Re: Can D simulated by H terminate normally?

On 5/2/2024 8:48 PM, Richard Damon wrote:

On 5/2/24 2:46 PM, olcott wrote:

On 5/2/2024 6:03 AM, Richard Damon wrote:

On 5/2/24 12:21 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.

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Which is just meaningless gobbledygook by your definitions.
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It means that
>
int H(ptr m, ptr d) {
    return 0;
}
>

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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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D is never simulated by H is not the same as D is simulated by H.
D is simulated by H entails that 1 to N steps are simulated and
then the simulation is aborted or the simulation is never aborted.
>
>

 Then simulate EXACTLY 1 step and abort.

The infinite set of every possible implementation of H includes
simulating 1 to ∞ steps of D(D) every case where less than ∞ steps
are simulated H aborts its simulation of D(D).
Competent software engineers that know the C programming language
know that D(D) never reaches past its own line 03 for every H/D pair.

 It makes all but the most trivial programs "non-halting."
 Again, a TOY.,

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer