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

On 3/8/25 11:04 AM, olcott wrote:

On 3/8/2025 8:22 AM, joes wrote:

Am Sat, 08 Mar 2025 08:06:04 -0600 schrieb olcott:

On 3/8/2025 3:17 AM, joes wrote:

Am Fri, 07 Mar 2025 21:40:56 -0600 schrieb olcott:

On 3/7/2025 8:23 PM, dbush wrote:

On 3/7/2025 9:19 PM, olcott wrote:

On 3/7/2025 7:52 PM, dbush wrote:

On 3/7/2025 8:49 PM, olcott wrote:

On 3/7/2025 10:25 AM, Fred. Zwarts wrote:

Op 07.mrt.2025 om 16:17 schreef olcott:

On 3/7/2025 2:59 AM, Fred. Zwarts wrote:

Op 06.mrt.2025 om 21:13 schreef olcott:

On 3/6/2025 3:13 AM, Fred. Zwarts wrote:

Op 06.mrt.2025 om 04:53 schreef olcott:

On 3/5/2025 9:31 PM, dbush wrote:

On 3/5/2025 10:17 PM, olcott wrote:

On 3/5/2025 7:10 PM, dbush wrote:

>

In other words, you know that what you're working on has
nothing to do with the halting problem, but you don't
care.

In other words I WILL NOT TOLERATE ANY BULLSHIT
DEFLECTION.
You have proven that you know these things pretty well SO
QUIT THE SHIT!

You want people to accept that HHH(DD) does in fact report
that changing the code of HHH to an unconditional simulator
and running HHH(DD) will not halt.

DD correctly emulated by HHH cannot possibly reach its own
"ret" instruction and terminate normally.

Yes, we agree that HHH fails to reach the 'ret' instruction,

Despicably dishonest attempt at the straw-man deception.

No rebuttal. So, we agree that HHH fails to reach the 'ret'
instruction.

Not at all. Trying to get away with changing the subject WILL
NOT BE TOLERATED.

If you do not agree that HHH fails to reach the 'ret' instruction
(that world-class simulators do reach, just as the direct
execution does), show how it reaches the 'ret' instruction.

*set X*
When-so-ever any input to any simulating termination analyzer
calls the simulator that is simulating itself

Not an issue, since termination analyzers don't exist.

I thought that you demonstrated knowledge of these things.
Maybe I was wrong.

We know termination analyzers don't exist because no algorithm exists
that 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

Automated Termination Analysis of C Programs
https://publications.rwth-aachen.de/record/972440/files/972440.pdf
AProVE seems to be the leading authority on what you say DOES NOT
EXIST

It isn't claimed to be total. Have you tried running it on itself or on
a program similar to DD (instead of calling HHH, ...)?

Termination analyzers are not required to be infallible.

Neither do they disprove the undecidability of halting.
>

 The actual behavior that Halt7.c specifies combined with
the semantics of the x86 language conclusively proves:
 DD correctly emulated by HHH cannot possibly reach
its own "ret" instruction and terminate normally
because DD calls HHH(DD) in recursive emulation.
 

No, it shows that if HHH is the program given in it, then that HHH does not do a correct emulation and thus gets the wrong answer.
It can be edited to show that a DIFFERENT program DD, that calls and is emulated by a DIFFERENT HHH that doesn't abort will not return.
But this shows that if HHH returns an answer, and it is emulating the DD that calls itself, it could not have done a correct emulation.
Sorry, you are just proving you don't understand what you are talking aobut, and you have admitted your whole work is a FRAUD based on lying about basic definitions of Terms-of-art, so nothing you say should be believed, as there is no way to know what you actually mean by it.