Sujet : Re: Can D simulated by H terminate normally? ---
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 03. May 2024, 03:48:45
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v11fpt$2tlr1$1@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
User-Agent : Mozilla Thunderbird
On 5/2/24 2:48 PM, olcott wrote:
On 5/2/2024 6:04 AM, Richard Damon wrote:
On 5/2/24 12:24 AM, olcott wrote:
On 5/1/2024 7:28 PM, Richard Damon wrote:
On 5/1/24 11:51 AM, olcott wrote:
Every D simulated by H that cannot possibly stop running unless
aborted by H does specify non-terminating behavior to H. When
H aborts this simulation that does not count as D halting.
>
Which is just meaningless gobbledygook by your definitions.
>
It means that
>
int H(ptr m, ptr d) {
return 0;
}
>
>
Your H is not simulating D at all thus not
"Every D simulated by H" quoted above
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Yes it is, it is just aborting the simulation before it started.
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"Interpreting" {D is simulated by H} as {D is NEVER simulated by H}
does not seem honest to me.
But ALL steps simulated by H were correctly simulated, weren't they?
Just shows you don't understand the essential nature of the problem, perhaps taken slightly too extreme.
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