Re: DDD correctly emulated by HHH --- Correct Emulation Defined 2 addendum

Am Sat, 22 Mar 2025 08:43:03 -0500 schrieb olcott:

typedef void (*ptr)();
int HHH(ptr P);
int main()
{
   HHH(Infinite_Recursion);
}
There is no program DDD in the above code.

There is also no Infinite_Recursion.

Since no Turing machine M can ever compute the mapping from the behavior
of any directly executed TM2 referring to the behavior of the directly
executed DDD has always been incorrect. Halt Deciders always report on
the behavior that their input finite string specifies.

Please explain what behaviour the description of a TM "specifies",
and which TM the input describes.

In every case that does not involve pathological self-reference the
behavior that the finite string specifies is coincidentally the same
behavior as the direct execution of the corresponding machine. The
actual measure, however, has always been the behavior that the finite
string input specifies.

...which is the direct execution. Not much of a coincidence.

Prior to my work on simulating termination analyzers the behavior of the
counter-example input to the conventional halting problem proofs was
unknown. It was unknown because it was previously assumed the input DD
could actually do the opposite of whatever value that HHH returned.

It isn't unknown. It depends on HHH's return value, which is wrong either
way. And of course you can program a simple if() on that.

When we define the  termination analyzer's purpose is to
report on the behavior that its input specifies
(thus defining what a correct emulation is)

No, a simulator can't make up its own rules.

then we see that DD cannot possibly reach past its
own first instruction in any finite number of steps of correct
emulation.

It can, but it isn't simulated.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.