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

Am Sun, 21 Jul 2024 08:08:53 -0500 schrieb olcott:

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:

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?

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.

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

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,

A simulated program doesn't actually run - it is only simulated.

The SIMULATION is an observation of the program, that if it stops
doesn't affect the actual behavior of the program in question.
>

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

When the simulation stops running the whole program exits to the
operating system.

Yes. That doesn't mean that DDD itself would terminate.

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(); }.

That IS exactly the input.

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.

The basis is the direct behaviour.

--
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.