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

Am Tue, 18 Jun 2024 21:51:56 -0500 schrieb olcott:

On 6/18/2024 9:41 PM, Richard Damon wrote:

On 6/18/24 10:30 PM, olcott wrote:

On 6/18/2024 9:16 PM, Richard Damon wrote:

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

On 6/18/2024 12:06 PM, joes wrote:

When such a UTM has been adapted to only simulate the first ten states
of its input TMD, then every simulated TMD with more than ten states
did not terminate normally.

Terminating is a property of the actual machine, and not a simulation
of it.

This.

Thus according to your faulty reasoning when the source-code of a C
program is simulated by interpreter this is mere nonsense gibberish
having nothing to do what the behavior that this source-code specifies.

YOUR partial decider makes everything halt, even that which doesn't.
So yes, it can't simulate infinite loops.

You could say the SIMULATION didn't terminate normally, but you can't
say the machine didn't or even the Turing Machine Description, as you
could give that exact same TMD to a real UTM and find out the actual
behaviof or the input.

Sure you can otherwise interpreters of source-code would be a bogus
concept.

When I write an infinite loop, I want it to be interpreted as an
infinite loop. Your H0 is bogus.

You just have lost track of the defintions of what is REALITY (the
actual behavior of the machine) and what is just imagination.

Not I but you.

The map is not the territory. The simulation is not the machine.

Every C programmer that knows what an x86 emulator is knows that when
H0 emulates the machine language of Infinite_Loop, Infinite_Recursion,
and DDD that it must abort these emulations so that itself can
terminate normally.

Which doesn't mean the program DDD needs to be abort to have it halt.

The verified that that it does need to be aborted contradicts your
nonsense to the contrary.

[The verification that it ... ?]
If H0 halts, so does DDD (which only calls it).

--
joes