Re: D simulated by H never halts no matter what H does V3

On 4/27/24 10:32 AM, olcott wrote:

On 4/27/2024 6:45 AM, Richard Damon wrote:

On 4/27/24 1:34 AM, olcott wrote:

On 4/26/2024 10:45 PM, Richard Damon wrote:

On 4/26/24 11:02 PM, olcott wrote:

On 4/26/2024 9:18 PM, Richard Damon wrote:

On 4/26/24 9:43 PM, olcott wrote:

On 4/26/2024 7:26 PM, Richard Damon wrote:

On 4/26/24 8:02 PM, olcott wrote:

On 4/26/2024 12:05 PM, olcott wrote:

On 4/26/2024 11:19 AM, Richard Damon wrote:

On 4/26/24 11:34 AM, olcott wrote:

On 4/26/2024 3:32 AM, Mikko wrote:

On 2024-04-25 14:15:20 +0000, olcott said:

01 int D(ptr x)  // ptr is pointer to int function
02 {
03   int Halt_Status = H(x, x);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 void main()
10 {
11   D(D);
12 }
>
That H(D,D) must report on the behavior of its caller is the
one that is incorrect.

>
What H(D,D) must report is independet of what procedure (if any)
calls it.
>

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Thus when H(D,D) correctly reports that its input D(D) cannot possibly
reach its own line 6 and halt no matter what H does then H can abort its
input and report that its input D(D) does not halt.

>
But since the program D(D) DOES reach its own line 6 when run, because H aborts its simulation and return 0 (since that is what you say this H will do), your statement is PROVEN TO BE A LIE, and you "logic" just a collection of contradictions.
>

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D simulated by H cannot possibly reach its own line 06 thus when we do
not use the strawman deception to refer to a different D then we know
that D simulated by H never halts.
>

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The fact that the D(D) executed in main does halt is none of H's
business because H is not allowed to report on the behavior of its
caller.
>

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In other words, H doesn't need to report on the Behavior of the Program described by its input because it isn't actually a Halt Decider, because you are just a LIAR.
>
>

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Anyone knowing the theory of computation knows that H is not allowed to
report on the behavior of its caller.
>
In computability theory and computational complexity theory, an
undecidable problem is a decision problem for which it is proved to be
impossible to construct an algorithm that always leads to a correct yes-
or-no answer. https://en.wikipedia.org/wiki/Undecidable_problem
>
The behavior of the simulated D(D) before H aborts its simulation is
different than the behavior of the executed D(D) after H has aborted
its simulation.
>
Every time that a simulated input would never stop running unless
aborted the simulating termination analyzer must abort this simulation
to prevent its own infinite execution.
>
H(D,D) is a case of this H1(D,D) is not a case of this even though
the only difference between H and H1 is that D calls H and D does
not call H1.
>
D simulated by H would never stop running unless aborted and cannot
possibly reach its own line 06 and halt no matter what H does.
>
Thus whenever we do not use the strawman deception to refer to a
different D we know that D simulated by H specifies a non-halting
sequence of configurations to H.
>

>
*This might be a more succinct way of summing that up*
When you understand that D simulated by H cannot possibly reach past its own line 03 (thus cannot possibly halt) no matter what H does and

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But since H does whatever H does, if H aborts and returns 0, the the direct execution of D, which is what actually matters, DOES get to that point.
>

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That is another much less useful way to make a universally correct
termination analyzer:
>
int H(ptr x, ptr y)
{
   printf("The input program does whatever it does!\n");
   return 777; // code for it does what it does
}

>
But that doesn't make H answer the question.
>
I guess you don't understand what I am saying.
>
You said "no matter what H does", but that is a MEANINGLESS statement, because H will do what H is programmed to do, so we don't need to look at other behavior, but just the behavior that H ac
>

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It can be verified through ordinary software engineering that D(D)
simulated H cannot possibly reach past its own line 03.

>
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Yes, but that is just a lying RED HERRING, as the question isn't about what H's simulation of the input does, but what the program the input actually represents does when run.
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YOu are just effectively admitting that you are nothing but a stupid liar that doesn't know what he is talking about.
>

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It can be verified through computer science that this means that D(D) simulated H by never reaches its own final state and halts whether
H aborts its simulation or not.

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Which, since this H DOES abort its simulation is trying to introduce a red herring.
>

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D(D) simulated by H never halts
D(D) simulated by H never halts
D(D) simulated by H never halts
D(D) simulated by H never halts
D(D) simulated by H never halts

>
STRAWMAN

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A method that I used very effectively on another forum and made much
progress with is utterly insisting that we cannot move on to any
additional point until we have 100% complete mutual agreement on
the current point.
>
(1) If we (possibly falsely assume) that H(D,D) is supposed to report
on the behavior of D(D) simulated by H and
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(2) we understand that D(D) simulated by H cannot possibly reach past
its own line 3 thus cannot reach its own final state at line 6 and halt
>
then (within this possibly false assumption) (1) and the understanding of (2) it <is> correct for H to abort its simulation and report halting.
>

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Let go to actual DEFINITIONS shall we.
>
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What do you define the Halting Problem to be?
>

 THIS IS A STIPULATED DEFINITION (like a given" in Geometry)
Simulating termination analyzer H determines whether or not D(D)
simulated by H can possibly reach its final state at its own
line 06 and halt whether or not H aborts its simulation.
 

In other words, you are ADMITTING, by stipulating a definition that differs from the standard definition, that you are NOT working on the actual halting Problem.

A stipulative definition is a type of definition in which a new or
currently existing term is given a new specific meaning for the purposes
of argument or discussion in a given context. When the term already
exists, this definition may, but does not necessarily, contradict the
dictionary (lexical) definition of the term.
https://en.wikipedia.org/wiki/Stipulative_definition
 

Right, so, for EVERYTHING you have said, you are admitting that it is based on a STIPULATED definition that differs from the definition of the actual theory, so NOTHING you have said has ANY bearing to the actual problem, and all claim that they do are just LIES.
As you have been told many times, if you want to try to define an alternate problem, like your POOP, then you are free to do so, but you must realize that it will say NOTHING about the original problem.
If you can show that your alternate problem does something useful, then maybe you have something, but since all you even attempt to do is to say that one proof of the undeciability of the original problem doesn't work, that isn't going to get any people excited.
*IF* you could show that there exists some broad class of interesting problems that can be POOP decided, but not Halt Decide, you might have something, but since you never actually establish any such class, only that this ONE program (that other halt deciders can get right) can be decided by your POOP decider "correctly".
The biggest point is that since you don't actually refute the uncomputability of the actual standardly define Halting Function, your POOP does NOTHING about all the other proofs that are based on it.