Re: H(D,D) cannot even be asked about the behavior of D(D) V2 ---ignoring all other replies

On 6/15/24 12:31 PM, olcott wrote:

On 6/15/2024 11:26 AM, Richard Damon wrote:

On 6/15/24 12:19 PM, olcott wrote:

On 6/15/2024 11:11 AM, joes wrote:

Am Sat, 15 Jun 2024 10:54:54 -0500 schrieb olcott:

On 6/15/2024 10:12 AM, Richard Damon wrote:

On 6/15/24 11:07 AM, olcott wrote:

On 6/15/2024 10:00 AM, Richard Damon wrote:

On 6/15/24 10:37 AM, olcott wrote:

On 6/13/2024 8:24 PM, Richard Damon wrote:

On 6/13/24 11:32 AM, olcott wrote:

>

*Then do as I originally requested and provide ALL OF THE STEPS*

The mapping, for this H and D, is:
(D,D) -> 1

I am asking for a mapping from the machine language finite string of the
input to H(D,D) to each of the individual steps of the behavior of D(D).

>
*THIS SEEMS WAY WAY OVER YOUR HEAD*
It is contingent upon you to show the exact steps of how H computes
the mapping from the x86 machine language finite string input to
H(D,D) using the finite string transformation rules specified by
the semantics of the x86 programming language that reaches the
behavior of the directly executed D(D)

>
Why is that contingent on anyone but the programmer who claims to be able to write such a decider?
>
>

The first six steps of this mapping are when instructions
at the machine address range of [00000cfc] to [00000d06]
are simulated/executed.
>
After that the behavior of D correctly simulated by H diverges
from the behavior of D(D) because the call to H(D,D) by D
correctly simulated by H cannot possibly return to D.

>
Which just shows that H can not actually correctly simulate all of the behavior of the input, but it always gives up and just guesses at an answer, which is wrong.
>

>
_D()
[00000cfc](01) 55          push ebp
[00000cfd](02) 8bec        mov ebp,esp
[00000cff](03) 8b4508      mov eax,[ebp+08]
[00000d02](01) 50          push eax       ; push D
[00000d03](03) 8b4d08      mov ecx,[ebp+08]
[00000d06](01) 51          push ecx       ; push D
[00000d07](05) e800feffff  call 00000b0c  ; call H
[00000d0c](03) 83c408      add esp,+08
[00000d0f](02) 85c0        test eax,eax
[00000d11](02) 7404        jz 00000d17
[00000d13](02) 33c0        xor eax,eax
[00000d15](02) eb05        jmp 00000d1c
[00000d17](05) b801000000  mov eax,00000001
[00000d1c](01) 5d          pop ebp
[00000d1d](01) c3          ret
Size in bytes:(0034) [00000d1d]
>
>
>

D calls H, which by definition terminates. From its return value we fork
to either enter an endless loop or halt. Then H (which is not allowed to
simulate D if that doesn't halt) gives the result we just used to fork.
>

>

>

 ONLY addressing V3

Which is why you will never be able to publsih, because they won't let you do that.
You STILL are fundamentally ignoring the basic definitions of the system, and thus are just LYING about what you are doing.