Re: DD correctly emulated by HHH --- Totally ignoring invalid rebuttals ---PSR---

On 3/8/2025 9:19 AM, olcott wrote:

On 3/8/2025 2:54 AM, Fred. Zwarts wrote:

Op 08.mrt.2025 om 02:09 schreef olcott:

On 3/7/2025 9:33 AM, dbush wrote:

On 3/7/2025 10:30 AM, olcott wrote:

On 3/7/2025 9:15 AM, Fred. Zwarts wrote:

Op 07.mrt.2025 om 15:49 schreef olcott:

On 3/7/2025 2:02 AM, joes wrote:

Am Thu, 06 Mar 2025 20:59:49 -0600 schrieb olcott:

On 3/6/2025 6:37 PM, Richard Damon wrote:

On 3/6/25 3:16 PM, olcott wrote:

On 3/6/2025 3:17 AM, Fred. Zwarts wrote:

Op 06.mrt.2025 om 05:46 schreef olcott:

On 3/5/2025 5:36 PM, Richard Damon wrote:

On 3/5/25 4:03 PM, dbush wrote:

On 3/5/2025 3:55 PM, olcott wrote:

On 3/5/2025 10:14 AM, joes wrote:

Am Wed, 05 Mar 2025 08:10:00 -0600 schrieb olcott:

On 3/5/2025 6:19 AM, Richard Damon wrote:

On 3/5/25 12:09 AM, olcott wrote:

>

I WILL NOT TOLERATE ANY OTHER ORDER

In other words, you CAN'T handle any other order, even though
logically requried, because you need to hide your fraud.

My proof requires a specific  prerequisite order.
One cannot learn algebra before one has learned to count to
ten. DD correctly emulated by HHH cannot possibly reach its own
"ret" instruction and terminate normally.
Is the first step of the mandatory prerequisite order of my
proof

What is the next step?

*DD correctly emulated by HHH cannot possibly reach*
*its own "ret" instruction and terminate normally*
It has taken two years to create this first step such that it is
the the simplest way to state the key element of the whole proof
and make this element impossible to correctly refute.
EVERY ATTEMPT TO CHANGE THE SUBJECT AWAY FROM THIS POINT IS
DISHONEST.

So what's the next step?
>

Before agreeing on an answer, it is first required to agree on the
question.

Which is the problem, since you don't have the correct question.
If HHH is a Halt Decider / Termination analyzer, the ONLY behavior
that matters is the behavior of the directly executed program whose
description is provided.

That is a stupid thing to say.
HHH computes the mapping to a return value on the basis of what its
finite string INPUT specifies.

Yes, that is the directly executed program.
>

THIS IS WHAT IT SPECIFIES *DD correctly emulated by HHH cannot
possibly reach its own "ret" instruction and terminate normally*

No, DD doesn't specify anything about what is to simulate it.
>

Yes, that is what HHH reports: I cannot complete the simulation up to
the end. No more, no less.
There are easier ways to make a program to report the failure of a
simulation.

The finite string of DD correctly emulated by HHH specifies recursive
emulation that cannot possibly reach its own "ret" instruction BECAUSE
IT SPECIFIES RECURSINVE EMULATION.

No, HHH aborts.
>

But the HHH that decides are returns can't be that HHH, so the DD given
to that HHH doesn't call the correctly emulating HHH, so you whole
argument is shown to be the fraud you have admitted to.

That seems to be a little incoherent so I cannot tell what you are
saying yet you are at least attempting to use reasoning.
I am just saying what the actual x86 machine code actually specifies
therefore any rebuttal is necessarily incorrect.

>

And the actual code of DD specifies that it halts.
>

*Straw-man deception*
>
DD correctly emulated by HHH cannot possibly
reach its own "ret" instruction and terminate normally
because DD calls HHH(DD) in recursive emulation.
>

Strawman. HHH fails to reach the 'ret' instruction, so HHH fails to do a compete simulation.

>
Simulating termination analyzer HHH

>
So you're saying it maps the halting function?
>
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
>
>

>
DD correctly emulated by HHH cannot possibly
reach its own "ret" instruction and terminate normally
because DD calls HHH(DD) in recursive emulation.
>

>
Yes. And what is the advantage of a simulator HHH that reports that it cannot possibly complete the simulation of a proven halting program up to the end, because it gets stuck in simulating *itself*?

 It is ridiculously stupid to expect a simulating
termination analyzer to

Do anything except map the halting function:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed