int DD()
{
� int Halt_Status = HHH(DD);
� if (Halt_Status)
��� HERE: goto HERE;
� return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
On 6/22/2025 4:32 AM, Mikko wrote:
On 2025-06-21 22:48:01 +0000, olcott said:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
If you like to play with ChatGPT you should ask it for
a solution to the halting problem.
I gave you the link showing exactly what ChatGPT agreed to. https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
On 6/22/2025 6:23 AM, Richard Damon wrote:
On 6/21/25 6:48 PM, olcott wrote:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
Just shows your natural stupidity in believing a lie you convinced the
artificial inteligence to say.
Sorry, I guess you natural stupidity extends to the failure to
understand that AI's are programmed to have no problem with LYING, but
are really just glorified "Yes Men", who will say what your prompt
directs them to say.
MIT Technology Review https://www.technologyreview.com/2024/03/04/1089403/large-language- models-amazing-but-nobody-knows-why/
On 6/21/2025 11:01 PM, Alan Mackenzie wrote:
In comp.theory olcott <[email protected]> wrote:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
That's neither here nor there. The plain fact is you have NOT refuted
any proof technique. How could you, you don't even understand what is
meant by proof?
A proof is any sequence of steps such that its conclusion
can be correctly determined to be necessarily true.
You have merely deluded yourself, spending year after year failing to
understand the simplest elements of computation theory, something a
maths or computer science undergraduate grasps in a very few hours at
most.
This stuff simply isn't your thing. Instead, you could and should do
something worthwhile with however much longer you have to live.
Then you could find one mistake, so far no one has.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
On 6/22/2025 4:09 PM, Richard Damon wrote:
On 6/22/25 10:46 AM, olcott wrote:
On 6/22/2025 6:23 AM, Richard Damon wrote:
On 6/21/25 6:48 PM, olcott wrote:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
Just shows your natural stupidity in believing a lie you convinced
the artificial inteligence to say.
Sorry, I guess you natural stupidity extends to the failure to
understand that AI's are programmed to have no problem with LYING,
but are really just glorified "Yes Men", who will say what your
prompt directs them to say.
MIT Technology Review
https://www.technologyreview.com/2024/03/04/1089403/large-language-
models-amazing-but-nobody-knows-why/
Which doesn't make a claim that their answers are without error.
I was not rebutting this and you know it.
On 6/22/2025 4:32 AM, Mikko wrote:
On 2025-06-21 22:48:01 +0000, olcott said:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
If you like to play with ChatGPT you should ask it for
a solution to the halting problem.
I gave you the link showing exactly what ChatGPT agreed to. https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
On 6/21/2025 11:01 PM, Alan Mackenzie wrote:
In comp.theory olcott <[email protected]> wrote:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
That's neither here nor there. The plain fact is you have NOT refuted
any proof technique. How could you, you don't even understand what is
meant by proof?
A proof is any sequence of steps such that its conclusion
can be correctly determined to be necessarily true.
On 6/23/2025 2:20 AM, Mikko wrote:
On 2025-06-22 14:42:13 +0000, olcott said:
On 6/22/2025 4:32 AM, Mikko wrote:
On 2025-06-21 22:48:01 +0000, olcott said:
int DD()
{
�� int Halt_Status = HHH(DD);
�� if (Halt_Status)
���� HERE: goto HERE;
�� return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
If you like to play with ChatGPT you should ask it for
a solution to the halting problem.
I gave you the link showing exactly what ChatGPT agreed to.
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
What an artificial idiot agrees to is not interesting.
A solution to the halting problem would be.
On 6/21/2025 6:21 PM, Mike Terry wrote:
Is that the ChatGPT that does the peer reviewing for submissions
for Proceedings of the London Mathematical Society?
Seems like you're nearly there at last!
Mike.
You said:
Which version of ChatGPT are you?
ChatGPT said:
You're currently chatting with ChatGPT using the GPT-4-turbo model (often referred to as GPT-4o, the
"omni" model). This is the most advanced model available as of June 2025 and is designed for high
performance across text, vision, and audio tasks.
On 6/23/2025 2:37 AM, Mikko wrote:
On 2025-06-22 14:38:56 +0000, olcott said:
On 6/21/2025 11:01 PM, Alan Mackenzie wrote:
In comp.theory olcott <[email protected]> wrote:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
That's neither here nor there. The plain fact is you have NOT refuted >>>> any proof technique. How could you, you don't even understand what is >>>> meant by proof?
A proof is any sequence of steps such that its conclusion
can be correctly determined to be necessarily true.
False. There are other requirements. Every sentence of the sequence,
not just the last one, must either be a premise or follow from
earlier ones with an acceptable inference rule.
There is a subset of proofs that have this requirement.
They typically are of the form that a conclusion is
proved definitely true within a set of assumptions.
Another form of this same proof only has expressions
of language known to be true as its premises.
I always interpret expressions of language according
to the literal base meaning of their words.
On 6/24/2025 3:39 AM, Mikko wrote:
On 2025-06-23 16:37:53 +0000, olcott said:
I always interpret expressions of language according
to the literal base meaning of their words.
I interprete the above to mean that the author of those words is stupid.
Counter factual, my IQ is in the top 3%
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
On 6/24/2025 11:43 AM, Alan Mackenzie wrote:
olcott <[email protected]> wrote:
On 6/24/2025 3:39 AM, Mikko wrote:
On 2025-06-23 16:37:53 +0000, olcott said:
I always interpret expressions of language according
to the literal base meaning of their words.
I interprete the above to mean that the author of those words is stupid.
Counter factual, my IQ is in the top 3%
Pull the other one!
Given your demonstrated lack of understanding of abstraction, of what a
proof is, of so many other things, it is clear to all the regulars in
this group that your IQ is not "in the top 3%", or anywhere near it.
It would seem to me you are, yet again, in the words of Sir Robert
Armstrong, being economical with the truth.
*I really did get that IQ on the Mensa entrance exam*
That I am unwilling to accept that textbooks on computer
science are inherently infallible is the broader minded
perspective of philosophy of computation.
This requires much more intelligence than simply memorizing a set of
rules and then mindlessly following these rules.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
It is an easily verified fact that no *input* to any partial halt
decider (PHD) can possibly do the opposite of what its corresponding PHD decides. In all of the years of all of these proofs no such *input* was
ever presented.
On 6/24/2025 12:39 PM, Alan Mackenzie wrote:
olcott <[email protected]> wrote:
On 6/24/2025 11:43 AM, Alan Mackenzie wrote:
olcott <[email protected]> wrote:
On 6/24/2025 3:39 AM, Mikko wrote:
On 2025-06-23 16:37:53 +0000, olcott said:
I always interpret expressions of language according
to the literal base meaning of their words.
I interprete the above to mean that the author of those words is
stupid.
Counter factual, my IQ is in the top 3%
Pull the other one!
Given your demonstrated lack of understanding of abstraction, of
what a proof is, of so many other things, it is clear to all the
regulars in this group that your IQ is not "in the top 3%", or
anywhere near it.
It would seem to me you are, yet again, in the words of Sir Robert
Armstrong, being economical with the truth.
*I really did get that IQ on the Mensa entrance exam*
OK, let us be charitable, and suggest that that exam was a very long
time ago, and that your general intelligence has declined
substantially in the interval.
That I am unwilling to accept that textbooks on computer
science are inherently infallible is the broader minded
perspective of philosophy of computation.
That's an inaccurate summary. You're clearly unable to understand these
textbooks. If you were able, you'd see that the things they say are
necessarily correct, according to clear reasoning from obvious axioms.
Whether you'd accept these books if you could understand them is more
the question.
It is an easily verified fact that no *input* to any
partial halt decider (PHD) can possibly do the opposite
of what its corresponding PHD decides.
.... In all of the years of all of these proofs no such *input* was
ever presented.
This requires much more intelligence than simply memorizing a set of
rules and then mindlessly following these rules.
It does. Somebody simply memorizing these rules would be unable to pass
his exams and graduate. To graduate in computer science, and certainly
in mathematics, requires an ability fluently to manipulate abstractions,
something which cannot be done without fundamental understanding.
You clearly lack this ability and that understanding.
I don't believe there's any substance behind your "in the top 3%" claim.
It's at complete odds with what we see you writing in this newsgroup.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
On 6/24/2025 12:57 PM, joes wrote:So common that nobody would suggest such. You are the king of strawmen.
Am Tue, 24 Jun 2025 12:46:01 -0500 schrieb olcott:It is common knowledge the no Turing Machine can take another directly executed Turing Machine as an input.
It is an easily verified fact that no *input* to any partial halt
decider (PHD) can possibly do the opposite of what its corresponding
PHD decides. In all of the years of all of these proofs no such
*input* was ever presented.
You should clarify that you don't even think programs can be passed as
input.
On 6/24/2025 3:39 AM, Mikko wrote:
On 2025-06-23 16:37:53 +0000, olcott said:
I always interpret expressions of language according
to the literal base meaning of their words.
I interprete the above to mean that the author of those words is stupid.
Counter factual, my IQ is in the top 3%
On 6/24/2025 4:07 PM, joes wrote:
Am Tue, 24 Jun 2025 13:06:22 -0500 schrieb olcott:
On 6/24/2025 12:57 PM, joes wrote:So common that nobody would suggest such. You are the king of strawmen.
Am Tue, 24 Jun 2025 12:46:01 -0500 schrieb olcott:It is common knowledge the no Turing Machine can take another directly
It is an easily verified fact that no *input* to any partial halt
decider (PHD) can possibly do the opposite of what its corresponding >>>>> PHD decides. In all of the years of all of these proofs no such
*input* was ever presented.
You should clarify that you don't even think programs can be passed as >>>> input.
executed Turing Machine as an input.
*From the bottom of page 319 has been adapted to this* https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
if Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt
Ĥ applied to ⟨Ĥ⟩ does not have embedded_H reporting on
the behavior specified by its input ⟨Ĥ⟩ ⟨Ĥ⟩ it has embedded_H reporting on its own behavior.
Since Turing Machines cannot take directly executing
Turing Machines as inputs this means that the directly
executed Ĥ applied to ⟨Ĥ⟩ is not in the domain of
Ĥ.embedded_H, *thus no contradiction is ever formed*
On 6/24/2025 3:47 AM, Mikko wrote:
On 2025-06-23 15:16:14 +0000, olcott said:
On 6/23/2025 2:37 AM, Mikko wrote:
On 2025-06-22 14:38:56 +0000, olcott said:
On 6/21/2025 11:01 PM, Alan Mackenzie wrote:
In comp.theory olcott <[email protected]> wrote:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
That's neither here nor there. The plain fact is you have NOT refuted >>>>>> any proof technique. How could you, you don't even understand what is >>>>>> meant by proof?
A proof is any sequence of steps such that its conclusion
can be correctly determined to be necessarily true.
False. There are other requirements. Every sentence of the sequence,
not just the last one, must either be a premise or follow from
earlier ones with an acceptable inference rule.
There is a subset of proofs that have this requirement.
They typically are of the form that a conclusion is
proved definitely true within a set of assumptions.
Another form of this same proof only has expressions
of language known to be true as its premises.
If the set of the premises is not the same it is not the same proof.
When a proof has known facts all of its premises thenn
its conclusion is proven definitely true when it is proven.
On 6/25/2025 2:26 AM, Mikko wrote:
On 2025-06-24 14:55:11 +0000, olcott said:
On 6/24/2025 3:47 AM, Mikko wrote:
On 2025-06-23 15:16:14 +0000, olcott said:
On 6/23/2025 2:37 AM, Mikko wrote:
On 2025-06-22 14:38:56 +0000, olcott said:
On 6/21/2025 11:01 PM, Alan Mackenzie wrote:
In comp.theory olcott <[email protected]> wrote:
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://chatgpt.com/s/t_6857335b37a08191a077d57039fa4a76
ChatGPT agrees that I have correctly refuted every
halting problem proof technique that relies on the above
pattern.
That's neither here nor there. The plain fact is you have NOT refuted
any proof technique. How could you, you don't even understand what is
meant by proof?
A proof is any sequence of steps such that its conclusion
can be correctly determined to be necessarily true.
False. There are other requirements. Every sentence of the sequence, >>>>>> not just the last one, must either be a premise or follow from
earlier ones with an acceptable inference rule.
There is a subset of proofs that have this requirement.
They typically are of the form that a conclusion is
proved definitely true within a set of assumptions.
Another form of this same proof only has expressions
of language known to be true as its premises.
If the set of the premises is not the same it is not the same proof.
When a proof has known facts all of its premises thenn
its conclusion is proven definitely true when it is proven.
Nevertheless, the proofa are not the same if their sets of premises
are not the same.
Proofs can be semantically equivalent when these proofs
bother to pay attention to the semantics.
On 6/25/2025 2:21 AM, Mikko wrote:
On 2025-06-24 21:41:37 +0000, olcott said:
On 6/24/2025 4:07 PM, joes wrote:
Am Tue, 24 Jun 2025 13:06:22 -0500 schrieb olcott:*From the bottom of page 319 has been adapted to this*
On 6/24/2025 12:57 PM, joes wrote:So common that nobody would suggest such. You are the king of strawmen. >>>
Am Tue, 24 Jun 2025 12:46:01 -0500 schrieb olcott:It is common knowledge the no Turing Machine can take another directly >>>>> executed Turing Machine as an input.
It is an easily verified fact that no *input* to any partial halt >>>>>>> decider (PHD) can possibly do the opposite of what its corresponding >>>>>>> PHD decides. In all of the years of all of these proofs no such
*input* was ever presented.
You should clarify that you don't even think programs can be passed as >>>>>> input.
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
if Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt
Ĥ applied to ⟨Ĥ⟩ does not have embedded_H reporting on
the behavior specified by its input ⟨Ĥ⟩ ⟨Ĥ⟩ it has embedded_H
reporting on its own behavior.
As made clear in the source text, embedded_H does the same as
H when given the same input. The only difference is that if
that same behaviour reaches its qy state then H halts there
but Ĥ runs forever in a tight loop.
*You are not getting the main point*
The fact that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to ⟨Ĥ.qn⟩ is not contradicted by the fact that Ĥ.embedded_H itself halts.
Because Ĥ.embedded_H cannot possibly take any directly
executing TM as its input that makes the behavior of
Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H.
Since Turing Machines cannot take directly executing
Turing Machines as inputs this means that the directly
executed Ĥ applied to ⟨Ĥ⟩ is not in the domain of
Ĥ.embedded_H, *thus no contradiction is ever formed*
False. That Turing Machines cannot take directly executing
Turing Machnes as inputs is irrelevant.
Directly executing TM's are not in the domain of any
halt decider.
The definition of
"halting decider" requires that the decider thakes a
description of a Turing machine and a an input to it.
Yes.
From the construction of Ĥ follows that the domain of Ĥ is
the same as the required domain of a halt decider. As the
Maybe, IDK. What I do know is that
Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H.
proof proves H does not do what a halting decider is
required to do
It is required to take a directly executing TM as input.
and
It is not allowed to take a directly executing TM as input.
when the input is <Ĥ> <Ĥ>, contradicting
the claim that H is a halting decider.
*It never has been doing any such thing*
When Ĥ.embedded_H is a simulating partial halt decider
then Ĥ.embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ specifies recursive simulation that cannot possibly reach its own simulated
final halt state of ⟨Ĥ.qn⟩.
On 6/26/2025 5:00 AM, Mikko wrote:
On 2025-06-25 15:26:28 +0000, olcott said:
On 6/25/2025 2:21 AM, Mikko wrote:
On 2025-06-24 21:41:37 +0000, olcott said:
On 6/24/2025 4:07 PM, joes wrote:
Am Tue, 24 Jun 2025 13:06:22 -0500 schrieb olcott:*From the bottom of page 319 has been adapted to this*
On 6/24/2025 12:57 PM, joes wrote:So common that nobody would suggest such. You are the king of strawmen. >>>>>
Am Tue, 24 Jun 2025 12:46:01 -0500 schrieb olcott:It is common knowledge the no Turing Machine can take another directly >>>>>>> executed Turing Machine as an input.
It is an easily verified fact that no *input* to any partial halt >>>>>>>>> decider (PHD) can possibly do the opposite of what its corresponding >>>>>>>>> PHD decides. In all of the years of all of these proofs no such >>>>>>>>> *input* was ever presented.
You should clarify that you don't even think programs can be passed as >>>>>>>> input.
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
if Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt
Ĥ applied to ⟨Ĥ⟩ does not have embedded_H reporting on
the behavior specified by its input ⟨Ĥ⟩ ⟨Ĥ⟩ it has embedded_H >>>>> reporting on its own behavior.
As made clear in the source text, embedded_H does the same as
H when given the same input. The only difference is that if
that same behaviour reaches its qy state then H halts there
but Ĥ runs forever in a tight loop.
*You are not getting the main point*
The fact that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to ⟨Ĥ.qn⟩ is >>> not contradicted by the fact that Ĥ.embedded_H itself halts.
That is not the main point.
It is the *only* reason why
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
is incorrectly construed as being incorrect.
The main point is that Ĥ ⟨Ĥ⟩ halts if
iH ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn but not if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to
Ĥ.qy. Anything said about embedded_H is merely an intermediate
step in the proof of the main point if not totally irrelevant.
Because Ĥ.embedded_H cannot possibly take any directly
executing TM as its input that makes the behavior of
Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H.
Irrelevant. It can and does take the same input as H and from that
computes the same as H. That is all that is needed for the proof.
Since Turing Machines cannot take directly executing
Turing Machines as inputs this means that the directly
executed Ĥ applied to ⟨Ĥ⟩ is not in the domain of
Ĥ.embedded_H, *thus no contradiction is ever formed*
False. That Turing Machines cannot take directly executing
Turing Machnes as inputs is irrelevant.
Directly executing TM's are not in the domain of any
halt decider.
The definition of a halt decider requires that a halt decider
correctly predicts whether a direct execution halts
That has always been incorrect because no Turing machine
can ever take any directly executing Turing machine as
its input all of these directly executed Turing machines
are outside of the domain of any partial halt decider.
Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
is correct because ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by
Ĥ.embedded_H cannot possibly reach its own simulated
final halt state ⟨Ĥ.qn⟩.
On 6/28/2025 6:38 AM, Mikko wrote:
On 2025-06-27 15:22:50 +0000, olcott said:
On 6/26/2025 5:00 AM, Mikko wrote:
On 2025-06-25 15:26:28 +0000, olcott said:
On 6/25/2025 2:21 AM, Mikko wrote:
On 2025-06-24 21:41:37 +0000, olcott said:
On 6/24/2025 4:07 PM, joes wrote:
Am Tue, 24 Jun 2025 13:06:22 -0500 schrieb olcott:
On 6/24/2025 12:57 PM, joes wrote:So common that nobody would suggest such. You are the king of strawmen.
Am Tue, 24 Jun 2025 12:46:01 -0500 schrieb olcott:It is common knowledge the no Turing Machine can take another directly
It is an easily verified fact that no *input* to any partial halt >>>>>>>>>>> decider (PHD) can possibly do the opposite of what its corresponding
PHD decides. In all of the years of all of these proofs no such >>>>>>>>>>> *input* was ever presented.
You should clarify that you don't even think programs can be passed as
input.
executed Turing Machine as an input.
*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
if Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt
Ĥ applied to ⟨Ĥ⟩ does not have embedded_H reporting on
the behavior specified by its input ⟨Ĥ⟩ ⟨Ĥ⟩ it has embedded_H >>>>>>> reporting on its own behavior.
As made clear in the source text, embedded_H does the same as
H when given the same input. The only difference is that if
that same behaviour reaches its qy state then H halts there
but Ĥ runs forever in a tight loop.
*You are not getting the main point*
The fact that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to ⟨Ĥ.qn⟩ is
not contradicted by the fact that Ĥ.embedded_H itself halts.
That is not the main point.
It is the *only* reason why
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
is incorrectly construed as being incorrect.
It is neither correct nor incorrect. There are no requirements about Ĥ.
The above shows that Ĥ.embedded_H decided not halting.
This is either correct or incorrect depending on the
criterion measure.
If Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is to report on the behavior
that its inputs specify then transitioning to Ĥ.qn
is correct.
When it is understood that the directly executing
Ĥ applied to ⟨Ĥ⟩ is not in the domain of Ĥ.embedded_H
then the behavior of Ĥ applied to ⟨Ĥ⟩ does not contradict
the reporting of non-halting.
The main point is that Ĥ ⟨Ĥ⟩ halts if
iH ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn but not if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to
Ĥ.qy. Anything said about embedded_H is merely an intermediate
step in the proof of the main point if not totally irrelevant.
Because Ĥ.embedded_H cannot possibly take any directly
executing TM as its input that makes the behavior of
Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H.
Irrelevant. It can and does take the same input as H and from that
computes the same as H. That is all that is needed for the proof.
Since Turing Machines cannot take directly executing
Turing Machines as inputs this means that the directly
executed Ĥ applied to ⟨Ĥ⟩ is not in the domain of
Ĥ.embedded_H, *thus no contradiction is ever formed*
False. That Turing Machines cannot take directly executing
Turing Machnes as inputs is irrelevant.
Directly executing TM's are not in the domain of any
halt decider.
The definition of a halt decider requires that a halt decider
correctly predicts whether a direct execution halts
That has always been incorrect because no Turing machine
can ever take any directly executing Turing machine as
its input all of these directly executed Turing machines
are outside of the domain of any partial halt decider.
No, it is not incorrect. It is what the words mean.
Likewise with the definition of a circle as having four
equal length sides.
The requirement that a halt decider
report on the behavior of the direct execution of a machine
is contradicted by the fact that no Turing Machine can take
a directly executing machine as its input.
Just because no one has ever noticed this before does not
mean that I am wrong.
Partial halt deciders are only held
accountable for *inputs* in their domain. Their own directly
executed selves are not *inputs* in their domain.
Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
is correct because ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by
Ĥ.embedded_H cannot possibly reach its own simulated
final halt state ⟨Ĥ.qn⟩.
Correct or incorrect does not apply to Ĥ as there are no requirements.
The requirement is that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
determine the halt status that its input specifies.
*Here is the whole Linz proof* https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
On 6/29/2025 3:25 AM, Mikko wrote:
On 2025-06-28 12:56:04 +0000, olcott said:
On 6/28/2025 6:38 AM, Mikko wrote:
On 2025-06-27 15:22:50 +0000, olcott said:The above shows that Ĥ.embedded_H decided not halting.
On 6/26/2025 5:00 AM, Mikko wrote:
On 2025-06-25 15:26:28 +0000, olcott said:
On 6/25/2025 2:21 AM, Mikko wrote:
On 2025-06-24 21:41:37 +0000, olcott said:
On 6/24/2025 4:07 PM, joes wrote:
Am Tue, 24 Jun 2025 13:06:22 -0500 schrieb olcott:
On 6/24/2025 12:57 PM, joes wrote:So common that nobody would suggest such. You are the king of strawmen.
Am Tue, 24 Jun 2025 12:46:01 -0500 schrieb olcott:It is common knowledge the no Turing Machine can take another directly
It is an easily verified fact that no *input* to any partial halt >>>>>>>>>>>>> decider (PHD) can possibly do the opposite of what its corresponding
PHD decides. In all of the years of all of these proofs no such >>>>>>>>>>>>> *input* was ever presented.
You should clarify that you don't even think programs can be passed as
input.
executed Turing Machine as an input.
*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞ >>>>>>>>> if Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt
Ĥ applied to ⟨Ĥ⟩ does not have embedded_H reporting on >>>>>>>>> the behavior specified by its input ⟨Ĥ⟩ ⟨Ĥ⟩ it has embedded_H
reporting on its own behavior.
As made clear in the source text, embedded_H does the same as
H when given the same input. The only difference is that if
that same behaviour reaches its qy state then H halts there
but Ĥ runs forever in a tight loop.
*You are not getting the main point*
The fact that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to ⟨Ĥ.qn⟩ is
not contradicted by the fact that Ĥ.embedded_H itself halts.
That is not the main point.
It is the *only* reason why
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
is incorrectly construed as being incorrect.
It is neither correct nor incorrect. There are no requirements about Ĥ. >>>
This is either correct or incorrect depending on the
criterion measure.
No, there is a third possibility: it is irrelevant if the criteria
don't say anything about that.
If Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is to report on the behavior
that its inputs specify then transitioning to Ĥ.qn
is correct.
If. The proof of unprovability does not specify any requirement
about that.
When it is understood that the directly executing
Ĥ applied to ⟨Ĥ⟩ is not in the domain of Ĥ.embedded_H
then the behavior of Ĥ applied to ⟨Ĥ⟩ does not contradict
the reporting of non-halting.
Whatever embedded_H reports does not not contradict anyting specified
in the proof of uncomputability of halting.
The main point is that Ĥ ⟨Ĥ⟩ halts if
iH ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn but not if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to
Ĥ.qy. Anything said about embedded_H is merely an intermediate
step in the proof of the main point if not totally irrelevant.
Because Ĥ.embedded_H cannot possibly take any directly
executing TM as its input that makes the behavior of
Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H.
Irrelevant. It can and does take the same input as H and from that >>>>>> computes the same as H. That is all that is needed for the proof.
Since Turing Machines cannot take directly executing
Turing Machines as inputs this means that the directly
executed Ĥ applied to ⟨Ĥ⟩ is not in the domain of
Ĥ.embedded_H, *thus no contradiction is ever formed*
False. That Turing Machines cannot take directly executing
Turing Machnes as inputs is irrelevant.
Directly executing TM's are not in the domain of any
halt decider.
The definition of a halt decider requires that a halt decider
correctly predicts whether a direct execution halts
That has always been incorrect because no Turing machine
can ever take any directly executing Turing machine as
its input all of these directly executed Turing machines
are outside of the domain of any partial halt decider.
No, it is not incorrect. It is what the words mean.
Likewise with the definition of a circle as having four
equal length sides.
The concept defined by that definition is good and well-defined but
the word "circle" is already reserved for another geometric concept
by earlier geometers. The usual meaning of "hating" in context of
Turing machines has no prior meaning because Turing machine is a
recent innovation and the meaning is compatible with the traditional
meaning of "halting" in Common Language. Therefore your "likewise"
is false.
The requirement that a halt decider
report on the behavior of the direct execution of a machine
is contradicted by the fact that no Turing Machine can take
a directly executing machine as its input.
The requirement says "predicts", not reports, though both words
mean the same in this context.
But "the requirement is not contradicted" is a category error.
When it is required that a Turing Machine halt decider is to
report on the behavior of another directly executing Turing
machine this requirement is incorrect.
This requirement is incorrect because Turing machines only
take finite string inputs they do not take directly executing
Turing Machines as inputs.
It usually makes no difference because almost all of the time
the finite string machine description serves as a proxy and
has the same behavior.
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