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

Op 11.jun.2024 om 19:12 schreef olcott:

On 6/11/2024 6:47 AM, Richard Damon wrote:

On 6/11/24 12:31 AM, olcott wrote:

On 6/10/2024 10:32 PM, Richard Damon wrote:

On 6/10/24 10:06 PM, olcott wrote:

On 6/10/2024 6:16 AM, Richard Damon wrote:

On 6/9/24 11:54 PM, olcott wrote:

*No one has verified the actual facts of this for THREE YEARS*
*No one has verified the actual facts of this for THREE YEARS*
*No one has verified the actual facts of this for THREE YEARS*

>
So, I guess you are admitting that you claim it as a verified fact is just a LIE.
>

>
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).

>
So, I guess you are admitting that this means that "D correctly simulated by H" is NOT a possible equivalent statement for the behavior of the direct execution of the input as required by the Halting Problem, so you admit you have been LYING every time you imply that it is.
>

>
_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*

>
No, H can, and must, simulate the call instruction correctly.
>

>
*Ah so you finally admit that the directly executed D(D) that*
*cannot possibly reach this instruction *is not* the behavior*
*of D correctly simulated by H that reaches this instruction*
*and simulates H simulating H*
>

>
No, I admit that THIS H didn't do it,

>
*This H does do it*
D is correctly simulated by H and H simulates itself simulating D
as the above line of code requires.
>
The directly executed D(D) can't possibly reach that line of code
thus proving that it has different behavior than D correctly
simulated by H.
>

>
WHy do you say the directly executed D(D) Can't reach its return statement?
>

 That is my second big mistake that I am aware of in the last year.
 *No one has verified the actual facts of this for THREE YEARS*
*No one has verified the actual facts of this for THREE YEARS*
*No one has verified the actual facts of this for THREE YEARS*
 On 5/29/2021 2:26 PM, olcott wrote:
https://groups.google.com/g/comp.theory/c/dTvIY5NX6b4/m/cHR2ZPgPBAAJ
 _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]
 It is impossible for D correctly simulated by H to ever reach
its simulated final state at its own machine address [00000d1d].

Because your H, that is required to halt, does not reach its final state.

 People disagree with this by changing the subject to D not simulated
by H as all. They have been indoctrinated into believing that this
strawman deception is correct yet

No, the do not disagree that the simulation does not reach the final state of D. They disagree that it is a correct simulation. Your H has the requirement to halt, but its simulation does not return to D. By your claim that D does not reach its final state, you also claim that H does not reach its final state, so, H is incorrect.

 cannot possibly show the detailed steps of how D correctly simulated
by H can possibly reach its own simulated machine address of [00000d1d].
*Here are the steps that prove that I am correct*
 (1) Executed H simulates the first seven instructions of D.
 (2) Simulated D calls simulated H(D,D) to simulate itself again.
 (3) Simulated H simulates the first seven instructions of simulated
     simulated D.
 (4) Simulated simulated D simulated by simulated H calls
     simulated simulated H(D,D) to simulate itself again.
 *HERE ARE ALL OF CONCRETE DETAILS OF THAT*
*Begin Local Halt Decider Simulation at Machine Address:cfc*
[00000cfc][00211839][0021183d](01)  55          push ebp      ; begin D
[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 by directly executed H*
   machine   stack     stack     machine          assembly
  address   address   data      code             language
  ========  ========  ========  ===============  =============
[00000cfc][0025c261][0025c265](01)  55          push ebp      ; begin D
[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
*This call to H would be simulated by simulated executed H*
*Infinitely Nested Simulation Detected Simulation Stopped*
 

No infinite recursion is detected, because the simulation is aborted prematurely. In fact no H is capable to simulate itself up to its final state. You are only showing that your H is incapable to do a correct simulation of itself. It will always miss the crucial part of itself, namely, the abort.