Re: Simulating termination analyzers by dummies --- criteria is met

On 6/24/2024 2:52 PM, joes wrote:

Am Mon, 24 Jun 2024 08:46:56 -0500 schrieb olcott:

On 6/24/2024 2:22 AM, Mikko wrote:

On 2024-06-23 13:13:42 +0000, olcott said:

On 6/23/2024 2:57 AM, Mikko wrote:

On 2024-06-22 14:11:28 +0000, olcott said:

On 6/22/2024 8:27 AM, Richard Damon wrote:

On 6/22/24 9:04 AM, olcott wrote:

 

In particular, you can't. You have insisted that your "decider" or
"anlyzer" (or whatever word you happen to use) H or HH (or hwatever
name you happen to use) must return false because a non-input (where
instead of the actually called function another function that does
not halt is called) does not halt.

 

Which is all we need to know about H in ordet to determine that it is
not a decider.
>

void DDD()
{
    H0(DDD);
}
The call from DDD to H0(DDD) when DDD is correctly emulated by H0 cannot
possibly return.

Why not? H0 is a decider AND simulator, so it can simulate itself
terminating.
 

That is a stupid thing to say.
When H0 is called in recursive emulation the semantics of the x86
language do not allow H0 to simply ignore this and still terminate.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer