Re: D correctly simulated by H proved for THREE YEARS ---

On 6/10/2024 3:35 AM, joes wrote:

Am Sun, 09 Jun 2024 22:54:52 -0500 schrieb olcott:

On 5/29/2021 2:26 PM, olcott wrote:
https://groups.google.com/g/comp.theory/c/dTvIY5NX6b4/m/cHR2ZPgPBAAJ
>
THE ONLY POSSIBLE WAY for D simulated by H to have the same behavior as
the directly executed D(D) is for the instructions of D to be
incorrectly simulated by H (details provided below).
>
_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]
>
In order for D simulated by H to have the same behavior as the directly
executed D(D) H must ignore the instruction at machine address
[00000d07]. *That is an incorrect simulation of D*

 >

I don't understand. Does D(D) ignore the call to H(D,D)?
>

H does not ignore that instruction and simulates itself simulating D.
The simulated H outputs its own execution trace of D.
The directly executed D(D) reaps the benefit of D correctly
simulated by H proving that *its input never halts*
Begin Local Halt Decider Simulation at Machine Address:cfc
..[00000cfc][00211839][0021183d](01)  55          push ebp
..[00000cfd][00211839][0021183d](02)  8bec        mov ebp,esp
..[00000cff][00211839][0021183d](03)  8b4508      mov eax,[ebp+08]
..[00000d02][00211835][00000cfc](01)  50          push eax      ; push D
..[00000d03][00211835][00000cfc](03)  8b4d08      mov ecx,[ebp+08]
..[00000d06][00211831][00000cfc](01)  51          push ecx      ; push D
..[00000d07][0021182d][00000d0c](05)  e800feffff  call 00000b0c ; call H
This call to H is simulated H.
We can tell that it is the simulated H is providing this
trace because it has a different virtual machine stack.
*The simulated H derives this execution trace of D*
     machine   stack     stack     machine          assembly
     address   address   data      code             language
     ========  ========  ========  ===============  =============
..[00000cfc][0025c261][0025c265](01)  55          push ebp
..[00000cfd][0025c261][0025c265](02)  8bec        mov ebp,esp
..[00000cff][0025c261][0025c265](03)  8b4508      mov eax,[ebp+08]
..[00000d02][0025c25d][00000cfc](01)  50          push eax      ; push D
..[00000d03][0025c25d][00000cfc](03)  8b4d08      mov ecx,[ebp+08]
..[00000d06][0025c259][00000cfc](01)  51          push ecx      ; push D
..[00000d07][0025c255][00000d0c](05)  e800feffff  call 00000b0c ; call H
Infinitely Nested Simulation Detected Simulation Stopped
Because the H(D,D) that D(D) calls correctly recognizes the its input
DOES NOT HALT, it correctly aborts the simulation of this input causing
the directly executed D(D) to halt.
I proved that D simulated by H can only have the same behavior as the
directly executed D(D) when D is simulated by H incorrectly.
This requires D simulated by H to skip over the machine address
[00000d07] and not call H(D,D) to simulate itself again.
D simulated by H does not do that. It simulates itself simulating D.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer