Re: D correctly simulated by H proved for THREE YEARS --- verified fact for 3 years

Am Mon, 10 Jun 2024 09:39:48 -0500 schrieb olcott:

On 6/10/2024 3:35 AM, Alan Mackenzie wrote:

In comp.theory olcott <polcott333@gmail.com> wrote:

On 6/9/2024 8:36 PM, Richard Damon wrote:

On 6/9/24 9:23 PM, olcott wrote:

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).

That doesn't make any sense. Surely the direct execution must be correct.

_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*

D(D) does not ignore the call to H(D, D), whether it is simulated or not.

H does not ignore that instruction and simulates itself simulating D.
The simulated H outputs its own execution trace of D.

--
joes