Re: Hypothetical possibilities -- I reread this again more carefully

On 7/21/2024 6:37 AM, Richard Damon wrote:

On 7/21/24 12:15 AM, olcott wrote:

On 7/20/2024 10:51 PM, Richard Damon wrote:

On 7/20/24 11:14 PM, olcott wrote:

On 7/20/2024 8:46 PM, Richard Damon wrote:

On 7/20/24 9:23 PM, olcott wrote:

On 7/20/2024 8:01 PM, Richard Damon wrote:

On 7/20/24 8:21 PM, olcott wrote:

On 7/20/2024 7:05 PM, Richard Damon wrote:

On 7/20/24 7:06 PM, olcott wrote:

On 7/20/2024 6:00 PM, Richard Damon wrote:

On 7/20/24 6:47 PM, olcott wrote:

On 7/20/2024 5:11 PM, Richard Damon wrote:

On 7/20/24 5:21 PM, olcott wrote:

On 7/20/2024 4:06 PM, joes wrote:

Am Sat, 20 Jul 2024 15:05:53 -0500 schrieb olcott:

On 7/20/2024 2:50 PM, Richard Damon wrote:

On 7/20/24 3:09 PM, olcott wrote:

On 7/20/2024 2:00 PM, Fred. Zwarts wrote:

Op 20.jul.2024 om 17:28 schreef olcott:

>

(a) Termination Analyzers / Partial Halt Deciders must halt this is
a design requirement.
(b) Every simulating termination analyzer HHH either aborts the
simulation of its input or not.
(c) Within the hypothetical case where HHH does not abort the
simulation of its input {HHH, emulated DDD and executed DDD}
never stop running.
This violates the design requirement of (a) therefore HHH must abort
the simulation of its input.

You missed a couple details:
A terminating input shouldn't be aborted, or at least not classified
as not terminating. Terminating inputs needn't be aborted; they and the
simulator halt on their own.
>

And when it aborts, the simulation is incorrect. When HHH aborts and
halts, it is not needed to abort its simulation, because it will halt
of its own.

So you are trying to get away with saying that no HHH ever needs to
abort the simulation of its input and HHH will stop running?

Pretty much.

It is the fact that HHH DOES abort its simulation that makes it not
need to.

No stupid it is not a fact that every HHH that can possibly exist aborts
its simulation.

I thought they all halt after a finite number of steps?
>

>
void DDD()
{
    HHH(DDD);
    return;
}
>
DDD correctly simulated by pure function HHH cannot
possibly reach its own return instruction.
>

>
Wrong.
>

>
You know that you are lying about this as you admit below:

>
Nope, YOU just don't what the words mean, and reckless disregard the teaching you have been getting, which makes your errors not just honest mistakes but reckless pathological lies.
>

>

It may be that the simulation by HHH never reaches that point,

>

but if HHH aborts its simuliaton and returns (as required for it to be a decider) then the behavior of DDD

>
Simulated by HHH is to Die, stop running, no longer function.

>
Nope, HHH is NOT the "Machine" that determines what the code does, so can not "Kill" it.
>

>
So you are trying to get away with the lie
that an aborted simulation keeps on running.
>

>
No, but the BEHAVIOR of the program does, and that is what matters.

>
So you agree that DDD correctly simulated by any pure function
HHH cannot possibly reach its own return instruction?
>
>

>
No, I will let you claim (without proof, so we can argue tha later) that the simulation by HHH of DDD does not reach the return, but the behavior of the DDD simuliated by HHH continues,

>
We are talking about real hardware here not figments
of your imagination.
>

>
No, you are not. The "Hardware" would be the actual CPU chip which never stops the program when it is running. A Simulator is just a piece of software running on it, and what it does can't affect the behavior of the actual CPU running the program.
>
>

When an actual x86 emulator stops emulating its input
this emulated input immediately stops running.
>

>
Nope, that is you stupidity where you confuse the observation for the facts.
>
It has been told to you MANY times, but it seems that you just can not understand it.
>
The SIMULATION is an observation of the program, that if it stops doesn't affect the actual behavior of the program in question.
>

>
*If the simulator stops simulating then the simulated stops running*

>
No, the SIMULA*TION* stops running, the SIMULATED (which is the actual program) behaviof continues.
>

>
Yes you are clueless.
>
void DDD()
{
   HHH(DDD);
   return;
}
>
When the simulation stops running the whole program
exits to the operating system.

  Which just shows you don't understand the MEANING of the behavior of a program.
 

void DDD()
{
   HHH(DDD);
   return;
}
int main()
{
   HHH(DDD);
}
Because
(a) We know that it is a logical impossibility for any decider HHH
to report on the halt status of any input that does the opposite
of whatever it reports.
(b) We know that a decider is not allowed to report on the behavior
computation that itself is contained within. Deciders only take finite
string inputs. They do not take executing processes as inputs. Thus HHH
is not allowed to report on the behavior of this int main() { DDD(); }.
Therefore we map the finite string input to HHH(DDD) to the behavior
that it species on the basis of DDD correctly emulated by any pure
function HHH that can possibly exist.
This is the behavior of the finite string input to HHH(DDD) correctly
emulated by any pure function HHH according to the semantics of its
machine language instructions. This includes the recursive simulation
specified by the call from DDD to HHH(DDD).
 From this we know that DDD correctly emulated by any pure function HHH
cannot possibly reach its own return instruction and halt.
You keep trying to get away with the strawman deception
of changing the subject to (b) and (b) is not allowed.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer