In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
=20
typedef void (*ptr)();
int HHH(ptr P);
=20
void DDD()
{
=C2=A0=C2=A0 HHH(DDD);
=C2=A0=C2=A0 return;
}
=20
int main()
{
=C2=A0=C2=A0 HHH(DDD);
=C2=A0=C2=A0 DDD();
}
=20
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern.=C2=A0
When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
=C2=A0> In comp.theory olcott <polcott333@gmail.com
=C2=A0> wrote:
=C2=A0>
=C2=A0>> I know that DDD .... simulated by HHH cannot
=C2=A0>> possibly reach its own simulated "return" statement
=C2=A0>> final halt state because the execution trace
=C2=A0>> conclusively proves this.
=C2=A0>
=C2=A0> Everybody else knows this, too, and nobody has
=C2=A0> said otherwise. *The conclusion is that the*
=C2=A0> *simulation by HHH is incorrect*
=C2=A0>
=20
*That last sentence is an intentional falsehood*
On 27/06/2025 20:36, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
Well, people here use the term "simulation" in a number of ways, right?
Like I've pointed out to you on several occasions. Maybe you should consider a simple misunderstanding over terminology before assuming bad intentions.
So what does Alan mean by "the simulation by HHH is incorrect",
exactly? (And why do you think it is incorrect?)
That is a question for PO, rather than Alan, since it is PO who claims
Alan is lying...
Mike.
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern.
HHH(DDD) will run in infinite loop, which conforms to Halting Problem proof. >>
When
HHH detects such a pattern it aborts its simulation
and returns 0.
This looks like a problem specification, but you said "Halting Problem" is >> incorrect. Peter Olcott's Own Problem is never clear.
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped the below
in his reply but YOU:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped the below
in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
You Are the epitome of bad faith and dishonesty.
This may cost you your actual soul: Revelations 21:8.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern.
HHH(DDD) will run in infinite loop, which conforms to Halting Problem >>>> proof.
When
HHH detects such a pattern it aborts its simulation
and returns 0.
And here is the part that you dishonestly said was erased when in fact
it was not.
This proves that it is YOU who are lying and will say anything to push
your agenda.
This looks like a problem specification, but you said "Halting
Problem" is
incorrect. Peter Olcott's Own Problem is never clear.
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped the
below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove you wrong.
The definition of HHH is missing.
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped the
below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove you wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
Your mere rhetoric to the contrary does not actually change
these verified facts. Are you able to stick with correct
reasoning or is mere rhetoric all that you have?
Your dishonesty knows no bounds. And as you yourself self:
On 5/25/2025 3:04 PM, olcott wrote:
You Are the epitome of bad faith and dishonesty.
This may cost you your actual soul: Revelations 21:8.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern.
HHH(DDD) will run in infinite loop, which conforms to Halting
Problem proof.
When
HHH detects such a pattern it aborts its simulation
and returns 0.
And here is the part that you dishonestly said was erased when in
fact it was not.
This proves that it is YOU who are lying and will say anything to
push your agenda.
Your failure to reply to the above, and in fact your dishonest erasing
of it constitutes your admission of lying to push your agenda.
This looks like a problem specification, but you said "Halting
Problem" is
incorrect. Peter Olcott's Own Problem is never clear.
On 6/28/2025 11:30 AM, olcott wrote:
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped the
below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove you
wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
No, you didn't give a definition of HHH, just gave a vauge incomplete description of what you think it does.
On 6/28/2025 10:39 AM, dbush wrote:
On 6/28/2025 11:30 AM, olcott wrote:
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped the
below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove you
wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
No, you didn't give a definition of HHH, just gave a vauge incomplete
description of what you think it does.
It completely defines the generic notion of a simulating
termination analyzer and
It specifies every detail about HHH that is required
to correctly determine whether or not DDD correctly
simulated by HHH can possibly reach its own simulated
"return" statement final halt state.
On 5/4/2025 11:03 PM, dbush wrote:
On 5/4/2025 10:05 PM, olcott wrote:
On 5/4/2025 7:23 PM, Richard Damon wrote:
But HHH doesn't correct emulated DD by those rules, as those rules
do not allow HHH to stop its emulation,
Sure they do you freaking moron...
Then show where in the Intel instruction manual that the execution of
any instruction other than a HLT is allowed to stop instead of
executing the next instruction.
Failure to do so in your next reply, or within one hour of your next
post on this newsgroup, will be taken as you official on-the-record
admission that there is no such allowance and that HHH does NOT
correctly simulate DD.
Let the record show that Peter Olcott made the following post in this newsgroup after the above message:
On 5/4/2025 11:04 PM, olcott wrote:
D *WOULD NEVER STOP RUNNING UNLESS*
indicates that professor Sipser was agreeing
to hypotheticals AS *NOT CHANGING THE INPUT*
You are taking
*WOULD NEVER STOP RUNNING UNLESS*
to mean *NEVER STOPS RUNNING* that is incorrect.
And has made no attempt after over 9 hours to show where in the Intel instruction manual that execution is allowed to stop after any
instruction other than HLT.
Therefore, as per the above criteria:
LET THE RECORD SHOW
That Peter Olcott
Has *officially* admitted
That DD is NOT correctly simulated by HHH
What is is a verified fact that you openly and blatantly lied about wij
erasing your description when it was in fact you yourself that did it.
On 6/28/2025 11:46 AM, olcott wrote:
On 6/28/2025 10:39 AM, dbush wrote:
On 6/28/2025 11:30 AM, olcott wrote:
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped the >>>>>>> below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove you
wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
No, you didn't give a definition of HHH, just gave a vauge incomplete
description of what you think it does.
It completely defines the generic notion of a simulating
termination analyzer and
It specifies every detail about HHH that is required
to correctly determine whether or not DDD correctly
simulated by HHH can possibly reach its own simulated
"return" statement final halt state.
But why would you say that when you've admitted on the record (see
below) that DDD is not in fact correctly simulated by HHH?
On 6/27/2025 7:14 PM, Mike Terry wrote:
On 27/06/2025 20:36, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
Well, people here use the term "simulation" in a number of ways, right?
*There is only one correct way*
void DDD()
{
HHH(DDD);
return;
}
The correct simulation of DDD by HHH means that HHH simulates
DDD and then emulates itself simulating DDD a number of times
according to the semantics of C.
Everyone here knows that DDD correctly simulated by HHH
cannot possibly reach its own simulated "return" statement.
Most everyone on comp.theory lies about this.
Like I've pointed out to you on several occasions. Maybe you should consider a simple
misunderstanding over terminology before assuming bad intentions.
After three years on this same point a simple misunderstanding
has been ruled out.
So what does Alan mean by "the simulation by HHH is incorrect", exactly? (And why do you think it
is incorrect?)
He is simply lying. Most of the reviews of my work are
counter-factual dogmatic assertions utterly bereft of
any supporting reasoning.
On 28/06/2025 16:00, olcott wrote:
On 6/27/2025 7:14 PM, Mike Terry wrote:
On 27/06/2025 20:36, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
Well, people here use the term "simulation" in a number of ways, right?
*There is only one correct way*
That's patently rubbish.
This is one of your major cognitive mistakes - believing that the
squiggles which make up words have an absolute meaning independently of
the people using them to (hopefully) communicate.
void DDD()
{
HHH(DDD);
return;
}
The correct simulation of DDD by HHH means that HHH simulates
DDD and then emulates itself simulating DDD a number of times
according to the semantics of C.
"a number of times"???
a) Does a correct simulation need to simulate until its target
computation terminates?
b) Does a simulation need to also correctly decide halting for its
target computation?
[My personal answers would be "no" and "no" - I would consider what
others may insist on calling a partial simulation to a correct example
of "simulation" (but of course incomplete), provided the steps of the simulation are correctly calculated; and I consider deciding the halt
status to be the job of a halt "decider" rather than a simulator [of
course a halt decider may /utilise/ simulation as a technique in
reaching its decision]. Other ways of looking at this are equally valid
and we need to understand each other's positions.]
Everyone here knows that DDD correctly simulated by HHH
cannot possibly reach its own simulated "return" statement.
That's what Alan said above. Read his words: "Everybody else knows
this, too,
and nobody has said otherwise." You need to read what people
write and keep track of the conversation...
Most everyone on comp.theory lies about this.
Well, not Alan (whom you accused), or me, or Richard or Mikko or Fred
(I expect I could include others but can't be bothered to research the question).
So who lies about it exactly, and you had better provide a (recent) link
to them lying! Or just admit you haven't a clue what people are
actually saying to you...
Like I've pointed out to you on several occasions. Maybe you should
consider a simple misunderstanding over terminology before assuming
bad intentions.
After three years on this same point a simple misunderstanding
has been ruled out.
For a "normal" correspondant perhaps, but you forget your own
disabilities - you are incapable of understanding what people say to
you, including whether they are agreeing with you (in strictly limited parts) or not.
So what does Alan mean by "the simulation by HHH is incorrect",
exactly? (And why do you think it is incorrect?)
He is simply lying. Most of the reviews of my work are
counter-factual dogmatic assertions utterly bereft of
any supporting reasoning.
Lol, you haven't even answered the question of what you think Alan meant
by his statement,
or why that is a lie. So how can you be sure he is
lying? All you've talked about above is the bit Alan explicitly said
he /agreed/ with.
Mike.
On 6/28/2025 11:22 AM, Mike Terry wrote:
On 28/06/2025 16:00, olcott wrote:
On 6/27/2025 7:14 PM, Mike Terry wrote:
On 27/06/2025 20:36, olcott wrote:*There is only one correct way*
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
Well, people here use the term "simulation" in a number of ways, right? >>>
That's patently rubbish.
This is one of your major cognitive mistakes - believing that the squiggles which make up words
have an absolute meaning independently of the people using them to (hopefully) communicate.
void DDD()
{
HHH(DDD);
return;
}
The correct simulation of DDD by HHH means that HHH simulates
DDD and then emulates itself simulating DDD a number of times
according to the semantics of C.
"a number of times"???
Do you know what a number is?
In this case I am referring to every non-negative integer.
a) Does a correct simulation need to simulate until its target computation terminates?
There is no target computation here we only have the
behavior that the input to HHH(DDD) specifies.
On 6/28/2025 10:52 AM, dbush wrote:
On 6/28/2025 11:46 AM, olcott wrote:
On 6/28/2025 10:39 AM, dbush wrote:
On 6/28/2025 11:30 AM, olcott wrote:
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped the >>>>>>>> below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove you >>>>>> wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
No, you didn't give a definition of HHH, just gave a vauge
incomplete description of what you think it does.
It completely defines the generic notion of a simulating
termination analyzer and
It specifies every detail about HHH that is required
to correctly determine whether or not DDD correctly
simulated by HHH can possibly reach its own simulated
"return" statement final halt state.
But why would you say that when you've admitted on the record (see
below) that DDD is not in fact correctly simulated by HHH?
*I am not going to tolerate any misdirection to any other points*
void DDD()
{
HHH(DDD);
return;
}
The only point relevant to this forum is that DDD correctly
simulated by HHH
On 5/4/2025 11:03 PM, dbush wrote:
On 5/4/2025 10:05 PM, olcott wrote:
On 5/4/2025 7:23 PM, Richard Damon wrote:
But HHH doesn't correct emulated DD by those rules, as those rules
do not allow HHH to stop its emulation,
Sure they do you freaking moron...
Then show where in the Intel instruction manual that the execution of
any instruction other than a HLT is allowed to stop instead of
executing the next instruction.
Failure to do so in your next reply, or within one hour of your next
post on this newsgroup, will be taken as you official on-the-record
admission that there is no such allowance and that HHH does NOT
correctly simulate DD.
Let the record show that Peter Olcott made the following post in this newsgroup after the above message:
On 5/4/2025 11:04 PM, olcott wrote:
D *WOULD NEVER STOP RUNNING UNLESS*
indicates that professor Sipser was agreeing
to hypotheticals AS *NOT CHANGING THE INPUT*
You are taking
*WOULD NEVER STOP RUNNING UNLESS*
to mean *NEVER STOPS RUNNING* that is incorrect.
And has made no attempt after over 9 hours to show where in the Intel instruction manual that execution is allowed to stop after any
instruction other than HLT.
Therefore, as per the above criteria:
LET THE RECORD SHOW
That Peter Olcott
Has *officially* admitted
That DD is NOT correctly simulated by HHH
On 6/28/2025 12:03 PM, olcott wrote:
On 6/28/2025 10:52 AM, dbush wrote:
On 6/28/2025 11:46 AM, olcott wrote:
On 6/28/2025 10:39 AM, dbush wrote:
On 6/28/2025 11:30 AM, olcott wrote:
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped >>>>>>>>> the below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove
you wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
No, you didn't give a definition of HHH, just gave a vauge
incomplete description of what you think it does.
It completely defines the generic notion of a simulating
termination analyzer and
It specifies every detail about HHH that is required
to correctly determine whether or not DDD correctly
simulated by HHH can possibly reach its own simulated
"return" statement final halt state.
But why would you say that when you've admitted on the record (see
below) that DDD is not in fact correctly simulated by HHH?
*I am not going to tolerate any misdirection to any other points*
void DDD()
{
HHH(DDD);
return;
}
The only point relevant to this forum is that DDD correctly
simulated by HHH
Is something that you have admitted on the record doesn't happen:
On 6/28/2025 12:03 PM, olcott wrote:
On 6/28/2025 10:52 AM, dbush wrote:
On 6/28/2025 11:46 AM, olcott wrote:
On 6/28/2025 10:39 AM, dbush wrote:
On 6/28/2025 11:30 AM, olcott wrote:
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped >>>>>>>>> the below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove
you wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
No, you didn't give a definition of HHH, just gave a vauge
incomplete description of what you think it does.
It completely defines the generic notion of a simulating
termination analyzer and
It specifies every detail about HHH that is required
to correctly determine whether or not DDD correctly
simulated by HHH can possibly reach its own simulated
"return" statement final halt state.
But why would you say that when you've admitted on the record (see
below) that DDD is not in fact correctly simulated by HHH?
*I am not going to tolerate any misdirection to any other points*
void DDD()
{
HHH(DDD);
return;
}
The only point relevant to this forum is that DDD correctly
simulated by HHH
Is something that you have admitted on the record doesn't happen:
On 6/28/2025 12:13 PM, dbush wrote:
On 6/28/2025 12:03 PM, olcott wrote:
On 6/28/2025 10:52 AM, dbush wrote:
On 6/28/2025 11:46 AM, olcott wrote:
On 6/28/2025 10:39 AM, dbush wrote:
On 6/28/2025 11:30 AM, olcott wrote:
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
The dishonest one here is YOU, as it was not wij who snipped >>>>>>>>>> the below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here.
Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove >>>>>>>> you wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
No, you didn't give a definition of HHH, just gave a vauge
incomplete description of what you think it does.
It completely defines the generic notion of a simulating
termination analyzer and
It specifies every detail about HHH that is required
to correctly determine whether or not DDD correctly
simulated by HHH can possibly reach its own simulated
"return" statement final halt state.
But why would you say that when you've admitted on the record (see
below) that DDD is not in fact correctly simulated by HHH?
*I am not going to tolerate any misdirection to any other points*
void DDD()
{
HHH(DDD);
return;
}
The only point relevant to this forum is that DDD correctly
simulated by HHH
Is something that you have admitted on the record doesn't happen:
*This is the only point that I will address*
*Any attempt at changing the subject will*
*be construed as dishonest*
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH
On 28/06/2025 17:47, olcott wrote:
On 6/28/2025 11:22 AM, Mike Terry wrote:
On 28/06/2025 16:00, olcott wrote:
On 6/27/2025 7:14 PM, Mike Terry wrote:
On 27/06/2025 20:36, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
Well, people here use the term "simulation" in a number of ways,
right?
*There is only one correct way*
That's patently rubbish.
This is one of your major cognitive mistakes - believing that the
squiggles which make up words have an absolute meaning independently
of the people using them to (hopefully) communicate.
void DDD()
{
HHH(DDD);
return;
}
The correct simulation of DDD by HHH means that HHH simulates
DDD and then emulates itself simulating DDD a number of times
according to the semantics of C.
"a number of times"???
Do you know what a number is?
In this case I am referring to every non-negative integer.
a) Does a correct simulation need to simulate until its target
computation terminates?
There is no target computation here we only have the
behavior that the input to HHH(DDD) specifies.
ok I see you intend to be completely non-responsive, just as you were in
the "diverging simulations" thread.
I'll just point out for the record that you haven't provided any justification for accusing Alan or anybody else of lying to you.
Indeed, your responses just show that you have no clue what other people
are saying to you!
Mike.
In comp.theory olcott <polcott333@gmail.com> wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has said otherwise.
The conclusion is that the simulation by HHH is incorrect.
I know that DDD correctly simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
On 6/28/2025 1:27 PM, olcott wrote:
On 6/28/2025 12:13 PM, dbush wrote:
On 6/28/2025 12:03 PM, olcott wrote:
On 6/28/2025 10:52 AM, dbush wrote:
On 6/28/2025 11:46 AM, olcott wrote:
On 6/28/2025 10:39 AM, dbush wrote:
On 6/28/2025 11:30 AM, olcott wrote:
On 6/28/2025 10:21 AM, dbush wrote:
On 6/28/2025 11:17 AM, olcott wrote:
On 6/28/2025 10:14 AM, dbush wrote:
On 6/28/2025 11:04 AM, olcott wrote:
On 6/28/2025 2:43 AM, wij wrote:
On Fri, 2025-06-27 at 14:36 -0500, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
The definition of HHH is missing.
The definition is specified in this part that you
dishonestly erased:
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When >>>>>>>>>>>> > HHH detects such a pattern it aborts its simulation >>>>>>>>>>>> > and returns 0.
The dishonest one here is YOU, as it was not wij who snipped >>>>>>>>>>> the below in his reply but YOU:
I stop at the first counter-factual mistake so I stop here. >>>>>>>>>> Everything else is ignored.
In other words, you INTENTIONALLY don't read things that prove >>>>>>>>> you wrong.
On 6/28/2025 2:43 AM, wij wrote:
The definition of HHH is missing.
On 6/27/2025 2:36 PM, olcott wrote:
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
It is a verified fact that the definition of HHH was
provided thus the claim that it was not provided is
counter factual.
No, you didn't give a definition of HHH, just gave a vauge
incomplete description of what you think it does.
It completely defines the generic notion of a simulating
termination analyzer and
It specifies every detail about HHH that is required
to correctly determine whether or not DDD correctly
simulated by HHH can possibly reach its own simulated
"return" statement final halt state.
But why would you say that when you've admitted on the record (see
below) that DDD is not in fact correctly simulated by HHH?
*I am not going to tolerate any misdirection to any other points*
void DDD()
{
HHH(DDD);
return;
}
The only point relevant to this forum is that DDD correctly
simulated by HHH
Is something that you have admitted on the record doesn't happen:
*This is the only point that I will address*
*Any attempt at changing the subject will*
*be construed as dishonest*
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH
False, as it doesn't meet the requirements to be one:
On 28/06/2025 17:47, olcott wrote:
On 6/28/2025 11:22 AM, Mike Terry wrote:
On 28/06/2025 16:00, olcott wrote:
On 6/27/2025 7:14 PM, Mike Terry wrote:
On 27/06/2025 20:36, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
Well, people here use the term "simulation" in a number of ways,
right?
*There is only one correct way*
That's patently rubbish.
This is one of your major cognitive mistakes - believing that the
squiggles which make up words have an absolute meaning independently
of the people using them to (hopefully) communicate.
void DDD()
{
HHH(DDD);
return;
}
The correct simulation of DDD by HHH means that HHH simulates
DDD and then emulates itself simulating DDD a number of times
according to the semantics of C.
"a number of times"???
Do you know what a number is?
In this case I am referring to every non-negative integer.
a) Does a correct simulation need to simulate until its target
computation terminates?
There is no target computation here we only have the
behavior that the input to HHH(DDD) specifies.
ok I see you intend to be completely non-responsive,
just as you were in
the "diverging simulations" thread.
I'll just point out for the record that you haven't provided any justification for accusing Alan or anybody else of lying to you.
Indeed, your responses just show that you have no clue what other people
are saying to you!
Mike.
On 27/06/2025 20:36, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
Well, people here use the term "simulation" in a number of ways, right? Like I've pointed out to you on several occasions. Maybe you should consider a simple misunderstanding over terminology before assuming bad intentions.
So what does Alan mean by "the simulation by HHH is incorrect",
exactly? (And why do you think it is incorrect?)
That is a question for PO, rather than Alan, since it is PO who claims
Alan is lying...
Mike.
On 27/06/2025 20:36, olcott wrote:
I am only here for the validation of the behavior
of DDD correctly simulated by HHH.
I have included proof that the people on comp.theory
lied about this at the bottom.
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
DDD();
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
On 6/27/2025 12:27 PM, Alan Mackenzie wrote:
In comp.theory olcott <polcott333@gmail.com
wrote:
I know that DDD .... simulated by HHH cannot
possibly reach its own simulated "return" statement
final halt state because the execution trace
conclusively proves this.
Everybody else knows this, too, and nobody has
said otherwise. *The conclusion is that the*
*simulation by HHH is incorrect*
*That last sentence is an intentional falsehood*
Well, people here use the term "simulation" in a number of ways, right? Like I've pointed out to you on several occasions. Maybe you should consider a simple misunderstanding over terminology before assuming bad intentions.
So what does Alan mean by "the simulation by HHH is incorrect",
exactly? (And why do you think it is incorrect?)
That is a question for PO, rather than Alan, since it is PO who claims
Alan is lying...
Mike.
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