Re: How Partial Simulations correctly determine non-halting ---Ben's 10/2022 analysis

Am Sat, 08 Jun 2024 08:04:14 -0500 schrieb olcott:

On 6/8/2024 1:42 AM, Mikko wrote:

On 2024-06-07 22:26:05 +0000, olcott said:

On 6/7/2024 4:00 PM, joes wrote:

Am Fri, 07 Jun 2024 09:47:35 -0500 schrieb olcott:

On 6/7/2024 1:30 AM, Mikko wrote:

On 2024-06-06 15:31:36 +0000, olcott said:

On 6/6/2024 10:14 AM, Mikko wrote:

On 2024-06-06 13:53:58 +0000, olcott said:

On 6/6/2024 5:15 AM, Mikko wrote:

On 2024-06-05 13:29:28 +0000, olcott said:

On 6/5/2024 2:37 AM, Mikko wrote:

On 2024-06-04 18:02:03 +0000, olcott said:

Any finite string can be an input to some Turing machine.
Can you prove that a Turing machine is not a finite string?

By definition Turing Machines are not finite strings in the conventional
model. In my x86utm model of computation x86 machine language <is> the
input to another function written in the x86 language.

In your model, the machine code is also finite.

Your own attempts of a conter-proof are not about Turing machines but C
programs. C programs are finite strings, so a C program is a valid
input to a C program (and a Turing machine, too).

They are about Turing Machines yet cannot be sufficiently understood
with less than the 100% compete precision of the x86 language. They
x86utm model is required to prove that false assumptions about the
nature of correct simulation are false assumptions.

You can't hide behind an x86 implementation. The same arguments hold.
Which assumptions are false?

I have all of the details of the machine code and C code for HH and DDD.
I can't to the same thing for embedded_H and ⟨Ĥ⟩ ⟨Ĥ⟩ so we have to learn
by analogy.

Why can't you do that? The simulator can simulate itself.

This means that H is not allowed to report on the behavior of the
directly executed P(P).

So on which program does it report then?

DD(DD) is just like infinite recursion that gets terminated at its
second recursive call. DD(DD) halts only because HH(DD,DD)
correctly determines that its input DOES NOT HALT.
If HH(DD,DD) did not correctly determine that its input DOES NOT HALT
then DD(DD) would never halt.

That doesn't make sense. The function halts because a simulator says it
doesn't?

--
joes