Re: Proof that DD correctly simulated by HH provides the correct halt status criteria

On 6/9/2024 2:23 AM, Mikko wrote:

On 2024-06-08 12:27:09 +0000, olcott said:
 

On 6/8/2024 12:53 AM, Mikko wrote:

On 2024-06-07 21:48:57 +0000, olcott said:
 

*That no counter-example to the following exists proves that it is true*

 Not wihout a proof that no counter-example exists.
 

 I incorporate by reference
(a) The x86 language
(b) The notion of an x86 emulator
 (c) I provide this complete function
 void DDD(int (*x)())
{
   HH(x, x);
}
 _DDD()
[00001de2] 55         push ebp
[00001de3] 8bec       mov ebp,esp
[00001de5] 8b4508     mov eax,[ebp+08]
[00001de8] 50         push eax         ; push DDD
[00001de9] 8b4d08     mov ecx,[ebp+08]
[00001dec] 51         push ecx         ; push DDD
[00001ded] e890f5ffff call 00001382    ; call HH
[00001df2] 83c408     add esp,+08
[00001df5] 5d         pop ebp
[00001df6] c3         ret
Size in bytes:(0021) [00001df6]
 Then I state that No DDD correctly emulated by any
x86 emulator H can possibly reach its own [00001df6]
instruction.
 To anyone having this mandatory prerequisite knowledge
(perhaps not you) every x86 emulation of DDD by any
x86 emulator H continually repeats the first seven lines
of DDD until it crashes due to out-of-memory error.
 Try and show that the above sequence is incorrect, you
cannot because it is correct.

 OK. Tell us when you have a proof with that incorporated.
 

 That <is> a complete proof to anyone that has all of the mandatory prerequisite knowledge.