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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: DD correctly simulated by HH cannot possible halt --- Try to
 prove otherwise --- x86 DD
Date: Mon, 3 Jun 2024 10:10:21 +0200
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Op 03.jun.2024 om 05:20 schreef olcott:
> On 6/2/2024 10:13 PM, Richard Damon wrote:
>> On 6/2/24 10:54 PM, olcott wrote:
>>> IT HAS ALWAYS BEEN ABOUT THE BEHAVIOR THAT THE INPUT SPECIFIES.
>>> That you did get confused by the Linz text proves that you do
>>> get confused. Previously it looked just like willful deception.
>>
>> Which is, for a Halt Decider, exactly and only the behavior of the 
>> Turing Machine the input describes.
>>
>> PERIOD.
>>
>> Anything else is just a LIE.
>>
>>>
>>>> You don't seem to understand that you can't just "redefine" the 
>>>> system to meet your desires.
>>>>
>>>
>>> Deciders compute the mapping FROM THEIR INPUTS.
>>> DD correctly simulated by HH specifies NON-HALTING.
>>
>> No, Running DD(DD) and seeing that it will never, after an unbounded 
>> number of steps, indicate it is non-halting.
>>
>> DEFINITION.
>>
>>>
>>> Deciders compute the mapping FROM THEIR INPUTS.
>>> DD correctly simulated by HH specifies NON-HALTING.
>>
>> Right, and the input is a representation of a Turing Machine and its 
>> input, whose behavior the decider is to decide on.
>>
>>>
>>> Deciders compute the mapping FROM THEIR INPUTS.
>>> DD correctly simulated by HH specifies NON-HALTING.
>>>
>>
>> And that is the machine the input describes.
>>
>> ANYTHING ELSE IS JUST A LIE.
>>
>>> You can't get away with implicitly saying that you
>>> just don't "believe in" UTMs.
>>
>> I do, and a UTM is DEFINED as a machine that exactly reproduces the 
>> behavior of the machine described by its input.
>>
> 
> *If that was true then you prove that this statement is false*
> *We can see that the following DD cannot possibly halt when*
> *correctly simulated by every HH that can possibly exist*
> 
> typedef int (*ptr)();  // ptr is pointer to int function in C
> 00       int HH(ptr p, ptr i);
> 01       int DD(ptr p)
> 02       {
> 03         int Halt_Status = HH(p, p);
> 04         if (Halt_Status)
> 05           HERE: goto HERE;
> 06         return Halt_Status;
> 07       }
> 
> _DD()
> [00001c22] 55         push ebp
> [00001c23] 8bec       mov ebp,esp
> [00001c25] 51         push ecx
> [00001c26] 8b4508     mov eax,[ebp+08]
> [00001c29] 50         push eax        ; push DD 1c22
> [00001c2a] 8b4d08     mov ecx,[ebp+08]
> [00001c2d] 51         push ecx        ; push DD 1c22
> [00001c2e] e80ff7ffff call 00001342   ; call HH
> [00001c33] 83c408     add esp,+08
> [00001c36] 8945fc     mov [ebp-04],eax
> [00001c39] 837dfc00   cmp dword [ebp-04],+00
> [00001c3d] 7402       jz 00001c41
> [00001c3f] ebfe       jmp 00001c3f
> [00001c41] 8b45fc     mov eax,[ebp-04]
> [00001c44] 8be5       mov esp,ebp
> [00001c46] 5d         pop ebp
> [00001c47] c3         ret
> Size in bytes:(0038) [00001c47]
> 

It is clear (both from the C code as from the compiled code) that at the 
moment that the outer simulator reaches the place where DD calls HH then 
HH finds *itself* in a 'repeated state'. At that same moment DD is *not* 
(yet) in a repeated state. (DD may get in a repeated state if the called 
HH would start a new simulation.) So, the only conclusion is that it is 
HH that displays repetition. And because of that DD will repeat as well.

By using simulation, a new halting problem is introduced, that has 
nothing to do with the pathological part of DD n lines 04, 05 and 06. 
This pathological part is not even processed by the simulation.