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

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
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. ---
 Unless you clarify your altered definitions, H is what H is and that just becomes the conclusion.
 

>

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).
>

>

 

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