Re: 197 page execution trace of DDD correctly simulated by HHH

Am Fri, 28 Jun 2024 10:25:36 -0500 schrieb olcott:

On 6/28/2024 8:14 AM, joes wrote:

Am Thu, 27 Jun 2024 12:30:38 -0500 schrieb olcott:

When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as
non-halting by returning 0 to its caller.

To the caller DDD, which then returns to its own caller H0, which
returns „halting” to main… hold on.

Where do you disagree?

Simulating termination analyzers must report on the behavior that
their finite string input specifies thus H0 must report that DDD
correctly emulated by H0 remains stuck in recursive simulation.

H0 must not report on itself, only on DDD. Which you’ve proven halts.
We don’t care how H0 deviates (i.e. is incorrect) in its simulation.
That would be main {H0(H0(DDD))}.

The behavior of the directly executed DDD() is irrelevant because that
is not the behavior of the input.

What is the difference here?

Deciders compute the mapping from
their actual finite string input to an output by a sequence of finite
string transformations.

And should get the right answer.

In this case the sequence is the line-by-line execution trace of the
behavior of DDD correctly emulated by HHH.

No, the sequence is the behaviour of DDD, period.

The behavior of this input must include and cannot ignore the recursive
emulation specified by the fact that DDD is calling its own emulator.

Yes, and the behaviour of H0 is that it produces the exact same behaviour
as DDD.