT merely needs to compute the mapping >>>>>>>>>>>> from any
arbitrary input pair of finite strings to its accept or >>>>>>>>>>>> reject state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of
finite string
inputs. If any of these finite strings contains a character >>>>>>>>>> that is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was >>>>>>>>>> previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question?
IS_PRIME would have a single finite string of ASCII digits as
its only input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
And computes ONE mapping, out of the uncountable infinte many
mappings of strings -> the set of answers for each input.
must do is compute each mapping that it is presented with, one
finite set of finite strings at a time.
So????
We aren't talking about IS_SUM being non-computable,
Then the "you" of "we" is off topic. Countability was supposed to be
an alternate proof that halting is undecidable.
The computability of a single mapping show nothing about the
countability arguement.
No one says that NO mappings are computable, just that most are not.
No one can show even one mapping of a finite set of string inputs to
IS_SUM that is not computable, thus the conclusion that there are more
than one counter-example is totally refuted.
????
You seem to think that is_sum computes multiple mappings.
*A* mapping, as being talked about here, is a complete listing of the required output for every possible input.
IS_SUM computes *ONE* mapping of input to output.
Yes, it can computer the
You clearly don't understand categorical logic.
Categorically exhaustive reasoning is my key innovation it utterly
eliminates the error of omission.
No, it is your key fallicy. You seem to think that a detail analysis of
one case tells you all you need about every case that you don't look at.
That is just
If it is impossible to show as many as a single counter-example then we
can correctly generalize this to say that more than one counter example
also does not exist.
No, you are just using the fallacy of the Red Herring, because you
demand a counter-example of something that isn't the problem.
How does your "IS_SUM" machine answer both the question of "is 8 the sum
of two primes?" and "is 8 a perfect number?" (assuming the first is what
you are defininig IS_SUM to be, not sure what else you mean, sometime it seems like IS_SUM is computing the sum of two numbers give to is, which
isn't an "IS" question.
Note, these two question both take the SAME input (8), but have
different outputs,
For IS_SUM_OF_PRIMES, the answer is yes, because 8 = 3 + 5
For IS_PERFECT, the answer is no, because 8 has factors to 1, 2, 4 which
only total to 7, not 8.
You claim same machine, in is the same, output will be the same, thus
must give the wrong answer to one of them.
Again, your use of the fallacy of proof by example.
On 1/30/2023 6:10 AM, Richard Damon wrote:
On 1/30/23 12:08 AM, olcott wrote:
On 1/29/2023 8:52 PM, Richard Damon wrote:
On 1/29/23 9:27 PM, olcott wrote:
On 1/29/2023 8:06 PM, Richard Damon wrote:Right, countablility of mappings.
On 1/29/23 8:48 PM, olcott wrote:
On 1/29/2023 7:41 PM, Richard Damon wrote:
On 1/29/23 8:28 PM, olcott wrote:IS_SUM need not ever count any uncountable set. All that IS_SUM
On 1/29/2023 7:10 PM, Richard Damon wrote:
On 1/29/23 7:28 PM, olcott wrote:
On 1/28/2023 9:47 PM, Richard Damon wrote:
On 1/28/23 4:29 PM, olcott wrote:
On 1/28/2023 3:13 PM, Richard Damon wrote:
On 1/28/23 3:59 PM, olcott wrote:
On 1/28/2023 2:42 PM, Richard Damon wrote:
Yes, you can build SOME Deciders from a Sumation. >>>>>>>>>>>>>>>>
Not All Deciders can be built from a Sumation. >>>>>>>>>>>>>>>>
One TM is *ALL* deciders.
Really?
It can be both a is_prime and is_perfect decider?
Maybe your problem is you don't understand at all what a >>>>>>>>>>>>>> decider is?
You just don't understand the basics of category theory, >>>>>>>>>>>>>>>> and just fall into the fallacy of Proof by Example. >>>>>>>>>>>>>>>>
Just because one subset of a set happens to have a >>>>>>>>>>>>>>>> property doesn't mean that property applies to the whole >>>>>>>>>>>>>>>> class.
Try and provide a single counter-example where IS_SUM >>>>>>>>>>>>>>> gets the wrong answer.
If the question is Is the number prime?
If there are no counter-examples that prove that IS_SUM >>>>>>>>>>>>>>> gets the wrong
answer then this logically entails that IS_SUM always >>>>>>>>>>>>>>> gets the correct
answer.
But only to the one question it was built for.
Likewise for DOES_HALT, there are zero countability issues >>>>>>>>>>>>> with
DOES_HALT. DOES_HALT merely needs to compute the mapping >>>>>>>>>>>>> from any
arbitrary input pair of finite strings to its accept or >>>>>>>>>>>>> reject state.
Can you PROVE that this is doable?
We can change IS_SUM to allow any arbitrary finite set of >>>>>>>>>>> finite string
inputs. If any of these finite strings contains a character >>>>>>>>>>> that is not
an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it >>>>>>>>>>> was
previously specified.
Ok, how do you use IS_SUM to answer the IS_PRIME question? >>>>>>>>>>
IS_PRIME would have a single finite string of ASCII digits as >>>>>>>>> its only input and you already know this is computable.
IS_SUM takes the countably infinite number of inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
IS_SUM takes a finite set of finite string inputs
And computes ONE mapping, out of the uncountable infinte many
mappings of strings -> the set of answers for each input.
must do is compute each mapping that it is presented with, one
finite set of finite strings at a time.
So????
We aren't talking about IS_SUM being non-computable,
Then the "you" of "we" is off topic. Countability was supposed to
be an alternate proof that halting is undecidable.
The computability of a single mapping show nothing about the
countability arguement.
No one says that NO mappings are computable, just that most are not.
No one can show even one mapping of a finite set of string inputs to
IS_SUM that is not computable, thus the conclusion that there are more
than one counter-example is totally refuted.
????
You seem to think that is_sum computes multiple mappings.
*A* mapping, as being talked about here, is a complete listing of the
required output for every possible input.
IS_SUM computes *ONE* mapping of input to output.
Yes, it can computer the
You clearly don't understand categorical logic.
Categorically exhaustive reasoning is my key innovation it utterly
eliminates the error of omission.
No, it is your key fallicy. You seem to think that a detail analysis
of one case tells you all you need about every case that you don't
look at.
That is just
If it is impossible to show as many as a single counter-example then we
can correctly generalize this to say that more than one counter example
also does not exist.
No, you are just using the fallacy of the Red Herring, because you
demand a counter-example of something that isn't the problem.
If no counter example can be provided that a decision problem is
undecidable then this proves that it is decidable.
How does your "IS_SUM" machine answer both the question of "is 8 the
sum of two primes?" and "is 8 a perfect number?" (assuming the first
is what you are defininig IS_SUM to be, not sure what else you mean,
sometime it seems like IS_SUM is computing the sum of two numbers give
to is, which isn't an "IS" question.
IS_SUM determines if its first finite string of ASCII digits is the sum
of its remaining finite list of finite strings of ASCII digits. Maybe
you need to take notes. There is also an adaptation of IS_SUM that takes arbitrary finite strings.
Note, these two question both take the SAME input (8), but have
different outputs,
For IS_SUM_OF_PRIMES, the answer is yes, because 8 = 3 + 5
For IS_PERFECT, the answer is no, because 8 has factors to 1, 2, 4
which only total to 7, not 8.
You claim same machine, in is the same, output will be the same, thus
must give the wrong answer to one of them.
One decision problem per TM. A finite number of different decision
problems could be selected for computation by a single TM when the
ordinal number of the decision problem is the first input parameter.
Every decision problem that operates on a finite set of finite strings
is not uncomputable for any countability reasons.
The most straight forward to validate the Goldbach Conjecture is to
simply test every element of the set of natural numbers. Since this is
not a finite list of inputs it is not computable. Mathematical induction
can be used to algorithmically compress the analysis of some infinite
sets. https://www.britannica.com/science/Goldbach-conjecture
Again, your use of the fallacy of proof by example.
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