On 6/22/2022 8:36 PM, Richard Damon wrote:
> On 6/22/22 9:29 PM, olcott wrote:
>> On 6/22/2022 8:14 PM, Richard Damon wrote:
>>> On 6/22/22 8:55 PM, olcott wrote:
>>>> On 6/22/2022 7:48 PM, Richard Damon wrote:
>>>>> On 6/22/22 8:37 PM, olcott wrote:
>>>>
>>>>>> First you agree that my words are perfectly correct within their
>>>>>> specified context
>>>>>
>>>>> Since you haven't actualy defined you context, and imply that it is
>>>>> the halting problem, where they can not be correct, that is not
>>>>> possible.
>>>>>>
>>>> First you agree that these words are 100% correct within the context
>>>> of software engineering totally ignoring the context of the halting
>>>> problem.
>>>>
>>>> #include <stdint.h>
>>>> #define u32 uint32_t
>>>>
>>>> #include <stdint.h>
>>>> typedef void (*ptr)();
>>>>
>>>> void P(ptr x)
>>>> {
>>>>    if (H(x, x))
>>>>      HERE: goto HERE;
>>>>    return;
>>>> }
>>>>
>>>> int main()
>>>> {
>>>>    Output("Input_Halts = ", H(P, P));
>>>> }
>>>>
>>>> _P()
>>>> [000010d2](01)  55              push ebp
>>>> [000010d3](02)  8bec            mov ebp,esp
>>>> [000010d5](03)  8b4508          mov eax,[ebp+08]
>>>> [000010d8](01)  50              push eax
>>>> [000010d9](03)  8b4d08          mov ecx,[ebp+08]
>>>> [000010dc](01)  51              push ecx
>>>> [000010dd](05)  e820feffff      call 00000f02
>>>> [000010e2](03)  83c408          add esp,+08
>>>> [000010e5](02)  85c0            test eax,eax
>>>> [000010e7](02)  7402            jz 000010eb
>>>> [000010e9](02)  ebfe            jmp 000010e9
>>>> [000010eb](01)  5d              pop ebp
>>>> [000010ec](01)  c3              ret
>>>> Size in bytes:(0027) [000010ec]
>>>>
>>>> Every sufficiently competent software engineer can easily verify
>>>> that the complete and correct x86 emulation of the input to H(P,P)
>>>> by H would never reach the "ret" instruction of P because both H and
>>>> P would remain stuck in infinitely recursive emulation.
>>>>
>>>
>>> So, if H actually is a program that does a COMPLETE and correct x86
>>> emulation of its input, then YES, as I have said many time before,
>>> this combination is non-halting.
>>>
>>> The fact that you need to keep going back to this, and seem to just
>>> be refusing to accept the conditions under which you have proved it
>>> just shows the problems with your thought process.
>>>
>>>> If H does correctly determine that this is the case in a finite
>>>> number of steps then H could reject its input on this basis. Here
>>>> are the details of exactly how H does this in a finite number of steps.
>>>
>>> Except that NOW H isn't the H we were just talking about, so you are
>>> just proving that you are either lying or an idiot.
>>>
>>> Remember, the first analysis had the CONDITION on it that H did a
>>> COMPLETE and correct x86 emulation.
>>>
>>> Once you remove that property form H, that conclusion no long holds
>>> and you are shown to be a lying idiot.
>>>
>>>>
>>>> typedef struct Decoded
>>>> {
>>>>    u32 Address;
>>>>    u32 ESP;          // Current value of ESP
>>>>    u32 TOS;          // Current value of Top of Stack
>>>>    u32 NumBytes;
>>>>    u32 Simplified_Opcode;
>>>>    u32 Decode_Target;
>>>> } Decoded_Line_Of_Code;
>>>>
>>>>   machine   stack     stack     machine    assembly
>>>>   address   address   data      code       language
>>>>   ========  ========  ========  =========  =============
>>>> [000010d2][00211e8a][00211e8e] 55         push ebp
>>>> [000010d3][00211e8a][00211e8e] 8bec       mov ebp,esp
>>>> [000010d5][00211e8a][00211e8e] 8b4508     mov eax,[ebp+08]
>>>> [000010d8][00211e86][000010d2] 50         push eax        // push P
>>>> [000010d9][00211e86][000010d2] 8b4d08     mov ecx,[ebp+08]
>>>> [000010dc][00211e82][000010d2] 51         push ecx        // push P
>>>> [000010dd][00211e7e][000010e2] e820feffff call 00000f02   // call H
>>>> Infinitely Recursive Simulation Detected Simulation Stopped
>>>>
>>>> // actual fully operational code in the x86utm operating system
>>>> u32 H(u32 P, u32 I)
>>>> {
>>>> HERE:
>>>>    u32 End_Of_Code;
>>>>    u32 Address_of_H;              // 2022-06-17
>>>>    u32 code_end                  = get_code_end(P);
>>>>    Decoded_Line_Of_Code *decoded = (Decoded_Line_Of_Code*)
>>>> Allocate(sizeof(Decoded_Line_Of_Code));
>>>>    Registers*  master_state      = (Registers*)
>>>> Allocate(sizeof(Registers));
>>>>    Registers*  slave_state       = (Registers*)
>>>> Allocate(sizeof(Registers));
>>>>    u32*        slave_stack       = Allocate(0x10000); // 64k;
>>>>    u32  execution_trace = (u32)Allocate(sizeof(Decoded_Line_Of_Code)
>>>> * 1000);
>>>>
>>>>    __asm lea eax, HERE             // 2022-06-18
>>>>    __asm sub eax, 6                // 2022-06-18
>>>>    __asm mov Address_of_H, eax     // 2022-06-18
>>>>    __asm mov eax, END_OF_CODE
>>>>    __asm mov End_Of_Code, eax
>>>>
>>>>    Output("Address_of_H:", Address_of_H); // 2022-06-11
>>>>    Init_slave_state(P, I, End_Of_Code, slave_state, slave_stack);
>>>>    Output("\nBegin Simulation   Execution Trace Stored at:",
>>>> execution_trace);
>>>>    if (Decide_Halting(&execution_trace, &decoded, code_end,
>>>> &master_state,
>>>>                       &slave_state, &slave_stack, Address_of_H, P, I))
>>>>        goto END_OF_CODE;
>>>>    return 0;  // Does not halt
>>>> END_OF_CODE:
>>>>    return 1; // Input has normally terminated
>>>> }
>>>>
>>>> H knows its own machine address and on this basis it can easily
>>>> examine its stored execution_trace of P and determine:
>>>> (a) P is calling H with the same arguments that H was called with.
>>>> (b) No instructions in P could possibly escape this otherwise
>>>> infinitely recursive emulation.
>>>> (c) H aborts its emulation of P before its call to H is invoked.
>>>>
>>>>
>>>
>>> (b) is NOT a correct rule. Thos has been pointed out before, and you
>>> have ignored it.
>>>
>> That you don't understand what I mean does not mean that it is an
>> incorrect rule.
>>
>> Here is an example where P does have instruction that could possibly
>> escape this otherwise infinitely recursive emulation:
>>
>>
>> void P(ptr x)
>> {
>> static count = 0;
>>    count++;
>>    if count > 3)
>>      return;
>>    if (H(x, x))
>>      HERE: goto HERE;
>>    return;
>> }
>>
>
> FALLACY of proof by example. I never said that (b) isn't sometimes true,
> just it isn't an always true condition. You fail at elementary logic.

Try and find a valid counter-example. Every attempt at rebuttal that is
not a valid counter-example is one form of deception or another.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer