Am Fri, 14 Mar 2025 11:33:02 -0500 schrieb olcott:

On 3/14/2025 11:01 AM, wij wrote:

On Fri, 2025-03-14 at 10:51 -0500, olcott wrote:

On 3/14/2025 10:04 AM, wij wrote:

On Fri, 2025-03-14 at 09:35 -0500, olcott wrote:

https://en.wikipedia.org/wiki/Halting_problem In computability
theory, the halting problem is the problem of determining, from a
description of an arbitrary computer program and an input, whether
the program will finish running, or continue to run forever.
That means: H(D)=1 if D() halts and H(D)=0 if D() does not halt.
But, it seems you don't understand English, as least as my level,
....

The only difference between HHH and HHH1 is that they are at different
locations in memory. DDD simulated by HHH1 has identical behavior to
DDD() directly executed in main().
The semantics of the finite string input DDD to HHH specifies that it
will continue to call HHH(DDD) in recursive simulation.
The semantics of the finite string input DDD to HHH1 specifies to
simulate to DDD exactly once.
When HHH(DDD) reports on the behavior that its input finite string
specifies it can only correctly report non-halting.
When HHH(DDD) is required to report on behavior other than the
behavior that its finite string specifies HHH is not a decider thus
not a halt decider.
All deciders are required to compute the mapping from their input
finite string to the semantic or syntactic property that this string
specifies. Deciders return true when this string specifies this
property otherwise they return false.

Are you solving The Halting Problem or not? Yes or No.

I have only correctly refuted the conventional halting problem proof.

Haha no.

Actually solving the halting problem requires a program that is ALL
KNOWING thus God like.

Serious question: are there actually other undecidable programs?

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

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On 3/14/2025 12:01 PM, dbush wrote:
<Snipped mostly garbage>

Good luck trying to show you've refuted a proof that you've gone on
record as admitting is correct.

Just because PO can neither get his claims straight or his head out of
his ass, is no reason to copy his style of argument:

Unless I missed it, he has (probably by mistake) admitted that the
theorem with original definitions of terms is probably correct. That,
however, doesn't mean some particular proof of that theorem is
necessarily a valid proving.

The history of these nonsense threads show that POs original claim was
that a particular proof was bungled. There followed a few thousand
rebuttals where people repeatedly point out 1) he didn't know what a
proof was, 2) he didn't understand what was to be proved, and 3) he
didn't even understand the basic terminology of the field. His brain was
not capable of grokking the issues.

Now his personality disorders begin to add to his limited comprehension problems. In particular his radical overestimate of his own intellect
and his needed to be worshiped. Rather than trying to learn more and
perhaps making a major, minor, or at least a fun contribution with hard
work, he embarks with a new two-part strategy: 1) claim the the
definitions of terms are wrong and 2) act out the role of a spoiled brat
and throw tantrums reminiscent of a disturbed 2 year old.

And, by the way, even if you can show (not that hard) that he is
completely fucked between the ears, it will not change a thing in his
limited mind.

He has made claims of being on a collision course with death due to
cancer. It may be true or it may be subtle crap that enhances his troll attraction, i.e., everybody wants to help him out so he doesn't die with trophies for Best Troll 1st Quarter 21st Century and winner DWIP (Dumb,
Whiny, Ignorant, Putz) All Time Usenet.
--
Jeff Barnett

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Am Fri, 14 Mar 2025 18:28:24 -0500 schrieb olcott:

On 3/14/2025 1:01 PM, dbush wrote:

On 3/14/2025 1:19 PM, olcott wrote:

On 3/14/2025 11:58 AM, wij wrote:

On Fri, 2025-03-14 at 11:33 -0500, olcott wrote:

On 3/14/2025 11:01 AM, wij wrote:

On Fri, 2025-03-14 at 10:51 -0500, olcott wrote:

On 3/14/2025 10:04 AM, wij wrote:

On Fri, 2025-03-14 at 09:35 -0500, olcott wrote:

Are you solving The Halting Problem or not? Yes or No.

I have only correctly refuted the conventional halting problem
proof. Actually solving the halting problem requires a program that
is ALL KNOWING thus God like.

I (GUR) had told you God cannot solve HP neither (maybe because the
problem is limited in a box)

When we define the HP as having H return a value corresponding to the
halting behavior of input D and input D can actually does the opposite
of whatever value that H returns, then we have boxed ourselves in to a
problem having no solution.

And that is EXACTLY what the halting problem is about: it is not
possible to construct an H where H(X,Y) reports whether X(Y) halts when
executed directly.
A problem that you have now EXPLICITLY agreed is unsolvable.  So...

Likewise:
What time is it (yes or no)? equally has no solution.
That BOGUS problem instances have no solution places no actual limit on anything.

It places a limit on deciding termination.

I coined the term "incorrect question" years ago. https://groups.google.com/g/sci.lang/c/lSdYexJ0ozo/m/aDN9-TYLHwIJ

(x) doubt
--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

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On 3/14/25 11:40 PM, olcott wrote:

On 3/14/2025 8:56 PM, dbush wrote:

On 3/14/2025 9:49 PM, olcott wrote:

On 3/14/2025 8:34 PM, dbush wrote:

On 3/14/2025 9:27 PM, olcott wrote:

On 3/14/2025 8:00 PM, dbush wrote:

On 3/14/2025 8:45 PM, olcott wrote:

On 3/14/2025 12:54 PM, dbush wrote:

On 3/14/2025 12:33 PM, olcott wrote:

On 3/14/2025 11:01 AM, wij wrote:

On Fri, 2025-03-14 at 10:51 -0500, olcott wrote:

On 3/14/2025 10:04 AM, wij wrote:

On Fri, 2025-03-14 at 09:35 -0500, olcott wrote:>>

void DDD()
{
     HHH(DDD);
     return;
}

DDD correctly simulated by HHH cannot possibly reach >>>>>>>>>>>>> its own "return" instruction in any finite number of >>>>>>>>>>>>> correctly simulated steps.

That you are clueless about the semantics of something >>>>>>>>>>>>> as simple as a tiny C function proves that you are not >>>>>>>>>>>>> competent to review my work.

https://en.wikipedia.org/wiki/Halting_problem
In computability theory, the halting problem is the problem >>>>>>>>>>>> of determining, from a description of
an
arbitrary computer program and an input, whether the program >>>>>>>>>>>> will finish running, or continue to
run
forever.

That means: H(D)=1 if D() halts and H(D)=0 if D() does not >>>>>>>>>>>> halt.

But, it seems you don't understand English, as least as my >>>>>>>>>>>> level, ....

void DDD()
{
    HHH(DDD);
    return;
}

The only difference between HHH and HHH1 is that they are >>>>>>>>>>> at different locations in memory. DDD simulated by HHH1
has identical behavior to DDD() directly executed in main(). >>>>>>>>>>>
The semantics of the finite string input DDD to HHH specifies >>>>>>>>>>> that it will continue to call HHH(DDD) in recursive simulation. >>>>>>>>>>>
The semantics of the finite string input DDD to HHH1 specifies >>>>>>>>>>> to simulate to DDD exactly once.

When HHH(DDD) reports on the behavior that its input finite >>>>>>>>>>> string specifies it can only correctly report non-halting. >>>>>>>>>>>
When HHH(DDD) is required to report on behavior other than >>>>>>>>>>> the behavior that its finite string specifies HHH is not >>>>>>>>>>> a decider thus not a halt decider.

All deciders are required to compute the mapping from
their input finite string to the semantic or syntactic property >>>>>>>>>>> that this string specifies. Deciders return true when this >>>>>>>>>>> string specifies this property otherwise they return false. >>>>>>>>>>>

Are you solving The Halting Problem or not? Yes or No.

I have only correctly refuted the conventional halting
problem proof.

And what exactly do you think this proof is proving?  More
specifically, what do you think the Linz proof is proving?

All of the proofs merely show that there cannot
possibly exist any halt decider that returns a
value corresponding to the behavior of any input
that is actually able to do the opposite of whatever
value is returned.

Not exactly.  What they prove is that no H exists that satisfies
these requirements:


Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:

A solution to the halting problem is an algorithm H that computes
the following mapping:

(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
directly

The executed directly part is bogus as I have
shown and your indoctrination blindly ignores.

But I want to know if any arbitrary X with input Y halts when
executed directly,

Even when some inputs are BOGUS.

Did I stutter?

I want to know if any arbitrary X with input Y halts when executed

If you reject "ls;dlfm skdofdfn 894&49.8244bewr" as a syntactically
incorrect input then you are being inconsistent when you fail to reject semantically incorrect inputs.

And when you reject a valid input, that was PROVEN to be valid from the
logic system.

All you have shown is that you don't know what you are talking about and
your statements are just based on LIES and thus not semantically correct.

directly.  If I had an H that could tell me that in *all* possible
cases, I could solve the Goldbach conjecture, among many other
unsolved problems.

Does an H exist that can tell me that or not?

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On 3/15/25 4:48 PM, olcott wrote:

On 3/15/2025 8:06 AM, dbush wrote:

On 3/14/2025 11:40 PM, olcott wrote:

On 3/14/2025 8:56 PM, dbush wrote:

On 3/14/2025 9:49 PM, olcott wrote:

On 3/14/2025 8:34 PM, dbush wrote:

On 3/14/2025 9:27 PM, olcott wrote:

On 3/14/2025 8:00 PM, dbush wrote:

On 3/14/2025 8:45 PM, olcott wrote:

On 3/14/2025 12:54 PM, dbush wrote:

On 3/14/2025 12:33 PM, olcott wrote:

On 3/14/2025 11:01 AM, wij wrote:

On Fri, 2025-03-14 at 10:51 -0500, olcott wrote:

On 3/14/2025 10:04 AM, wij wrote:

On Fri, 2025-03-14 at 09:35 -0500, olcott wrote:>> >>>>>>>>>>>>>>> void DDD()

{
     HHH(DDD);
     return;
}

DDD correctly simulated by HHH cannot possibly reach >>>>>>>>>>>>>>> its own "return" instruction in any finite number of >>>>>>>>>>>>>>> correctly simulated steps.

That you are clueless about the semantics of something >>>>>>>>>>>>>>> as simple as a tiny C function proves that you are not >>>>>>>>>>>>>>> competent to review my work.

https://en.wikipedia.org/wiki/Halting_problem
In computability theory, the halting problem is the >>>>>>>>>>>>>> problem of determining, from a description of
an
arbitrary computer program and an input, whether the >>>>>>>>>>>>>> program will finish running, or continue to
run
forever.

That means: H(D)=1 if D() halts and H(D)=0 if D() does not >>>>>>>>>>>>>> halt.

But, it seems you don't understand English, as least as my >>>>>>>>>>>>>> level, ....

void DDD()
{
    HHH(DDD);
    return;
}

The only difference between HHH and HHH1 is that they are >>>>>>>>>>>>> at different locations in memory. DDD simulated by HHH1 >>>>>>>>>>>>> has identical behavior to DDD() directly executed in main(). >>>>>>>>>>>>>
The semantics of the finite string input DDD to HHH specifies >>>>>>>>>>>>> that it will continue to call HHH(DDD) in recursive
simulation.

The semantics of the finite string input DDD to HHH1 specifies >>>>>>>>>>>>> to simulate to DDD exactly once.

When HHH(DDD) reports on the behavior that its input finite >>>>>>>>>>>>> string specifies it can only correctly report non-halting. >>>>>>>>>>>>>
When HHH(DDD) is required to report on behavior other than >>>>>>>>>>>>> the behavior that its finite string specifies HHH is not >>>>>>>>>>>>> a decider thus not a halt decider.

All deciders are required to compute the mapping from >>>>>>>>>>>>> their input finite string to the semantic or syntactic >>>>>>>>>>>>> property
that this string specifies. Deciders return true when this >>>>>>>>>>>>> string specifies this property otherwise they return false. >>>>>>>>>>>>>

Are you solving The Halting Problem or not? Yes or No. >>>>>>>>>>>>

I have only correctly refuted the conventional halting
problem proof.

And what exactly do you think this proof is proving?  More >>>>>>>>>> specifically, what do you think the Linz proof is proving?

All of the proofs merely show that there cannot
possibly exist any halt decider that returns a
value corresponding to the behavior of any input
that is actually able to do the opposite of whatever
value is returned.

Not exactly.  What they prove is that no H exists that satisfies >>>>>>>> these requirements:


Given any algorithm (i.e. a fixed immutable sequence of
instructions) X described as <X> with input Y:

A solution to the halting problem is an algorithm H that
computes the following mapping:

(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly >>>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when
executed directly

The executed directly part is bogus as I have
shown and your indoctrination blindly ignores.

But I want to know if any arbitrary X with input Y halts when
executed directly,

Even when some inputs are BOGUS.

Did I stutter?

I want to know if any arbitrary X with input Y halts when executed

If you reject "ls;dlfm skdofdfn 894&49.8244bewr" as a syntactically
incorrect input then you are being inconsistent when you fail to reject
semantically incorrect inputs.

No such thing.  All algorithms X are valid as are all inputs Y to
those algorithms.

That is false.
The algorithm that determines whether this sentence is
true or false does not exist because it is neither true
nor false: "This sentence is not true"

Until you understand that simplified example you will
continue to lack the capacity to understand the more
complex example.

Which shows part of your problem, the truth predicate doesn't determine
if a statement is true or false, but if it is true or untrue, which can
be false, or nonsense.

The problem is that we can show (in a sufficently powerful system) that
we can form a sentence that shows that the result of the truth predicate
must be nonsense, and thus it fails to be a needed predicate.

Of course, that seems to be just beyond your ability to understand, as
it requires an understanding of real mathematics, not just arithmetic.

All you are doing is proving your stupidity, and that your stupidity
extends to the point that makes you unable to see your stupidity.

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