Re: Can D simulated by H terminate normally?

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
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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

Yes it is, it is just aborting the simulation before it started.
After all, YOUR H stops simulating at a point based on no valid logic.
Now, if you want to try to define what your simulation actually means, and also have a tighter definition of your criteria, you could make a point.
Just make sure your H still meets it.
IF you what a more strict version, then H needs a tiny bit of actual code to look at the first state of program D, and if it is a final state return 1 else return 0.
So it becomes every program that doesn't start halted never halts.
It is still a toy problem.

 Your H is not simulating D at all thus not
"Every D simulated by H" quoted above
 Your H is not simulating D at all thus not
"Every D simulated by H" quoted above
 

is always correct, because THAT H can not possible simulate the input to the end before it aborts it, and that H is all that that H can be, or it isn't THAT H.
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Unless you clarify your altered definitions, H is what H is and that just becomes the conclusion.
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Then you can compare the definitions and try
to determine whether "abnormal termination" implies halting or non-halting
or neither. Note that "halting" is a freature of a Turing machine (a Turing
machine halts or does not halt) but "abnormal termination" seems to be
a feature of a particlar simulation (a simulation of a Truing machine
is or is not abnormally terminated).
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