Sujet : Re: Who knows that DDD correctly simulated by HHH cannot possibly reach its own return instruction final state? BUT ONLY that DDD
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 04. Aug 2024, 00:36:10
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8meta$3ma4t$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
User-Agent : Mozilla Thunderbird
On 8/3/2024 5:51 PM, Richard Damon wrote:
On 8/3/24 6:15 PM, olcott wrote:
On 8/3/2024 5:07 PM, Richard Damon wrote:
>
The problem is that every one of those emulation is of a *DIFFERENT* input, so they don't prove anything together except that each one didn't go far enough.
>
void DDD()
{
HHH(DDD);
return;
}
>
When each HHH correctly emulates 0 to infinity steps of
its corresponding DDD and none of them reach the "return"
halt state of DDD then even the one that emulated infinite
steps of DDD did not emulate enough steps?
>
>
Just says lying YOU.
You got any source for that other than yourself?
It is self-evident and you know it. I do have four
people (two with masters in CS) that attest to that.
*It is as simple as I can possibly make it*
I wonder how you think that you are not swearing your
allegiance to that father of lies?
Anyone that truly understands infinite recursion knows
that DDD correctly simulated by HHH cannot possibly reach
its own "return" final state.
Surpisingly (to me) Jeff Barnett set the record straight
on exactly what halting means.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer