Re: HHH maps its input to the behavior specified by it --- never reaches its halt state

On 8/8/2024 3:13 AM, Mikko wrote:

On 2024-08-08 04:41:11 +0000, olcott said:
 

On 8/7/2024 8:22 PM, Richard Damon wrote:

On 8/7/24 9:12 PM, olcott wrote:

On 8/7/2024 8:03 PM, Richard Damon wrote:

On 8/7/24 2:14 PM, olcott wrote:

On 8/7/2024 1:02 PM, joes wrote:

Am Wed, 07 Aug 2024 08:54:41 -0500 schrieb olcott:

On 8/7/2024 2:29 AM, Mikko wrote:

On 2024-08-05 13:49:44 +0000, olcott said:

>

I know what it means. But the inflected form "emulated" does not mean
what you apparently think it means. You seem to think that "DDD
emulated by HHH" means whatever HHH thinks DDD means but it does not.
DDD means what it means whether HHH emulates it or not.
>

In other words when DDD is defined to have a pathological relationship
to HHH we can just close our eyes and ignore it and pretend that it
doesn't exist?

It doesn't change anything about DDD. HHH was supposed to decide anything
and can't fulfill that promise. That doesn't mean that DDD is somehow
faulty, it's just a counterexample.
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
*HHH is required to report on the behavior of DDD*
Anyone that does not understand that HHH meets this criteria
has insufficient understanding.

>
But it doesn't, as a correct simulation of a DDD that calls an HHH that returns will stop running,

>
I really think that you must be a liar here because
you have known this for years:
>
On 8/2/2024 11:32 PM, Jeff Barnett wrote:
 > ...In some formulations, there are specific states
 >    defined as "halting states" and the machine only
 >    halts if either the start state is a halt state...
>
 > ...these and many other definitions all have
 >    equivalent computing prowess...
>
Anyone that knows C knows that DDD correctly simulated
by any HHH cannot possibly reach its "return" {halt state}.
>

>
But the problem is that you HHH ODESN'T correctly emulate the DDD it is given, because it aborts its emulation.
>

>

 Every one can see that Olcott is trying to get way
with ad-hominem instead of staying on topic.
 

void DDD()
{
   HHH(DDD);
   return;
}
Each HHH of every HHH that can possibly exist definitely
*emulates zero to infinity instructions correctly* In
none of these cases does the emulated DDD ever reach
its "return" instruction halt state.
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
There is no need to show any execution trace at the x86 level
every expert in the C language sees that the emulated DDD
cannot possibly reaches its "return" instruction halt state.
Every rebuttal that anyone can possibly make is necessarily
erroneous because the first paragraph is a tautology.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer