Re: DD specifies non-terminating behavior to HHH ---USPTO Incorporation by reference --- despicable dishonesty

On 2/24/25 7:06 PM, olcott wrote:

On 2/24/2025 6:16 AM, Richard Damon wrote:

On 2/23/25 10:36 PM, olcott wrote:

On 2/23/2025 8:50 PM, Richard Damon wrote:

On 2/23/25 12:32 PM, olcott wrote:

On 2/22/2025 8:41 PM, dbush wrote:

On 2/22/2025 7:45 PM, olcott wrote:

On 2/22/2025 6:02 PM, Richard Damon wrote:

On 2/22/25 11:52 AM, olcott wrote:

On 2/22/2025 5:05 AM, joes wrote:

Am Thu, 20 Feb 2025 18:25:27 -0600 schrieb olcott:

On 2/20/2025 4:38 PM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 2/20/2025 2:38 AM, Mikko wrote:

On 2025-02-20 00:31:33 +0000, olcott said:

>

I have given everyone here all of the complete source code for a few
years

True but irrelevant. OP did not specify that HHH means that
particular code.

Every post that I have been talking about for two or more years has
referred to variations of that same code.

Yes.  It would be a relief if you could move on to posting something
new and fresh.

As soon as people fully address rather than endlessly dodge my key
points I will be done.

Honestly, you're gonna die first, one way or the other.
>

Let's start with a root point.
All of the other points validate this root point.
*Simulating termination analyzer HHH correctly determines*
*the non-halt status of DD*

Since DD halts, that's dead in the water.
>

>
Despicably intentionally dishonest attempts at the straw-man
deception aside:
>
DD correctly simulated by HHH cannot possibly terminate
normally by reaching its own "return" instruction.
>

>
Only because that statement is based on a false premise.
>
Since HHH doesn't correctly simulate its input, your statement is just a fabrication of your imagination.

>
*Correct simulation means emulates the machine code as specified*
It cannot mean imagining a different sequence than the one that the machine code specifies. That most people here are clueless about
x86 machine code is far less than no rebuttal at all.
>
_DD()
[00002133] 55         push ebp      ; housekeeping
[00002134] 8bec       mov ebp,esp   ; housekeeping
[00002136] 51         push ecx      ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404     add esp,+04
[00002144] 8945fc     mov [ebp-04],eax
[00002147] 837dfc00   cmp dword [ebp-04],+00
[0000214b] 7402       jz 0000214f
[0000214d] ebfe       jmp 0000214d
[0000214f] 8b45fc     mov eax,[ebp-04]
[00002152] 8be5       mov esp,ebp
[00002154] 5d         pop ebp
[00002155] c3         ret
Size in bytes:(0035) [00002155]
>
When DD emulated by HHH calls HHH(DD) this call cannot
possibly return to the emulator, conclusively proving
that
>
DD correctly simulated by HHH cannot possibly terminate
normally by reaching its own "return" instruction.
>
Assuming that it does return is simply stupid.
>
>

>
Similarly, when no_numbers_greater_than_10 emulated by F calls F(0) this call cannot possibly return to the emulator, conclusively proving that

>
Not true. The stack eventually unwinds after ten emulations.

>
Just like a CORRECT emulation of DD would if the HHH doing the emulation didn't abort (but doing it by the hypothetical of NOT changing the HHH that DD calls, since that must be the original HHH).
>
Your problem is you have lied to yourself about what is a "correct emulation"

>
In other words you "believe" that the call from DD to HHH(DD)
returns when the above DD is emulated by HHH.

>
What happens in an incorrect emulation doesn't matter,

 One to infinity steps of the above DD emulated by HHH never
terminates normally. That you imagine that this code "should"
do something other than what it specifies is quite stupid.
 

INVALID, as you are changing your input, or the input isn't a program.
For any instance of the question, there can be one and only one HHH, and one and only one DD.
If the HHH(DD) returns 0, then there is a finite number of steps of correct emulation (more than what HHH did) that will show that DD is halting.
Thus, your claim is invalid, your logic proben to be unsound, and you to be proven stupid.
Sorry, Strawmen are not allowed, and changing the input with an attempt at slight-of-hand just proves you are nothing but a fraud.