Re: Simulating termination analyzers by dummies --- What does halting mean?

On 6/18/24 1:25 PM, olcott wrote:

On 6/18/2024 12:06 PM, joes wrote:
 void DDD()
{
   H0(DDD);
}
 DDD correctly simulated by any H0 cannot possibly halt.
 

DDD halts iff H0 halts.

 Halting is a technical term-of-the-art that corresponds
to terminates normally. Because Turing machines are
abstract mathematical objects there has been no notion
of abnormal termination for a Turing machine.

No "normally" as Turing Machine have no "abnormal terminatiom"
You just don't understand what they are.

 We can derive a notion of abnormal termination for Turing
machines from the standard terms-of-the-art.

How?

 Some TM's loop and thus never stop running, this is classical
non-halting behavior. UTM's simulate Turing machine descriptions.
This is the same thing as an interpreter interpreting the
source-code of a program.
 A UTM can be adapted so that it only simulates a fixed number
of iterations of an input that loops. When this UTM stops
simulating this Turing machine description we cannot correctly
say that this looping input halted.
 

And then are no longer UTMs, and YES, if a machine based on such am modifed UTM (so it is no long a UTM) when the UTM stops simulating, we can not say the input halted, nor can we say it didn't halt.
The not-a-UTM just came to a no-answer state.
The answer will be provided by useing an ACTUAL UTM that keeps on going, or the direct execution of the machine,
You are just stuck in your idea that Lies are sometimes ok.