Re: Can D simulated by H terminate normally? ---

On 5/3/24 8:17 AM, olcott wrote:

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

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.

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Which is just meaningless gobbledygook by your definitions.
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It means that
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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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"Interpreting" {D is simulated by H} as {D is NEVER simulated by H}
does not seem honest to me.
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But ALL steps simulated by H were correctly simulated, weren't they?
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 YOU DIDN'T MEET THIS SPEC AND YOU KNOW IT AND WEASEL WORDS WILL NOT HELP
 (a) It is a verified fact that D(D) simulated by H cannot
possibly reach past line 03 of D(D) simulated by H whether
H aborts its simulation or not.

Proven incorrect and not refuted so that is just a false statement.
Thus YOUR H doesn't meet your specification either.
Note, my comment was that IF we accept your definition of what is a correct answer, then by the same "logic" that you use, we can apply also to my H an make a claim with the same level of validity.

 

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Just shows you don't understand the essential nature of the problem, perhaps taken slightly too extreme.