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

On 6/8/2024 7:59 AM, Richard Damon wrote:

On 6/8/24 8:47 AM, olcott wrote:

On 6/8/2024 1:28 AM, Mikko wrote:

On 2024-06-08 02:43:57 +0000, Mike Terry said:
>

On 07/06/2024 17:34, joes wrote:

Am Fri, 07 Jun 2024 10:40:37 -0500 schrieb olcott:

On 6/7/2024 10:32 AM, joes wrote:

Am Fri, 07 Jun 2024 10:20:09 -0500 schrieb olcott:

On 6/7/2024 10:09 AM, Python wrote:

Le 07/06/2024 à 17:05, olcott a écrit :

On 6/7/2024 9:55 AM, Python wrote:

Le 07/06/2024 à 16:47, olcott a écrit :

>

The issue here is that I proved that DD correctly simulated by HH
has different behavior than the directly executed DD(DD) and
everyone's "rebuttal" to this proof is to simply ignore it.

When you actually try to form a rebuttal of the above you will see
that I am correct. So far everyone simply ignores the proof that I
am correct as their only rebuttal.

A {correct simulation} means that each instruction of the above x86
machine language of DD is correctly simulated by HH and simulated in
the correct order.

And to the end. Thus it can't behave differently than direct execution.

You must not be very good at programming if you believe that
Infinite_Recursion must be simulated "to the end"

As I said before, if there is no end the simulation can't end either.

>
PO chooses to leave his terminology ambiguous, so we have 1000 post threads perpetuated by little more than verbal misunderstandings.

>
Crackpots are sometines like that. When they start to see that they can't
convince they start to gradually replace non-convincing arguments with
trolling until, at some point, all they try to do is to troll. Unless,
of course, some external cause makes them stop.
>

>
*Nothing ambiguous here*

 Just not responsive.
 

That people lack the knowledge of common technical terms does
not mean that expressions containing these terms are ambiguous.
>
I incorporate by reference
(a) The x86 language
(b) The notion of an x86 emulator
>
(c) I provide this complete function

 Which does nothing by itself.
 

>
void DDD(int (*x)())
{
   HH(x, x);
}
>
_DDD()
[00001de2] 55         push ebp
[00001de3] 8bec       mov ebp,esp
[00001de5] 8b4508     mov eax,[ebp+08]
[00001de8] 50         push eax         ; push DD
[00001de9] 8b4d08     mov ecx,[ebp+08]
[00001dec] 51         push ecx         ; push DD
[00001ded] e890f5ffff call 00001382    ; call HH
[00001df2] 83c408     add esp,+08
[00001df5] 5d         pop ebp
[00001df6] c3         ret
Size in bytes:(0021) [00001df6]
>
Then I state that No DDD correctly emulated by any
x86 emulator H can possibly reach its own [00001df6]
instruction.

  Which just proves you don't understand what a proof is.
 Since that doesn't do anything to connect the truth-makers you referenced to the statement.
 

>
To anyone having this mandatory prerequisite knowledge
(perhaps not you) every x86 emulation of DDD by any
x86 emulator H continually repeats the first seven lines
of DDD until it crashes due to out-of-memory error.
>

 Which means you just proved that NO HH that meets your requirements can ever give an answer for this question.
 

There are more steps to this proof. I cannot conflate them
together and ever get closure.
The above HH is only an x86 emulator. Once it is understood
that DDD correctly simulated by any HH that can possibly
exist cannot possibly reach its own [00001df6] machine address
then we know that when HH sees the same thing that we see that
HH has its correct abort criteria.
*That is how partial simulation proves non-termination*
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer