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

On 7/20/24 11:31 PM, olcott wrote:

On 7/20/2024 10:14 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*
>
void DDD()
{
   HHH(DDD);
   return;
}
>
DDD *correctly simulated* by pure function HHH cannot
possibly reach its own return instruction.
>
>

 Only DDD correctly emulated by HHH maps the finite string
of the x86 machine code of DDD to the behavior that it
actually specifies.
 

No, a correct emulation doesn't stop until it reaches a final state, and an HHH that doesn't stop doesn't map the input to anything.
It is a FULLY CORRECT emulation of the input that shows that CORRECT mapping that HHH was supposed to compute, but doesn't as you LIE about it being correct.
First, the emulation needs ALL the code being emulated, which includes the code of HHH, so by your definition, you can't do it so are a lair.
When we fix that by including that code, we find that if HHH(DDD) aborts its  PARTIAL simulation to return an answer (and it must or it isn't a decider) then when we give that same input (which still calls that HHH that aborted and returned) to a CORRECT emulator, and see what happens, we see that DDD calls HHH(DDD) that emulates the string given to it for awhile, and then it aborts its simulation and returns to DDD which returns,
So, all you have done is admitted that you are just a lying idiot that doesn't know about what he is talking and haas wasted his life beliving his own lies.