https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/30/2021 2:24 PM, olcott wrote:
>
> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>
>
Refuting the Halting Problem Diagonalization Argument

Every machine that halts in a reject state is a halting computation. At
least two proofs ignore this when constructing Sipser's Figure 4.5.
Because these two proofs ignore this when they insert machine D into
Sipser's Figure 4.5 they do so incorrectly.

When machine D is inserted into both Figure 4.4 and Figure 4.5 correctly
then the contradiction goes away. Since Sipser implicitly assumes that
every blank entry of Figure 4.4 is a ~halt entry Figure 4.7a makes this
explicit.

    <M1>     <M2>     <M3>     <M4> ... <D>
M1  accept   ~halt    accept   ~halt    reject
M2  accept   accept   accept   accept   reject
M3  ~halt    ~halt    ~halt    ~halt    accept
M4  accept   accept   ~halt    ~halt    accept
....
D   reject   reject   accept   accept   reject

Figure 4.7a (corrected figure 4.6, inserting D into figure 4.4)

    <M1>     <M2>     <M3>     <M4> ... <D>
M1  accept   reject   accept   reject   accept
M2  accept   accept   accept   accept   accept
M3  reject   reject   reject   reject   accept
M4  accept   accept   reject   reject   accept
....
D   accept   accept   accept   accept   accept

Figure 4.7b (corrected figure 4.6, inserting D into figure 4.5)

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre minds." Einstein

Refuting the Halting Problem Diagonalization Argument

Every machine that halts in a reject state is a halting computation. At
least two proofs ignore this when constructing Sipser's Figure 4.5.
Because these two proofs ignore this when they insert machine D into
Sipser's Figure 4.5 they do so incorrectly.

When machine D is inserted into both Figure 4.4 and Figure 4.5 correctly
then the contradiction goes away. Since Sipser implicitly assumes that
every blank entry of Figure 4.4 is a ~halt entry Figure 4.7a makes this
explicit.

<M1> <M2> <M3> <M4> ... <D>
M1 accept ~halt accept ~halt reject
M2 accept accept accept accept reject
M3 ~halt ~halt ~halt ~halt accept
M4 accept accept ~halt ~halt accept
....
D reject reject accept accept reject

Figure 4.7a (corrected figure 4.6, inserting D into figure 4.4)

<M1> <M2> <M3> <M4> ... <D>
M1 accept reject accept reject accept
M2 accept accept accept accept accept
M3 reject reject reject reject accept
M4 accept accept reject reject accept
....
D accept accept accept accept accept

Figure 4.7b (corrected figure 4.6, inserting D into figure 4.5)

https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/30/21 4:37 PM, olcott wrote:
> On 5/30/2021 2:24 PM, olcott wrote:
>>
>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>
>>
> Refuting the Halting Problem Diagonalization Argument
>
> Every machine that halts in a reject state is a halting computation. At
> least two proofs ignore this when constructing Sipser's Figure 4.5.
> Because these two proofs ignore this when they insert machine D into
> Sipser's Figure 4.5 they do so incorrectly.
>
> When machine D is inserted into both Figure 4.4 and Figure 4.5 correctly
> then the contradiction goes away. Since Sipser implicitly assumes that
> every blank entry of Figure 4.4 is a ~halt entry Figure 4.7a makes this
> explicit.
>
>     <M1>     <M2>     <M3>     <M4> ... <D>
> M1  accept   ~halt    accept   ~halt    reject
> M2  accept   accept   accept   accept   reject
> M3  ~halt    ~halt    ~halt    ~halt    accept
> M4  accept   accept   ~halt    ~halt    accept
> ...
> D   reject   reject   accept   accept   reject
>
> Figure 4.7a (corrected figure 4.6, inserting D into figure 4.4)
>
>     <M1>     <M2>     <M3>     <M4> ... <D>
> M1  accept   reject   accept   reject   accept
> M2  accept   accept   accept   accept   accept
> M3  reject   reject   reject   reject   accept
> M4  accept   accept   reject   reject   accept
> ...
> D   accept   accept   accept   accept   accept
>
> Figure 4.7b (corrected figure 4.6, inserting D into figure 4.5)
>

You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
there is no end to the chart. The chart is a listing of EVERY Turing
Machine, ordered by some rule. In esssence, every possible Turing
Machine can be given a number, based on its position in the table. The
'end' of the chart would be the highest possible Natural Number (which
there isn't).

You have just assigned D to be the number "Highest Possible Natural
Number + 1", which isn't a number.

One Key to the diagonal theorem is that since it is a listing of EVERY
possible Turing Machine, if D actually WAS a Turing Machine, it could be
found somewhere in that listing. The fact that it CAN'T possibly be in
that listing says it can't be a Turing Machine. Since if H was a Turing
Machine, we could easily derive D from it, this means that H as defined
can't be a Turing Machine, i.e. H can't exist.

On 5/30/2021 5:57 PM, Richard Damon wrote:
> On 5/30/21 4:37 PM, olcott wrote:
>> On 5/30/2021 2:24 PM, olcott wrote:
>>>
>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>
>>>

> You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
> there is no end to the chart. The chart is a listing of EVERY Turing
> Machine, ordered by some rule.

I have rewritten the whole thing to make it more clear.

Please read the complete Sipser proof on the last page so that you don't
mistake what I am saying for what Sipser is saying.

All that I am doing is correcting Sipser's Figure 4.6 with my figure
4.4b and 4.5a. This one change refutes all of the halting problem
diagonal proofs.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/30/21 7:59 PM, olcott wrote:
> On 5/30/2021 5:57 PM, Richard Damon wrote:
>> On 5/30/21 4:37 PM, olcott wrote:
>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>
>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>
>>>>
>>>>
>
>> You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
>> there is no end to the chart. The chart is a listing of EVERY Turing
>> Machine, ordered by some rule.
>
> I have rewritten the whole thing to make it more clear.
>
> Please read the complete Sipser proof on the last page so that you don't
> mistake what I am saying for what Sipser is saying.
>
> All that I am doing is correcting Sipser's Figure 4.6 with my figure
> 4.4b and 4.5a. This one change refutes all of the halting problem
> diagonal proofs.
>
>

I don't think you read what I wrote.

The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines. There
is no 'end' to it to add something after it.

If D IS a real Turing Machine, there is no reason to add it to the 'end'
as it will already be in there somewhere, as it is the listing of ALL
Turing Machines.

It is logically incorrect to both say you NEED to add D to the list
(since if D is a Turing Machine, it will be already an element in the
list) or that you CAN add it to the list, since that says that it has a
ordinal number higher than any possible ordinal number.

Maybe your source of the description of Sipser's argument is incomplete,
but normally the diagonal proof starts with a description of the rule
that allows you to arrive at the numbering of any given Turing Machine.
One key factor of the proof is that you start with a mapping of the
Natural numbers to the Turing Machines that WILL assign a number to
EVERY Turing Machine, and a Turing Machine to every number.

The Fact that there is no entry for D in the list of Ms (due to the way
it is constructed with the diagonal) shows that D can NOT be an actual
Turing Machine, which implies that H can not be a actual Turing Machine,
i.e. the machine described as H can't actually exist.

On 5/30/2021 7:19 PM, Richard Damon wrote:
> On 5/30/21 7:59 PM, olcott wrote:
>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>> On 5/30/21 4:37 PM, olcott wrote:
>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>
>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>
>>>>>
>>>>>
>>
>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
>>> there is no end to the chart. The chart is a listing of EVERY Turing
>>> Machine, ordered by some rule.
>>
>> I have rewritten the whole thing to make it more clear.
>>
>> Please read the complete Sipser proof on the last page so that you don't
>> mistake what I am saying for what Sipser is saying.
>>
>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>> 4.4b and 4.5a. This one change refutes all of the halting problem
>> diagonal proofs.
>>
>>
>
> I don't think you read what I wrote.
>
> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines. There
> is no 'end' to it to add something after it.
>
> If D IS a real Turing Machine, there is no reason to add it to the 'end'
> as it will already be in there somewhere, as it is the listing of ALL
> Turing Machines.
>

I am merely using the Sipser convention of ellipses.

> It is logically incorrect to both say you NEED to add D to the list
> (since if D is a Turing Machine, it will be already an element in the
> list) or that you CAN add it to the list, since that says that it has a
> ordinal number higher than any possible ordinal number.
>
> Maybe your source of the description of Sipser's argument is incomplete,

I provided a scanned image of his actual proof on page 4.
Go back and do what I said and read what Sipser said on my page four.

https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/30/21 8:51 PM, olcott wrote:
> On 5/30/2021 7:19 PM, Richard Damon wrote:
>> On 5/30/21 7:59 PM, olcott wrote:
>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>
>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>
>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
>>>> there is no end to the chart. The chart is a listing of EVERY Turing
>>>> Machine, ordered by some rule.
>>>
>>> I have rewritten the whole thing to make it more clear.
>>>
>>> Please read the complete Sipser proof on the last page so that you don't
>>> mistake what I am saying for what Sipser is saying.
>>>
>>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>> diagonal proofs.
>>>
>>>
>>
>> I don't think you read what I wrote.
>>
>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines. There
>> is no 'end' to it to add something after it.
>>
>> If D IS a real Turing Machine, there is no reason to add it to the 'end'
>> as it will already be in there somewhere, as it is the listing of ALL
>> Turing Machines.
>>
>
> I am merely using the Sipser convention of ellipses.

No, you are not. There is one convention of ellipses for a finite
omission, where you do something like 1, 2, 3, 4, ... 99, 100.

With a terminal at the end, you can of course add more.

Sometimes, even if there is a finite number, the last element might not
be listed, and in that case, you could properly add something at the end.

But, in this case, the ellipses are pointing to an INFINITE sequence. It
is like: The counting numbers are: 1, 2, 3, 4, ...

There is NO implied end to the list, and you can't just 'add' something
to the end of the infinite sequence. The IS no number you could give
that item.

>
>> It is logically incorrect to both say you NEED to add D to the list
>> (since if D is a Turing Machine, it will be already an element in the
>> list) or that you CAN add it to the list, since that says that it has a
>> ordinal number higher than any possible ordinal number.
>>
>> Maybe your source of the description of Sipser's argument is incomplete,
>
> I provided a scanned image of his actual proof on page 4.
> Go back and do what I said and read what Sipser said on my page four.
>
> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>

Yes, and he said:

In these tables we list all TMs down the rows, M1, M2, ... and all their
descriptions across the columns, (MI ), (M2 ), ...

Let me ask you, how many TMs are their? Is their number finite or infinite?

If it is infinite, is there a space at the end of the table to add
something.

If it is a listing of *ALL* Turing Machines, why isn't D listed in it?

On 5/30/2021 8:42 PM, Richard Damon wrote:
> On 5/30/21 8:51 PM, olcott wrote:
>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>> On 5/30/21 7:59 PM, olcott wrote:
>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>
>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>
>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
>>>>> there is no end to the chart. The chart is a listing of EVERY Turing
>>>>> Machine, ordered by some rule.
>>>>
>>>> I have rewritten the whole thing to make it more clear.
>>>>
>>>> Please read the complete Sipser proof on the last page so that you don't
>>>> mistake what I am saying for what Sipser is saying.
>>>>
>>>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>> diagonal proofs.
>>>>
>>>>
>>>
>>> I don't think you read what I wrote.
>>>
>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines. There
>>> is no 'end' to it to add something after it.
>>>
>>> If D IS a real Turing Machine, there is no reason to add it to the 'end'
>>> as it will already be in there somewhere, as it is the listing of ALL
>>> Turing Machines.
>>>
>>
>> I am merely using the Sipser convention of ellipses.
>
> No, you are not. There is one convention of ellipses for a finite
> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>
> With a terminal at the end, you can of course add more.
>
> Sometimes, even if there is a finite number, the last element might not
> be listed, and in that case, you could properly add something at the end.
>
> But, in this case, the ellipses are pointing to an INFINITE sequence. It
> is like: The counting numbers are: 1, 2, 3, 4, ...
>
> There is NO implied end to the list, and you can't just 'add' something
> to the end of the infinite sequence. The IS no number you could give
> that item.
>
>
>>
>>> It is logically incorrect to both say you NEED to add D to the list
>>> (since if D is a Turing Machine, it will be already an element in the
>>> list) or that you CAN add it to the list, since that says that it has a
>>> ordinal number higher than any possible ordinal number.
>>>
>>> Maybe your source of the description of Sipser's argument is incomplete,
>>
>> I provided a scanned image of his actual proof on page 4.
>> Go back and do what I said and read what Sipser said on my page four.
>>
>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>
>
> Yes, and he said:
>
> In these tables we list all TMs down the rows, M1, M2, ... and all their
> descriptions across the columns, (MI ), (M2 ), ...
>
>
> Let me ask you, how many TMs are their? Is their number finite or infinite?
>
> If it is infinite, is there a space at the end of the table to add
> something.
>
> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?

It seems that you have set your mind on being disagreeable I will go
back to ignoring you. My Sipser diagonalization proof is clear enough
that I can start discussing it with actual computer scientists. Feel
free to disagree I won't be paying attention.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/30/21 9:53 PM, olcott wrote:
> On 5/30/2021 8:42 PM, Richard Damon wrote:
>> On 5/30/21 8:51 PM, olcott wrote:
>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>
>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>
>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
>>>>>> there is no end to the chart. The chart is a listing of EVERY Turing
>>>>>> Machine, ordered by some rule.
>>>>>
>>>>> I have rewritten the whole thing to make it more clear.
>>>>>
>>>>> Please read the complete Sipser proof on the last page so that you
>>>>> don't
>>>>> mistake what I am saying for what Sipser is saying.
>>>>>
>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>> diagonal proofs.
>>>>>
>>>>>
>>>>
>>>> I don't think you read what I wrote.
>>>>
>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines. There
>>>> is no 'end' to it to add something after it.
>>>>
>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>> 'end'
>>>> as it will already be in there somewhere, as it is the listing of ALL
>>>> Turing Machines.
>>>>
>>>
>>> I am merely using the Sipser convention of ellipses.
>>
>> No, you are not. There is one convention of ellipses for a finite
>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>
>> With a terminal at the end, you can of course add more.
>>
>> Sometimes, even if there is a finite number, the last element might not
>> be listed, and in that case, you could properly add something at the end.
>>
>> But, in this case, the ellipses are pointing to an INFINITE sequence. It
>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>
>> There is NO implied end to the list, and you can't just 'add' something
>> to the end of the infinite sequence. The IS no number you could give
>> that item.
>>
>>
>>>
>>>> It is logically incorrect to both say you NEED to add D to the list
>>>> (since if D is a Turing Machine, it will be already an element in the
>>>> list) or that you CAN add it to the list, since that says that it has a
>>>> ordinal number higher than any possible ordinal number.
>>>>
>>>> Maybe your source of the description of Sipser's argument is
>>>> incomplete,
>>>
>>> I provided a scanned image of his actual proof on page 4.
>>> Go back and do what I said and read what Sipser said on my page four.
>>>
>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>
>>>
>>
>> Yes, and he said:
>>
>> In these tables we list all TMs down the rows, M1, M2, ... and all their
>> descriptions across the columns, (MI ), (M2 ), ...
>>
>>
>> Let me ask you, how many TMs are their? Is their number finite or
>> infinite?
>>
>> If it is infinite, is there a space at the end of the table to add
>> something.
>>
>> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?
>
> It seems that you have set your mind on being disagreeable I will go
> back to ignoring you. My Sipser diagonalization proof is clear enough
> that I can start discussing it with actual computer scientists. Feel
> free to disagree I won't be paying attention.
>

If may be clear to you, but it is still wrong. Seems your typical
problem of not understanding infinities.

You CAN'T add something to the end of an infinity long sequence.

On 5/30/2021 9:20 PM, Richard Damon wrote:
> On 5/30/21 9:53 PM, olcott wrote:
>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>> On 5/30/21 8:51 PM, olcott wrote:
>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>
>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>
>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
>>>>>>> there is no end to the chart. The chart is a listing of EVERY Turing
>>>>>>> Machine, ordered by some rule.
>>>>>>
>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>
>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>> don't
>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>
>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>> diagonal proofs.
>>>>>>
>>>>>>
>>>>>
>>>>> I don't think you read what I wrote.
>>>>>
>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines. There
>>>>> is no 'end' to it to add something after it.
>>>>>
>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>> 'end'
>>>>> as it will already be in there somewhere, as it is the listing of ALL
>>>>> Turing Machines.
>>>>>
>>>>
>>>> I am merely using the Sipser convention of ellipses.
>>>
>>> No, you are not. There is one convention of ellipses for a finite
>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>
>>> With a terminal at the end, you can of course add more.
>>>
>>> Sometimes, even if there is a finite number, the last element might not
>>> be listed, and in that case, you could properly add something at the end.
>>>
>>> But, in this case, the ellipses are pointing to an INFINITE sequence. It
>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>
>>> There is NO implied end to the list, and you can't just 'add' something
>>> to the end of the infinite sequence. The IS no number you could give
>>> that item.
>>>
>>>
>>>>
>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>> (since if D is a Turing Machine, it will be already an element in the
>>>>> list) or that you CAN add it to the list, since that says that it has a
>>>>> ordinal number higher than any possible ordinal number.
>>>>>
>>>>> Maybe your source of the description of Sipser's argument is
>>>>> incomplete,
>>>>
>>>> I provided a scanned image of his actual proof on page 4.
>>>> Go back and do what I said and read what Sipser said on my page four.
>>>>
>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>
>>>>
>>>
>>> Yes, and he said:
>>>
>>> In these tables we list all TMs down the rows, M1, M2, ... and all their
>>> descriptions across the columns, (MI ), (M2 ), ...
>>>
>>>
>>> Let me ask you, how many TMs are their? Is their number finite or
>>> infinite?
>>>
>>> If it is infinite, is there a space at the end of the table to add
>>> something.
>>>
>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?
>>
>> It seems that you have set your mind on being disagreeable I will go
>> back to ignoring you. My Sipser diagonalization proof is clear enough
>> that I can start discussing it with actual computer scientists. Feel
>> free to disagree I won't be paying attention.
>>
>
> If may be clear to you, but it is still wrong. Seems your typical
> problem of not understanding infinities.
>
> You CAN'T add something to the end of an infinity long sequence.
>

You are telling me that when I use the same convention that Sipser uses
to specify infinite sets that I am wrong only because you insist on
being disagreeable.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 2021-05-30 20:20, Richard Damon wrote:
> On 5/30/21 9:53 PM, olcott wrote:
>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>> On 5/30/21 8:51 PM, olcott wrote:
>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>
>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>
>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart, because
>>>>>>> there is no end to the chart. The chart is a listing of EVERY Turing
>>>>>>> Machine, ordered by some rule.
>>>>>>
>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>
>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>> don't
>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>
>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>> diagonal proofs.
>>>>>>
>>>>>>
>>>>>
>>>>> I don't think you read what I wrote.
>>>>>
>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines. There
>>>>> is no 'end' to it to add something after it.
>>>>>
>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>> 'end'
>>>>> as it will already be in there somewhere, as it is the listing of ALL
>>>>> Turing Machines.
>>>>>
>>>>
>>>> I am merely using the Sipser convention of ellipses.
>>>
>>> No, you are not. There is one convention of ellipses for a finite
>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>
>>> With a terminal at the end, you can of course add more.
>>>
>>> Sometimes, even if there is a finite number, the last element might not
>>> be listed, and in that case, you could properly add something at the end.
>>>
>>> But, in this case, the ellipses are pointing to an INFINITE sequence. It
>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>
>>> There is NO implied end to the list, and you can't just 'add' something
>>> to the end of the infinite sequence. The IS no number you could give
>>> that item.
>>>
>>>
>>>>
>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>> (since if D is a Turing Machine, it will be already an element in the
>>>>> list) or that you CAN add it to the list, since that says that it has a
>>>>> ordinal number higher than any possible ordinal number.
>>>>>
>>>>> Maybe your source of the description of Sipser's argument is
>>>>> incomplete,
>>>>
>>>> I provided a scanned image of his actual proof on page 4.
>>>> Go back and do what I said and read what Sipser said on my page four.
>>>>
>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>
>>>>
>>>
>>> Yes, and he said:
>>>
>>> In these tables we list all TMs down the rows, M1, M2, ... and all their
>>> descriptions across the columns, (MI ), (M2 ), ...
>>>
>>>
>>> Let me ask you, how many TMs are their? Is their number finite or
>>> infinite?
>>>
>>> If it is infinite, is there a space at the end of the table to add
>>> something.
>>>
>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?
>>
>> It seems that you have set your mind on being disagreeable I will go
>> back to ignoring you. My Sipser diagonalization proof is clear enough
>> that I can start discussing it with actual computer scientists. Feel
>> free to disagree I won't be paying attention.
>>
>
> If may be clear to you, but it is still wrong. Seems your typical
> problem of not understanding infinities.
>
> You CAN'T add something to the end of an infinity long sequence.

I think the fundamental problem here is that Olcott doesn't understand
what an 'enumeration' is.

Unfortunately, I think he's gotten himself confused because the article
in Wikipedia which deals with diagonal proofs of the halting theorem
doesn't actually explain how diagonalization works but instead refers
one to the article on Cantor for an explanation.

But in Cantor's argument, diagonalization is used to demonstrate that an
enumeration of a particular set doesn't exist, whereas in the halting
problem, we start with something where it is known that an enumeration
does exist and the use diagonalization to demonstrate that a particular
Turing Machine isn't part of that enumeration.

For those in the set of people who do not exist in a state of perpetual
confusion, this difference isn't likely to cause problems, but that set
doesn't include Olcott.

Amdré

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

On 5/30/2021 10:59 PM, André G. Isaak wrote:
> On 2021-05-30 20:20, Richard Damon wrote:
>> On 5/30/21 9:53 PM, olcott wrote:
>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>
>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>
>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>> because
>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>> Turing
>>>>>>>> Machine, ordered by some rule.
>>>>>>>
>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>
>>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>>> don't
>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>
>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>>> diagonal proofs.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> I don't think you read what I wrote.
>>>>>>
>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines.
>>>>>> There
>>>>>> is no 'end' to it to add something after it.
>>>>>>
>>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>>> 'end'
>>>>>> as it will already be in there somewhere, as it is the listing of ALL
>>>>>> Turing Machines.
>>>>>>
>>>>>
>>>>> I am merely using the Sipser convention of ellipses.
>>>>
>>>> No, you are not. There is one convention of ellipses for a finite
>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>
>>>> With a terminal at the end, you can of course add more.
>>>>
>>>> Sometimes, even if there is a finite number, the last element might not
>>>> be listed, and in that case, you could properly add something at the
>>>> end.
>>>>
>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>> sequence. It
>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>
>>>> There is NO implied end to the list, and you can't just 'add' something
>>>> to the end of the infinite sequence. The IS no number you could give
>>>> that item.
>>>>
>>>>
>>>>>
>>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>>> (since if D is a Turing Machine, it will be already an element in the
>>>>>> list) or that you CAN add it to the list, since that says that it
>>>>>> has a
>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>
>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>> incomplete,
>>>>>
>>>>> I provided a scanned image of his actual proof on page 4.
>>>>> Go back and do what I said and read what Sipser said on my page four.
>>>>>
>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>
>>>>>
>>>>>
>>>>
>>>> Yes, and he said:
>>>>
>>>> In these tables we list all TMs down the rows, M1, M2, ... and all
>>>> their
>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>
>>>>
>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>> infinite?
>>>>
>>>> If it is infinite, is there a space at the end of the table to add
>>>> something.
>>>>
>>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?
>>>
>>> It seems that you have set your mind on being disagreeable I will go
>>> back to ignoring you. My Sipser diagonalization proof is clear enough
>>> that I can start discussing it with actual computer scientists. Feel
>>> free to disagree I won't be paying attention.
>>>
>>
>> If may be clear to you, but it is still wrong. Seems your typical
>> problem of not understanding infinities.
>>
>> You CAN'T add something to the end of an infinity long sequence.
>
> I think the fundamental problem here is that Olcott doesn't understand
> what an 'enumeration' is.
>

I am simply replacing Sipser's Figure 4.6 on page 2 with my
my 4.4b table and 4.5a tables on page 1. If people don't know what
ellipses are they might be quite confused. Sipser used twice as many
sets of ellipses, I didn't want to waste the space.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/30/21 11:04 PM, olcott wrote:
> On 5/30/2021 9:20 PM, Richard Damon wrote:
>> On 5/30/21 9:53 PM, olcott wrote:
>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>
>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>
>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>> because
>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>> Turing
>>>>>>>> Machine, ordered by some rule.
>>>>>>>
>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>
>>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>>> don't
>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>
>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>>> diagonal proofs.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> I don't think you read what I wrote.
>>>>>>
>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines.
>>>>>> There
>>>>>> is no 'end' to it to add something after it.
>>>>>>
>>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>>> 'end'
>>>>>> as it will already be in there somewhere, as it is the listing of ALL
>>>>>> Turing Machines.
>>>>>>
>>>>>
>>>>> I am merely using the Sipser convention of ellipses.
>>>>
>>>> No, you are not. There is one convention of ellipses for a finite
>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>
>>>> With a terminal at the end, you can of course add more.
>>>>
>>>> Sometimes, even if there is a finite number, the last element might not
>>>> be listed, and in that case, you could properly add something at the
>>>> end.
>>>>
>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>> sequence. It
>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>
>>>> There is NO implied end to the list, and you can't just 'add' something
>>>> to the end of the infinite sequence. The IS no number you could give
>>>> that item.
>>>>
>>>>
>>>>>
>>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>>> (since if D is a Turing Machine, it will be already an element in the
>>>>>> list) or that you CAN add it to the list, since that says that it
>>>>>> has a
>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>
>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>> incomplete,
>>>>>
>>>>> I provided a scanned image of his actual proof on page 4.
>>>>> Go back and do what I said and read what Sipser said on my page four.
>>>>>
>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>
>>>>>
>>>>>
>>>>
>>>> Yes, and he said:
>>>>
>>>> In these tables we list all TMs down the rows, M1, M2, ... and all
>>>> their
>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>
>>>>
>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>> infinite?
>>>>
>>>> If it is infinite, is there a space at the end of the table to add
>>>> something.
>>>>
>>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?
>>>
>>> It seems that you have set your mind on being disagreeable I will go
>>> back to ignoring you. My Sipser diagonalization proof is clear enough
>>> that I can start discussing it with actual computer scientists. Feel
>>> free to disagree I won't be paying attention.
>>>
>>
>> If may be clear to you, but it is still wrong. Seems your typical
>> problem of not understanding infinities.
>>
>> You CAN'T add something to the end of an infinity long sequence.
>>
>
> You are telling me that when I use the same convention that Sipser uses
> to specify infinite sets that I am wrong only because you insist on
> being disagreeable.
>

But Sipser never does what you did. He uses the notation of M1, M2, ...
to denote the infinite set of ALL Turing Machines. As such, if D was a
Turing Machine, it would already be in the set, no need to add it.

He points out that because of the diagonal proof, we can show that D
can't be the Turing Machine we assume it is, because it can't match any
of the elements of the Universal Set of Turing Machine.

You on the other hand, say that since D isn't in the list of All Turing
Machines, I will add it. That is an illogical operation. That is like
saying we have the set of all Cats, but this Dog isn't in it, so we will
add it to the set, but still consider it to be just the set of all Cats,
and now we can prove that some cats bark.

On 5/31/21 1:08 AM, olcott wrote:
> On 5/30/2021 10:59 PM, André G. Isaak wrote:
>> On 2021-05-30 20:20, Richard Damon wrote:
>>> On 5/30/21 9:53 PM, olcott wrote:
>>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>
>>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>>> because
>>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>>> Turing
>>>>>>>>> Machine, ordered by some rule.
>>>>>>>>
>>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>>
>>>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>>>> don't
>>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>>
>>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my
>>>>>>>> figure
>>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>>>> diagonal proofs.
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> I don't think you read what I wrote.
>>>>>>>
>>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines.
>>>>>>> There
>>>>>>> is no 'end' to it to add something after it.
>>>>>>>
>>>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>>>> 'end'
>>>>>>> as it will already be in there somewhere, as it is the listing of
>>>>>>> ALL
>>>>>>> Turing Machines.
>>>>>>>
>>>>>>
>>>>>> I am merely using the Sipser convention of ellipses.
>>>>>
>>>>> No, you are not. There is one convention of ellipses for a finite
>>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>>
>>>>> With a terminal at the end, you can of course add more.
>>>>>
>>>>> Sometimes, even if there is a finite number, the last element might
>>>>> not
>>>>> be listed, and in that case, you could properly add something at
>>>>> the end.
>>>>>
>>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>>> sequence. It
>>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>>
>>>>> There is NO implied end to the list, and you can't just 'add'
>>>>> something
>>>>> to the end of the infinite sequence. The IS no number you could give
>>>>> that item.
>>>>>
>>>>>
>>>>>>
>>>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>>>> (since if D is a Turing Machine, it will be already an element in
>>>>>>> the
>>>>>>> list) or that you CAN add it to the list, since that says that it
>>>>>>> has a
>>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>>
>>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>>> incomplete,
>>>>>>
>>>>>> I provided a scanned image of his actual proof on page 4.
>>>>>> Go back and do what I said and read what Sipser said on my page four.
>>>>>>
>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>
>>>>>>
>>>>>>
>>>>>
>>>>> Yes, and he said:
>>>>>
>>>>> In these tables we list all TMs down the rows, M1, M2, ... and all
>>>>> their
>>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>>
>>>>>
>>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>>> infinite?
>>>>>
>>>>> If it is infinite, is there a space at the end of the table to add
>>>>> something.
>>>>>
>>>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?
>>>>
>>>> It seems that you have set your mind on being disagreeable I will go
>>>> back to ignoring you. My Sipser diagonalization proof is clear enough
>>>> that I can start discussing it with actual computer scientists. Feel
>>>> free to disagree I won't be paying attention.
>>>>
>>>
>>> If may be clear to you, but it is still wrong. Seems your typical
>>> problem of not understanding infinities.
>>>
>>> You CAN'T add something to the end of an infinity long sequence.
>>
>> I think the fundamental problem here is that Olcott doesn't understand
>> what an 'enumeration' is.
>>
>
> I am simply replacing Sipser's Figure 4.6 on page 2 with my
> my 4.4b table and 4.5a tables on page 1. If people don't know what
> ellipses are they might be quite confused. Sipser used twice as many
> sets of ellipses, I didn't want to waste the space.
>

As I mentioned, ellipses have several related meanings, and you have to
use the right one when dealing with a notation.

In this case, they mean and so on to infinity. As such, you can't add to
the end of it, as you can never actually get to the end, only as far as
you need for a given case. It represents here a countably infinite set.

On Monday, 31 May 2021 at 04:37:37 UTC+8, olcott wrote:
> On 5/30/2021 2:24 PM, olcott wrote:
> >
> > https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
> >
> >
> Refuting the Halting Problem Diagonalization Argument
>
> Every machine that halts in a reject state is a halting computation. At
> least two proofs ignore this when constructing Sipser's Figure 4.5.
> Because these two proofs ignore this when they insert machine D into
> Sipser's Figure 4.5 they do so incorrectly.
>
> When machine D is inserted into both Figure 4.4 and Figure 4.5 correctly
> then the contradiction goes away. Since Sipser implicitly assumes that
> every blank entry of Figure 4.4 is a ~halt entry Figure 4.7a makes this
> explicit.
>
> <M1> <M2> <M3> <M4> ... <D>
> M1 accept ~halt accept ~halt reject
> M2 accept accept accept accept reject
> M3 ~halt ~halt ~halt ~halt accept
> M4 accept accept ~halt ~halt accept
> ...
> D reject reject accept accept reject
>
> Figure 4.7a (corrected figure 4.6, inserting D into figure 4.4)
>
> <M1> <M2> <M3> <M4> ... <D>
> M1 accept reject accept reject accept
> M2 accept accept accept accept accept
> M3 reject reject reject reject accept
> M4 accept accept reject reject accept
> ...
> D accept accept accept accept accept
>
> Figure 4.7b (corrected figure 4.6, inserting D into figure 4.5)
> --
> Copyright 2021 Pete Olcott
>
> "Great spirits have always encountered violent opposition from mediocre minds." Einstein

Cantor's diagonal argument is subject to interpretation, not a proof, similar to the
demonstration in limit that parallel rail tracks merge into one point at infinite distance,
not proof, only a magic show or delusive, misleading conclusion.
IMO, you should pick a better target to refute or create your own kind of proof.

On 5/31/2021 7:55 AM, wij wrote:
> On Monday, 31 May 2021 at 04:37:37 UTC+8, olcott wrote:
>> On 5/30/2021 2:24 PM, olcott wrote:
>>>
>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>
>>>
>> Refuting the Halting Problem Diagonalization Argument
>>
>> Every machine that halts in a reject state is a halting computation. At
>> least two proofs ignore this when constructing Sipser's Figure 4.5.
>> Because these two proofs ignore this when they insert machine D into
>> Sipser's Figure 4.5 they do so incorrectly.
>>
>> When machine D is inserted into both Figure 4.4 and Figure 4.5 correctly
>> then the contradiction goes away. Since Sipser implicitly assumes that
>> every blank entry of Figure 4.4 is a ~halt entry Figure 4.7a makes this
>> explicit.
>>
>> <M1> <M2> <M3> <M4> ... <D>
>> M1 accept ~halt accept ~halt reject
>> M2 accept accept accept accept reject
>> M3 ~halt ~halt ~halt ~halt accept
>> M4 accept accept ~halt ~halt accept
>> ...
>> D reject reject accept accept reject
>>
>> Figure 4.7a (corrected figure 4.6, inserting D into figure 4.4)
>>
>> <M1> <M2> <M3> <M4> ... <D>
>> M1 accept reject accept reject accept
>> M2 accept accept accept accept accept
>> M3 reject reject reject reject accept
>> M4 accept accept reject reject accept
>> ...
>> D accept accept accept accept accept
>>
>> Figure 4.7b (corrected figure 4.6, inserting D into figure 4.5)
>> --
>> Copyright 2021 Pete Olcott
>>
>> "Great spirits have always encountered violent opposition from mediocre minds." Einstein
>
> Cantor's diagonal argument is subject to interpretation, not a proof, similar to the
> demonstration in limit that parallel rail tracks merge into one point at infinite distance,
> not proof, only a magic show or delusive, misleading conclusion.
> IMO, you should pick a better target to refute or create your own kind of proof.
>

The halting problem diagonalization proof is not at all the same thing.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/31/2021 7:12 AM, Richard Damon wrote:
> On 5/30/21 11:04 PM, olcott wrote:
>> On 5/30/2021 9:20 PM, Richard Damon wrote:
>>> On 5/30/21 9:53 PM, olcott wrote:
>>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>
>>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>>> because
>>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>>> Turing
>>>>>>>>> Machine, ordered by some rule.
>>>>>>>>
>>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>>
>>>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>>>> don't
>>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>>
>>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my figure
>>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>>>> diagonal proofs.
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> I don't think you read what I wrote.
>>>>>>>
>>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines.
>>>>>>> There
>>>>>>> is no 'end' to it to add something after it.
>>>>>>>
>>>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>>>> 'end'
>>>>>>> as it will already be in there somewhere, as it is the listing of ALL
>>>>>>> Turing Machines.
>>>>>>>
>>>>>>
>>>>>> I am merely using the Sipser convention of ellipses.
>>>>>
>>>>> No, you are not. There is one convention of ellipses for a finite
>>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>>
>>>>> With a terminal at the end, you can of course add more.
>>>>>
>>>>> Sometimes, even if there is a finite number, the last element might not
>>>>> be listed, and in that case, you could properly add something at the
>>>>> end.
>>>>>
>>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>>> sequence. It
>>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>>
>>>>> There is NO implied end to the list, and you can't just 'add' something
>>>>> to the end of the infinite sequence. The IS no number you could give
>>>>> that item.
>>>>>
>>>>>
>>>>>>
>>>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>>>> (since if D is a Turing Machine, it will be already an element in the
>>>>>>> list) or that you CAN add it to the list, since that says that it
>>>>>>> has a
>>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>>
>>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>>> incomplete,
>>>>>>
>>>>>> I provided a scanned image of his actual proof on page 4.
>>>>>> Go back and do what I said and read what Sipser said on my page four.
>>>>>>
>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>
>>>>>>
>>>>>>
>>>>>
>>>>> Yes, and he said:
>>>>>
>>>>> In these tables we list all TMs down the rows, M1, M2, ... and all
>>>>> their
>>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>>
>>>>>
>>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>>> infinite?
>>>>>
>>>>> If it is infinite, is there a space at the end of the table to add
>>>>> something.
>>>>>
>>>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?
>>>>
>>>> It seems that you have set your mind on being disagreeable I will go
>>>> back to ignoring you. My Sipser diagonalization proof is clear enough
>>>> that I can start discussing it with actual computer scientists. Feel
>>>> free to disagree I won't be paying attention.
>>>>
>>>
>>> If may be clear to you, but it is still wrong. Seems your typical
>>> problem of not understanding infinities.
>>>
>>> You CAN'T add something to the end of an infinity long sequence.
>>>
>>
>> You are telling me that when I use the same convention that Sipser uses
>> to specify infinite sets that I am wrong only because you insist on
>> being disagreeable.
>>
>
> But Sipser never does what you did. He uses the notation of M1, M2, ...
> to denote the infinite set of ALL Turing Machines. As such, if D was a
> Turing Machine, it would already be in the set, no need to add it.
>

He does add it in his Figure 4.6 nitwit and he does add it incorrectly.
I already told you before you point out my errors read his freaking
proof. The mistakes that you say I am making is what he is doing.

> He points out that because of the diagonal proof, we can show that D
> can't be the Turing Machine we assume it is, because it can't match any
> of the elements of the Universal Set of Turing Machine.
>
> You on the other hand, say that since D isn't in the list of All Turing
> Machines, I will add it. That is an illogical operation. That is like
> saying we have the set of all Cats, but this Dog isn't in it, so we will
> add it to the set, but still consider it to be just the set of all Cats,
> and now we can prove that some cats bark.
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/31/2021 7:15 AM, Richard Damon wrote:
> On 5/31/21 1:08 AM, olcott wrote:
>> On 5/30/2021 10:59 PM, André G. Isaak wrote:
>>> On 2021-05-30 20:20, Richard Damon wrote:
>>>> On 5/30/21 9:53 PM, olcott wrote:
>>>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>>>
>>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>
>>>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>>>> because
>>>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>>>> Turing
>>>>>>>>>> Machine, ordered by some rule.
>>>>>>>>>
>>>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>>>
>>>>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>>>>> don't
>>>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>>>
>>>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my
>>>>>>>>> figure
>>>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>>>>> diagonal proofs.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>
>>>>>>>> I don't think you read what I wrote.
>>>>>>>>
>>>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines.
>>>>>>>> There
>>>>>>>> is no 'end' to it to add something after it.
>>>>>>>>
>>>>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>>>>> 'end'
>>>>>>>> as it will already be in there somewhere, as it is the listing of
>>>>>>>> ALL
>>>>>>>> Turing Machines.
>>>>>>>>
>>>>>>>
>>>>>>> I am merely using the Sipser convention of ellipses.
>>>>>>
>>>>>> No, you are not. There is one convention of ellipses for a finite
>>>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>>>
>>>>>> With a terminal at the end, you can of course add more.
>>>>>>
>>>>>> Sometimes, even if there is a finite number, the last element might
>>>>>> not
>>>>>> be listed, and in that case, you could properly add something at
>>>>>> the end.
>>>>>>
>>>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>>>> sequence. It
>>>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>>>
>>>>>> There is NO implied end to the list, and you can't just 'add'
>>>>>> something
>>>>>> to the end of the infinite sequence. The IS no number you could give
>>>>>> that item.
>>>>>>
>>>>>>
>>>>>>>
>>>>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>>>>> (since if D is a Turing Machine, it will be already an element in
>>>>>>>> the
>>>>>>>> list) or that you CAN add it to the list, since that says that it
>>>>>>>> has a
>>>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>>>
>>>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>>>> incomplete,
>>>>>>>
>>>>>>> I provided a scanned image of his actual proof on page 4.
>>>>>>> Go back and do what I said and read what Sipser said on my page four.
>>>>>>>
>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> Yes, and he said:
>>>>>>
>>>>>> In these tables we list all TMs down the rows, M1, M2, ... and all
>>>>>> their
>>>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>>>
>>>>>>
>>>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>>>> infinite?
>>>>>>
>>>>>> If it is infinite, is there a space at the end of the table to add
>>>>>> something.
>>>>>>
>>>>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in it?
>>>>>
>>>>> It seems that you have set your mind on being disagreeable I will go
>>>>> back to ignoring you. My Sipser diagonalization proof is clear enough
>>>>> that I can start discussing it with actual computer scientists. Feel
>>>>> free to disagree I won't be paying attention.
>>>>>
>>>>
>>>> If may be clear to you, but it is still wrong. Seems your typical
>>>> problem of not understanding infinities.
>>>>
>>>> You CAN'T add something to the end of an infinity long sequence.
>>>
>>> I think the fundamental problem here is that Olcott doesn't understand
>>> what an 'enumeration' is.
>>>
>>
>> I am simply replacing Sipser's Figure 4.6 on page 2 with my
>> my 4.4b table and 4.5a tables on page 1. If people don't know what
>> ellipses are they might be quite confused. Sipser used twice as many
>> sets of ellipses, I didn't want to waste the space.
>>
>
> As I mentioned, ellipses have several related meanings, and you have to
> use the right one when dealing with a notation.
>
> In this case, they mean and so on to infinity. As such, you can't add to
> the end of it, as you can never actually get to the end, only as far as
> you need for a given case. It represents here a countably infinite set.
>

Sipser creates Figure 4.6 incorrectly. I fix his mistake by creating
figures 4.4b and 4.5a that take the place of his incorrect figure 4.6.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/31/21 9:41 AM, olcott wrote:
> On 5/31/2021 7:12 AM, Richard Damon wrote:
>> On 5/30/21 11:04 PM, olcott wrote:
>>> On 5/30/2021 9:20 PM, Richard Damon wrote:
>>>> On 5/30/21 9:53 PM, olcott wrote:
>>>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>>>
>>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>
>>>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>>>> because
>>>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>>>> Turing
>>>>>>>>>> Machine, ordered by some rule.
>>>>>>>>>
>>>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>>>
>>>>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>>>>> don't
>>>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>>>
>>>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my
>>>>>>>>> figure
>>>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>>>>> diagonal proofs.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>
>>>>>>>> I don't think you read what I wrote.
>>>>>>>>
>>>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines.
>>>>>>>> There
>>>>>>>> is no 'end' to it to add something after it.
>>>>>>>>
>>>>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>>>>> 'end'
>>>>>>>> as it will already be in there somewhere, as it is the listing
>>>>>>>> of ALL
>>>>>>>> Turing Machines.
>>>>>>>>
>>>>>>>
>>>>>>> I am merely using the Sipser convention of ellipses.
>>>>>>
>>>>>> No, you are not. There is one convention of ellipses for a finite
>>>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>>>
>>>>>> With a terminal at the end, you can of course add more.
>>>>>>
>>>>>> Sometimes, even if there is a finite number, the last element
>>>>>> might not
>>>>>> be listed, and in that case, you could properly add something at the
>>>>>> end.
>>>>>>
>>>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>>>> sequence. It
>>>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>>>
>>>>>> There is NO implied end to the list, and you can't just 'add'
>>>>>> something
>>>>>> to the end of the infinite sequence. The IS no number you could give
>>>>>> that item.
>>>>>>
>>>>>>
>>>>>>>
>>>>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>>>>> (since if D is a Turing Machine, it will be already an element
>>>>>>>> in the
>>>>>>>> list) or that you CAN add it to the list, since that says that it
>>>>>>>> has a
>>>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>>>
>>>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>>>> incomplete,
>>>>>>>
>>>>>>> I provided a scanned image of his actual proof on page 4.
>>>>>>> Go back and do what I said and read what Sipser said on my page
>>>>>>> four.
>>>>>>>
>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> Yes, and he said:
>>>>>>
>>>>>> In these tables we list all TMs down the rows, M1, M2, ... and all
>>>>>> their
>>>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>>>
>>>>>>
>>>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>>>> infinite?
>>>>>>
>>>>>> If it is infinite, is there a space at the end of the table to add
>>>>>> something.
>>>>>>
>>>>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in
>>>>>> it?
>>>>>
>>>>> It seems that you have set your mind on being disagreeable I will go
>>>>> back to ignoring you. My Sipser diagonalization proof is clear enough
>>>>> that I can start discussing it with actual computer scientists. Feel
>>>>> free to disagree I won't be paying attention.
>>>>>
>>>>
>>>> If may be clear to you, but it is still wrong. Seems your typical
>>>> problem of not understanding infinities.
>>>>
>>>> You CAN'T add something to the end of an infinity long sequence.
>>>>
>>>
>>> You are telling me that when I use the same convention that Sipser uses
>>> to specify infinite sets that I am wrong only because you insist on
>>> being disagreeable.
>>>
>>
>> But Sipser never does what you did. He uses the notation of M1, M2, ...
>> to denote the infinite set of ALL Turing Machines. As such, if D was a
>> Turing Machine, it would already be in the set, no need to add it.
>>
>
> He does add it in his Figure 4.6 nitwit and he does add it incorrectly.
> I already told you before you point out my errors read his freaking
> proof. The mistakes that you say I am making is what he is doing.

No, he is doing something different here.

You went M1, M2, ... D

He went M1, M2, ... D ...

The difference here is you are trying to ADD D as a new member to the set.

Sipser, though maybe not explaining it well enough for you, is saying if
we FIND the D as one of the M#s, (not adding as a new one) that we get a
contradiction. We can not just 'insert' a new machine in the middle of
the sequence because that will totally change the definition of what D
needed to do.

When we try to find a M that D can be, we run into the problem that at
least one entry in the table can't match, thus it can not be that entry.

Thus D can't be in the list of all Turing machines and thus it can't be
a Turing machine. But, if H exists a a Turing Machine, then D will, so
that means that H can not exist.

>
>> He points out that because of the diagonal proof, we can show that D
>> can't be the Turing Machine we assume it is, because it can't match any
>> of the elements of the Universal Set of Turing Machine.
>>
>> You on the other hand, say that since D isn't in the list of All Turing
>> Machines, I will add it. That is an illogical operation. That is like
>> saying we have the set of all Cats, but this Dog isn't in it, so we will
>> add it to the set, but still consider it to be just the set of all Cats,
>> and now we can prove that some cats bark.
>>
>
>

On 5/31/21 10:55 AM, olcott wrote:
> On 5/31/2021 9:27 AM, Richard Damon wrote:
>> On 5/31/21 9:41 AM, olcott wrote:
>>> On 5/31/2021 7:12 AM, Richard Damon wrote:
>>>> On 5/30/21 11:04 PM, olcott wrote:
>>>>> On 5/30/2021 9:20 PM, Richard Damon wrote:
>>>>>> On 5/30/21 9:53 PM, olcott wrote:
>>>>>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>>>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>>>>>> because
>>>>>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>>>>>> Turing
>>>>>>>>>>>> Machine, ordered by some rule.
>>>>>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>>>>>
>>>>>>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>>>>>>> don't
>>>>>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>>>>>
>>>>>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my
>>>>>>>>>>> figure
>>>>>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>>>>>>> diagonal proofs.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>> I don't think you read what I wrote.
>>>>>>>>>>
>>>>>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines.
>>>>>>>>>> There
>>>>>>>>>> is no 'end' to it to add something after it.
>>>>>>>>>>
>>>>>>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>>>>>>> 'end'
>>>>>>>>>> as it will already be in there somewhere, as it is the listing
>>>>>>>>>> of ALL
>>>>>>>>>> Turing Machines.
>>>>>>>>>>
>>>>>>>>> I am merely using the Sipser convention of ellipses.
>>>>>>>> No, you are not. There is one convention of ellipses for a finite
>>>>>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>>>>>
>>>>>>>> With a terminal at the end, you can of course add more.
>>>>>>>>
>>>>>>>> Sometimes, even if there is a finite number, the last element
>>>>>>>> might not
>>>>>>>> be listed, and in that case, you could properly add something at the
>>>>>>>> end.
>>>>>>>>
>>>>>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>>>>>> sequence. It
>>>>>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>>>>>
>>>>>>>> There is NO implied end to the list, and you can't just 'add'
>>>>>>>> something
>>>>>>>> to the end of the infinite sequence. The IS no number you could give
>>>>>>>> that item.
>>>>>>>>
>>>>>>>>
>>>>>>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>>>>>>> (since if D is a Turing Machine, it will be already an element
>>>>>>>>>> in the
>>>>>>>>>> list) or that you CAN add it to the list, since that says that it
>>>>>>>>>> has a
>>>>>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>>>>>
>>>>>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>>>>>> incomplete,
>>>>>>>>> I provided a scanned image of his actual proof on page 4.
>>>>>>>>> Go back and do what I said and read what Sipser said on my page
>>>>>>>>> four.
>>>>>>>>>
>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>> Yes, and he said:
>>>>>>>>
>>>>>>>> In these tables we list all TMs down the rows, M1, M2, ... and all
>>>>>>>> their
>>>>>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>>>>>
>>>>>>>>
>>>>>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>>>>>> infinite?
>>>>>>>>
>>>>>>>> If it is infinite, is there a space at the end of the table to add
>>>>>>>> something.
>>>>>>>>
>>>>>>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in
>>>>>>>> it?
>>>>>>> It seems that you have set your mind on being disagreeable I will go
>>>>>>> back to ignoring you. My Sipser diagonalization proof is clear enough
>>>>>>> that I can start discussing it with actual computer scientists. Feel
>>>>>>> free to disagree I won't be paying attention.
>>>>>>>
>>>>>> If may be clear to you, but it is still wrong. Seems your typical
>>>>>> problem of not understanding infinities.
>>>>>>
>>>>>> You CAN'T add something to the end of an infinity long sequence.
>>>>>>
>>>>> You are telling me that when I use the same convention that Sipser uses
>>>>> to specify infinite sets that I am wrong only because you insist on
>>>>> being disagreeable.
>>>>>
>>>> But Sipser never does what you did. He uses the notation of M1, M2, ...
>>>> to denote the infinite set of ALL Turing Machines. As such, if D was a
>>>> Turing Machine, it would already be in the set, no need to add it.
>>>>
>>> He does add it in his Figure 4.6 nitwit and he does add it incorrectly.
>>> I already told you before you point out my errors read his freaking
>>> proof. The mistakes that you say I am making is what he is doing.
>> No, he is doing something different here.
>>
>> You went M1, M2, ... D
> *Corrections to Sipser's Figure 4.6 when he attempts
> **to insert machine D into his figure 4.5
> *
>
>     ⟨M1⟩      ⟨M2⟩      ⟨M3⟩       ⟨M4⟩ ... ⟨D⟩ ...
> M1  accept   ~halt    accept   ~halt    DC
> M2  accept   accept   accept   accept   DC
> M3  ~halt    ~halt    ~halt    ~halt    DC
> M4  accept   accept   ~halt    ~halt    DC
> .
> .
> .
> D   reject   reject   accept   accept   reject
> .
> .
> .
> *   Figure 4.4b (Insert D into Figure 4.4a adapted from Sipser 4.4) *
>
>
>     ⟨M1⟩      ⟨M2⟩      ⟨M3⟩      ⟨M4⟩ ...  ⟨D⟩ ...
> M1  _accept_   reject   accept   reject   DC
> M2  accept   _accept_   accept   accept   DC
> M3  reject   reject   _reject_   reject   DC
> M4  accept   accept   reject   _reject_   DC
> .
> .
> .
> D   accept   accept   accept   accept   _accept_
> .
> .
> .
> *Figure 4.5a (Insert D into Figure 4.5) *
>
> *
> *
>
> *Refuting the Sipser Halting Problem Diagonalization Argument*
>
> https://www.researchgate.net/publication/351947980_Refuting_the_Sipser_Halting_Problem_Diagonalization_Argument
>
>

On 5/31/2021 10:29 AM, Richard Damon wrote:
> On 5/31/21 10:55 AM, olcott wrote:
>> On 5/31/2021 9:27 AM, Richard Damon wrote:
>>> On 5/31/21 9:41 AM, olcott wrote:
>>>> On 5/31/2021 7:12 AM, Richard Damon wrote:
>>>>> On 5/30/21 11:04 PM, olcott wrote:
>>>>>> On 5/30/2021 9:20 PM, Richard Damon wrote:
>>>>>>> On 5/30/21 9:53 PM, olcott wrote:
>>>>>>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>>>>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>>>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>>>>>>> because
>>>>>>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>>>>>>> Turing
>>>>>>>>>>>>> Machine, ordered by some rule.
>>>>>>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>>>>>>
>>>>>>>>>>>> Please read the complete Sipser proof on the last page so that you
>>>>>>>>>>>> don't
>>>>>>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>>>>>>
>>>>>>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my
>>>>>>>>>>>> figure
>>>>>>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting problem
>>>>>>>>>>>> diagonal proofs.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>> I don't think you read what I wrote.
>>>>>>>>>>>
>>>>>>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL Machines.
>>>>>>>>>>> There
>>>>>>>>>>> is no 'end' to it to add something after it.
>>>>>>>>>>>
>>>>>>>>>>> If D IS a real Turing Machine, there is no reason to add it to the
>>>>>>>>>>> 'end'
>>>>>>>>>>> as it will already be in there somewhere, as it is the listing
>>>>>>>>>>> of ALL
>>>>>>>>>>> Turing Machines.
>>>>>>>>>>>
>>>>>>>>>> I am merely using the Sipser convention of ellipses.
>>>>>>>>> No, you are not. There is one convention of ellipses for a finite
>>>>>>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>>>>>>
>>>>>>>>> With a terminal at the end, you can of course add more.
>>>>>>>>>
>>>>>>>>> Sometimes, even if there is a finite number, the last element
>>>>>>>>> might not
>>>>>>>>> be listed, and in that case, you could properly add something at the
>>>>>>>>> end.
>>>>>>>>>
>>>>>>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>>>>>>> sequence. It
>>>>>>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>>>>>>
>>>>>>>>> There is NO implied end to the list, and you can't just 'add'
>>>>>>>>> something
>>>>>>>>> to the end of the infinite sequence. The IS no number you could give
>>>>>>>>> that item.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>>> It is logically incorrect to both say you NEED to add D to the list
>>>>>>>>>>> (since if D is a Turing Machine, it will be already an element
>>>>>>>>>>> in the
>>>>>>>>>>> list) or that you CAN add it to the list, since that says that it
>>>>>>>>>>> has a
>>>>>>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>>>>>>
>>>>>>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>>>>>>> incomplete,
>>>>>>>>>> I provided a scanned image of his actual proof on page 4.
>>>>>>>>>> Go back and do what I said and read what Sipser said on my page
>>>>>>>>>> four.
>>>>>>>>>>
>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>> Yes, and he said:
>>>>>>>>>
>>>>>>>>> In these tables we list all TMs down the rows, M1, M2, ... and all
>>>>>>>>> their
>>>>>>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>>>>>>> infinite?
>>>>>>>>>
>>>>>>>>> If it is infinite, is there a space at the end of the table to add
>>>>>>>>> something.
>>>>>>>>>
>>>>>>>>> If it is a listing of *ALL* Turing Machines, why isn't D listed in
>>>>>>>>> it?
>>>>>>>> It seems that you have set your mind on being disagreeable I will go
>>>>>>>> back to ignoring you. My Sipser diagonalization proof is clear enough
>>>>>>>> that I can start discussing it with actual computer scientists. Feel
>>>>>>>> free to disagree I won't be paying attention.
>>>>>>>>
>>>>>>> If may be clear to you, but it is still wrong. Seems your typical
>>>>>>> problem of not understanding infinities.
>>>>>>>
>>>>>>> You CAN'T add something to the end of an infinity long sequence.
>>>>>>>
>>>>>> You are telling me that when I use the same convention that Sipser uses
>>>>>> to specify infinite sets that I am wrong only because you insist on
>>>>>> being disagreeable.
>>>>>>
>>>>> But Sipser never does what you did. He uses the notation of M1, M2, ...
>>>>> to denote the infinite set of ALL Turing Machines. As such, if D was a
>>>>> Turing Machine, it would already be in the set, no need to add it.
>>>>>
>>>> He does add it in his Figure 4.6 nitwit and he does add it incorrectly.
>>>> I already told you before you point out my errors read his freaking
>>>> proof. The mistakes that you say I am making is what he is doing.
>>> No, he is doing something different here.
>>>
>>> You went M1, M2, ... D
>> *Corrections to Sipser's Figure 4.6 when he attempts
>> **to insert machine D into his figure 4.5
>> *
>>
>>     ⟨M1⟩      ⟨M2⟩      ⟨M3⟩       ⟨M4⟩ ... ⟨D⟩ ...
>> M1  accept   ~halt    accept   ~halt    DC
>> M2  accept   accept   accept   accept   DC
>> M3  ~halt    ~halt    ~halt    ~halt    DC
>> M4  accept   accept   ~halt    ~halt    DC
>> .
>> .
>> .
>> D   reject   reject   accept   accept   reject
>> .
>> .
>> .
>> *   Figure 4.4b (Insert D into Figure 4.4a adapted from Sipser 4.4) *
>>
>>
>>     ⟨M1⟩      ⟨M2⟩      ⟨M3⟩      ⟨M4⟩ ...  ⟨D⟩ ...
>> M1  _accept_   reject   accept   reject   DC
>> M2  accept   _accept_   accept   accept   DC
>> M3  reject   reject   _reject_   reject   DC
>> M4  accept   accept   reject   _reject_   DC
>> .
>> .
>> .
>> D   accept   accept   accept   accept   _accept_
>> .
>> .
>> .
>> *Figure 4.5a (Insert D into Figure 4.5) *
>>
>> *
>> *
>>
>> *Refuting the Sipser Halting Problem Diagonalization Argument*
>>
>> https://www.researchgate.net/publication/351947980_Refuting_the_Sipser_Halting_Problem_Diagonalization_Argument
>>
>>
>
> So your answer to the evidence that D can't be a Turing Machine is to
> just change the requirements on the machines. Right, That really works.
>
>
> Since by your 4.4b D rejects <D>, then the value that H must produce is
> *reject*, so you have just shown that H now fails at its job of Deciding.
>

On 30/05/2021 21:37, olcott wrote:
> On 5/30/2021 2:24 PM, olcott wrote:
>>
>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>

tldr; just the usual "PO totally failing to understand what he is
reading, and as a result confidently spouting nonsense as a result"...

FTR, PO is pretty much incapable of following abstract reasoning or
understanding abstact concepts. "Arguing" with him is totally
pointless. Do not assume that because he replies to a post and nobody
responds, that means he has said anything coherent or logical, or that
anybody else implicitly agrees with what he said.

>>
> Refuting the Halting Problem Diagonalization Argument
>
> Every machine that halts in a reject state is a halting computation. At
> least two proofs ignore this when constructing Sipser's Figure 4.5.
> Because these two proofs ignore this when they insert machine D into
> Sipser's Figure 4.5 they do so incorrectly.

Dude... Sipser explicitly says blank entries are "either non-halting or
reject" entries. He is not ignoring anything. You just can't follow
his argument.

I think you are confused at what you're looking at in Sipser. You
reference Sipser's Theorem 4.9, and it's clear you think that is a proof
of the undecidability of the Halting Problem. IT IS NOT - it's a proof
of the undecidability of the *Accepting* TM problem. The clue for you
should have been in the name ATM (rather than HTM). Didn't you wonder
What the A stood for?

>
> When machine D is inserted into both Figure 4.4 and Figure 4.5 correctly
> then the contradiction goes away. Since Sipser implicitly assumes that
> every blank entry of Figure 4.4 is a ~halt entry Figure 4.7a makes this
> explicit.

Rubbish. He explicitly says what blank entries mean.

>
>     <M1>     <M2>     <M3>     <M4> ... <D>
> M1  accept   ~halt    accept   ~halt    reject
> M2  accept   accept   accept   accept   reject
> M3  ~halt    ~halt    ~halt    ~halt    accept
> M4  accept   accept   ~halt    ~halt    accept
> ...
> D   reject   reject   accept   accept   reject
>
> Figure 4.7a (corrected figure 4.6, inserting D into figure 4.4)

Wrong. Sipser does not say or assume that M1(<M2>) does not halt. This
is all completely beyond you, I'm afraid...

>
>     <M1>     <M2>     <M3>     <M4> ... <D>
> M1  accept   reject   accept   reject   accept
> M2  accept   accept   accept   accept   accept
> M3  reject   reject   reject   reject   accept
> M4  accept   accept   reject   reject   accept
> ...
> D   accept   accept   accept   accept   accept
>
> Figure 4.7b (corrected figure 4.6, inserting D into figure 4.5)
>

Wrong. You don't understand what Sipser's figure 4.6 is showing, and
you clearly don't understand the definition of Sipser's D. As Sipser
points out, D /by definition/ computes the opposite of the diagonal
entries of his table 4.6. (..and hence the contradiction when you get
to the D(<D>) diagonal entry.)

Your document is completely hopeless (of course).

Mike.

On 5/31/2021 10:53 AM, Mike Terry wrote:
> On 30/05/2021 21:37, olcott wrote:
>> On 5/30/2021 2:24 PM, olcott wrote:
>>>
>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>
>
> tldr;  just the usual "PO totally failing to understand what he is
> reading, and as a result confidently spouting nonsense as a result"...
>
> FTR, PO is pretty much incapable of following abstract reasoning or
> understanding abstact concepts.  "Arguing" with him is totally
> pointless.  Do not assume that because he replies to a post and nobody
> responds, that means he has said anything coherent or logical, or that
> anybody else implicitly agrees with what he said.
>
>>>
>> Refuting the Halting Problem Diagonalization Argument
>>
>> Every machine that halts in a reject state is a halting computation.
>> At least two proofs ignore this when constructing Sipser's Figure 4.5.
>> Because these two proofs ignore this when they insert machine D into
>> Sipser's Figure 4.5 they do so incorrectly.
>
> Dude... Sipser explicitly says blank entries are "either non-halting or
> reject" entries.  He is not ignoring anything.  You just can't follow
> his argument.
>

When he inserts D into his figure 4.5 as his figure 4.6 he needed to
insert D into figure 4.4 first. When we insert D into figure 4.4 then we
have the correct basis for inserting D into Figure 4.5.

When we insert D into Figure 4.5 we must translate the reject values
from Figure 4.4 (with D inserted) into accept values because halting in
a reject state is a halting computation. In his figure 4.6 at ⟨D,⟨M1⟩⟩
and ⟨D,⟨M2⟩⟩ he makes the mistake of not translating reject values into
accept values.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/31/21 11:46 AM, olcott wrote:
> On 5/31/2021 10:29 AM, Richard Damon wrote:
>> On 5/31/21 10:55 AM, olcott wrote:
>>> On 5/31/2021 9:27 AM, Richard Damon wrote:
>>>> On 5/31/21 9:41 AM, olcott wrote:
>>>>> On 5/31/2021 7:12 AM, Richard Damon wrote:
>>>>>> On 5/30/21 11:04 PM, olcott wrote:
>>>>>>> On 5/30/2021 9:20 PM, Richard Damon wrote:
>>>>>>>> On 5/30/21 9:53 PM, olcott wrote:
>>>>>>>>> On 5/30/2021 8:42 PM, Richard Damon wrote:
>>>>>>>>>> On 5/30/21 8:51 PM, olcott wrote:
>>>>>>>>>>> On 5/30/2021 7:19 PM, Richard Damon wrote:
>>>>>>>>>>>> On 5/30/21 7:59 PM, olcott wrote:
>>>>>>>>>>>>> On 5/30/2021 5:57 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 5/30/21 4:37 PM, olcott wrote:
>>>>>>>>>>>>>>> On 5/30/2021 2:24 PM, olcott wrote:
>>>>>>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You DON'T, and in fact CAN'T 'add' D to the end of the chart,
>>>>>>>>>>>>>> because
>>>>>>>>>>>>>> there is no end to the chart. The chart is a listing of EVERY
>>>>>>>>>>>>>> Turing
>>>>>>>>>>>>>> Machine, ordered by some rule.
>>>>>>>>>>>>> I have rewritten the whole thing to make it more clear.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Please read the complete Sipser proof on the last page so
>>>>>>>>>>>>> that you
>>>>>>>>>>>>> don't
>>>>>>>>>>>>> mistake what I am saying for what Sipser is saying.
>>>>>>>>>>>>>
>>>>>>>>>>>>> All that I am doing is correcting Sipser's Figure 4.6 with my
>>>>>>>>>>>>> figure
>>>>>>>>>>>>> 4.4b and 4.5a. This one change refutes all of the halting
>>>>>>>>>>>>> problem
>>>>>>>>>>>>> diagonal proofs.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>> I don't think you read what I wrote.
>>>>>>>>>>>>
>>>>>>>>>>>> The list M1, M2, M3, M4 ... is the INFINITE list of ALL
>>>>>>>>>>>> Machines.
>>>>>>>>>>>> There
>>>>>>>>>>>> is no 'end' to it to add something after it.
>>>>>>>>>>>>
>>>>>>>>>>>> If D IS a real Turing Machine, there is no reason to add it
>>>>>>>>>>>> to the
>>>>>>>>>>>> 'end'
>>>>>>>>>>>> as it will already be in there somewhere, as it is the listing
>>>>>>>>>>>> of ALL
>>>>>>>>>>>> Turing Machines.
>>>>>>>>>>>>
>>>>>>>>>>> I am merely using the Sipser convention of ellipses.
>>>>>>>>>> No, you are not. There is one convention of ellipses for a finite
>>>>>>>>>> omission, where you do something like 1, 2, 3, 4, ... 99, 100.
>>>>>>>>>>
>>>>>>>>>> With a terminal at the end, you can of course add more.
>>>>>>>>>>
>>>>>>>>>> Sometimes, even if there is a finite number, the last element
>>>>>>>>>> might not
>>>>>>>>>> be listed, and in that case, you could properly add something
>>>>>>>>>> at the
>>>>>>>>>> end.
>>>>>>>>>>
>>>>>>>>>> But, in this case, the ellipses are pointing to an INFINITE
>>>>>>>>>> sequence. It
>>>>>>>>>> is like: The counting numbers are: 1, 2, 3, 4, ...
>>>>>>>>>>
>>>>>>>>>> There is NO implied end to the list, and you can't just 'add'
>>>>>>>>>> something
>>>>>>>>>> to the end of the infinite sequence. The IS no number you
>>>>>>>>>> could give
>>>>>>>>>> that item.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>>> It is logically incorrect to both say you NEED to add D to
>>>>>>>>>>>> the list
>>>>>>>>>>>> (since if D is a Turing Machine, it will be already an element
>>>>>>>>>>>> in the
>>>>>>>>>>>> list) or that you CAN add it to the list, since that says
>>>>>>>>>>>> that it
>>>>>>>>>>>> has a
>>>>>>>>>>>> ordinal number higher than any possible ordinal number.
>>>>>>>>>>>>
>>>>>>>>>>>> Maybe your source of the description of Sipser's argument is
>>>>>>>>>>>> incomplete,
>>>>>>>>>>> I provided a scanned image of his actual proof on page 4.
>>>>>>>>>>> Go back and do what I said and read what Sipser said on my page
>>>>>>>>>>> four.
>>>>>>>>>>>
>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Refutation_of_Halting_Problem_Diagonalization_Argument
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>> Yes, and he said:
>>>>>>>>>>
>>>>>>>>>> In these tables we list all TMs down the rows, M1, M2, ... and
>>>>>>>>>> all
>>>>>>>>>> their
>>>>>>>>>> descriptions across the columns, (MI ), (M2 ), ...
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Let me ask you, how many TMs are their? Is their number finite or
>>>>>>>>>> infinite?
>>>>>>>>>>
>>>>>>>>>> If it is infinite, is there a space at the end of the table to
>>>>>>>>>> add
>>>>>>>>>> something.
>>>>>>>>>>
>>>>>>>>>> If it is a listing of *ALL* Turing Machines, why isn't D
>>>>>>>>>> listed in
>>>>>>>>>> it?
>>>>>>>>> It seems that you have set your mind on being disagreeable I
>>>>>>>>> will go
>>>>>>>>> back to ignoring you. My Sipser diagonalization proof is clear
>>>>>>>>> enough
>>>>>>>>> that I can start discussing it with actual computer scientists.
>>>>>>>>> Feel
>>>>>>>>> free to disagree I won't be paying attention.
>>>>>>>>>
>>>>>>>> If may be clear to you, but it is still wrong. Seems your typical
>>>>>>>> problem of not understanding infinities.
>>>>>>>>
>>>>>>>> You CAN'T add something to the end of an infinity long sequence.
>>>>>>>>
>>>>>>> You are telling me that when I use the same convention that
>>>>>>> Sipser uses
>>>>>>> to specify infinite sets that I am wrong only because you insist on
>>>>>>> being disagreeable.
>>>>>>>
>>>>>> But Sipser never does what you did. He uses the notation of M1,
>>>>>> M2, ...
>>>>>> to denote the infinite set of ALL Turing Machines. As such, if D
>>>>>> was a
>>>>>> Turing Machine, it would already be in the set, no need to add it.
>>>>>>
>>>>> He does add it in his Figure 4.6 nitwit and he does add it
>>>>> incorrectly.
>>>>> I already told you before you point out my errors read his freaking
>>>>> proof. The mistakes that you say I am making is what he is doing.
>>>> No, he is doing something different here.
>>>>
>>>> You went M1, M2, ... D
>>> *Corrections to Sipser's Figure 4.6 when he attempts
>>> **to insert machine D into his figure 4.5
>>> *
>>>
>>>      ⟨M1⟩      ⟨M2⟩      ⟨M3⟩       ⟨M4⟩ ... ⟨D⟩ ...
>>> M1  accept   ~halt    accept   ~halt    DC
>>> M2  accept   accept   accept   accept   DC
>>> M3  ~halt    ~halt    ~halt    ~halt    DC
>>> M4  accept   accept   ~halt    ~halt    DC
>>> .
>>> .
>>> .
>>> D   reject   reject   accept   accept   reject
>>> .
>>> .
>>> .
>>> *   Figure 4.4b (Insert D into Figure 4.4a adapted from Sipser 4.4) *
>>>
>>>
>>>      ⟨M1⟩      ⟨M2⟩      ⟨M3⟩      ⟨M4⟩ ...  ⟨D⟩ ...
>>> M1  _accept_   reject   accept   reject   DC
>>> M2  accept   _accept_   accept   accept   DC
>>> M3  reject   reject   _reject_   reject   DC
>>> M4  accept   accept   reject   _reject_   DC
>>> .
>>> .
>>> .
>>> D   accept   accept   accept   accept   _accept_
>>> .
>>> .
>>> .
>>> *Figure 4.5a (Insert D into Figure 4.5) *
>>>
>>> *
>>> *
>>>
>>> *Refuting the Sipser Halting Problem Diagonalization Argument*
>>>
>>> https://www.researchgate.net/publication/351947980_Refuting_the_Sipser_Halting_Problem_Diagonalization_Argument
>>>
>>>
>>>
>>
>> So your answer to the evidence that D can't be a Turing Machine is to
>> just change the requirements on the machines. Right, That really works.
>>
>>
>> Since by your 4.4b D rejects <D>, then the value that H must produce is
>> *reject*, so you have just shown that H now fails at its job of Deciding.
>>
>
> What part of opposite of the diagonal do you not understand?
>
>      In the following figure, we added D to Figure 4.5.
>      By our assumption, H is a TM and so is D. Therefore
>      it must occur on the list M 1 , M2, ... of all TMs.
>      Note that D computes the opposite of the diagonal
>      entries... (Sipser:1997:167)
>
> Sipser, Michael 1997. Introduction to the Theory of Computation. Boston:
> PWS Publishing
>