Sujet : Re: Proof that executed HH(DD,DD) and simulated HH(DD,DD) simulate DD correctly
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logicDate : 05. Jun 2024, 03:48:11
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v3og4r$328ec$6@i2pn2.org>
References : 1
User-Agent : Mozilla Thunderbird
On 6/4/24 12:13 PM, olcott wrote:
Proof that executed HH(DD,DD) and simulated HH(DD,DD) simulate DD
correctly. This proof requires expert knowledge of the C programming
language and the x86 programming language.
With this expertise it is easy to confirm that both the directly
executed HH(DD,DD) and the simulated executed HH(DD,DD) simulate the
steps of DD exactly the way that the x86 machine language specifies.
The problem is you have two different classes of HH, that generate two different classes of DD (HH needs to simulate just an INSTANCE of your DD template, and can't simulate the template itself by your rules, and thus different HH get different DD that can have different answers).
The HH that never stop there simulation do show that their class of DD actually is non-halting, but fail to give an answer, so are not deciders.
The HH that are programmed to stop their simulation at some point may have correctly simulated PART of the behavior of DD, but not enough to actually show that it would be non-halting, and as can be proven, it WILL halt, as a correct FULL simulation of this class of DD will reach an end, as will the direct execution.
Your deception of trying to combine these two classes into one with all behaving the same is just a LIE. and proves how little you care about the actual truth of things.
If one also has expertise on the mapping from the C source code to the
x86 assembly language then one also confirms that the x86 version of
DD is exactly what the C source-code specifies.
So?
The problem is the input is incomplete, as to hactually HAVE behvaior, we need the code (C or x86) of HH.
As the simple template, we can only simulate the first 8 instructions, and then we get stuck.
The fact that you don't understand this means you must not understand very much about the behavior of C code.
01 int DD(int (*x)())
02 {
03 int Halt_Status = HH(x, x);
04 if (Halt_Status)
05 HERE: goto HERE;
06 return Halt_Status;
07 }
08
09 int main()
10 {
11 Output("Input_Halts = ", HH(DD,DD));
12 }
_DD()
[00001db2] 55 push ebp
[00001db3] 8bec mov ebp,esp
[00001db5] 51 push ecx
[00001db6] 8b4508 mov eax,[ebp+08]
[00001db9] 50 push eax ; push DD
[00001dba] 8b4d08 mov ecx,[ebp+08]
[00001dbd] 51 push ecx ; push DD
[00001dbe] e8bff5ffff call 00001382 ; call HH
[00001dc3] 83c408 add esp,+08
[00001dc6] 8945fc mov [ebp-04],eax
[00001dc9] 837dfc00 cmp dword [ebp-04],+00
[00001dcd] 7402 jz 00001dd1
[00001dcf] ebfe jmp 00001dcf
[00001dd1] 8b45fc mov eax,[ebp-04]
[00001dd4] 8be5 mov esp,ebp
[00001dd6] 5d pop ebp
[00001dd7] c3 ret
Size in bytes:(0038) [00001dd7]
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00001de2][00103292][00000000] 55 push ebp
[00001de3][00103292][00000000] 8bec mov ebp,esp
[00001de5][0010328e][00001db2] 68b21d0000 push 00001db2 ; push DD
[00001dea][0010328a][00001db2] 68b21d0000 push 00001db2 ; push DD
[00001def][00103286][00001df4] e88ef5ffff call 00001382 ; call HH
New slave_stack at:103336
Begin Local Halt Decider Simulation Execution Trace Stored at:11333e
[00001db2][0011332a][0011332e] 55 push ebp ; DD line 01
[00001db3][0011332a][0011332e] 8bec mov ebp,esp ; DD line 02
[00001db5][00113326][001032fa] 51 push ecx ; DD line 03
[00001db6][00113326][001032fa] 8b4508 mov eax,[ebp+08] ; DD line 04
[00001db9][00113322][00001db2] 50 push eax ; push DD
[00001dba][00113322][00001db2] 8b4d08 mov ecx,[ebp+08] ; DD line 06
[00001dbd][0011331e][00001db2] 51 push ecx ; push DD
[00001dbe][0011331a][00001dc3] e8bff5ffff call 00001382 ; call HH
New slave_stack at:14dd5e
[00001db2][0015dd52][0015dd56] 55 push ebp ; DD line 01
[00001db3][0015dd52][0015dd56] 8bec mov ebp,esp ; DD line 02
[00001db5][0015dd4e][0014dd22] 51 push ecx ; DD line 03
[00001db6][0015dd4e][0014dd22] 8b4508 mov eax,[ebp+08] ; DD line 04
[00001db9][0015dd4a][00001db2] 50 push eax ; push DD
[00001dba][0015dd4a][00001db2] 8b4d08 mov ecx,[ebp+08] ; DD line 06
[00001dbd][0015dd46][00001db2] 51 push ecx ; push DD
[00001dbe][0015dd42][00001dc3] e8bff5ffff call 00001382 ; call HH
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
[00001df4][00103292][00000000] 83c408 add esp,+08
[00001df7][0010328e][00000000] 50 push eax
[00001df8][0010328a][00000743] 6843070000 push 00000743
[00001dfd][0010328a][00000743] e8a0e9ffff call 000007a2
Input_Halts = 0
[00001e02][00103292][00000000] 83c408 add esp,+08
[00001e05][00103292][00000000] eb79 jmp 00001e80
[00001e80][00103292][00000000] 33c0 xor eax,eax
[00001e82][00103296][00000018] 5d pop ebp
[00001e83][0010329a][00000000] c3 ret
Number of Instructions Executed(16829) == 251 Pages