Re: Two dozen people were simply wrong --- Try to prove otherwise --- pinned down

Op 02.jun.2024 om 16:41 schreef olcott:

On 6/2/2024 4:03 AM, Fred. Zwarts wrote:

Op 01.jun.2024 om 21:51 schreef olcott:

On 6/1/2024 1:54 PM, Fred. Zwarts wrote:

Op 01.jun.2024 om 20:07 schreef olcott:

On 6/1/2024 12:56 PM, Richard Damon wrote:

On 6/1/24 1:44 PM, olcott wrote:

On 6/1/2024 12:33 PM, Richard Damon wrote:

On 6/1/24 1:27 PM, olcott wrote:

On 6/1/2024 12:22 PM, Richard Damon wrote:

On 6/1/24 12:38 PM, olcott wrote:

On 6/1/2024 11:27 AM, Richard Damon wrote:

On 6/1/24 12:13 PM, olcott wrote:

On 6/1/2024 10:56 AM, Richard Damon wrote:

On 6/1/24 11:30 AM, olcott wrote:

>
*I will not discuss any other points with you until after you either*
(a) Acknowledge that DD correctly simulated by HH and ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
     simulated by embedded_H remain stuck in recursive simulation for
     1 to ∞ of correct simulation or
>
(b) Correctly prove otherwise.

>
And until you answer the question of what that actually means, I will reply WHO CARES.
>

>
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       }
08
09       int main()
10       {
11         HH(DD,DD);
12         return 0;
13       }
>
Every DD correctly simulated by any HH of the infinite set of HH/DD
pairs that match the above template never reaches past its own simulated
line 03 in 1 to ∞ steps of correct simulation of DD by HH.
>
In this case HH is either a pure simulator that never halts or
HH is a pure function that stops simulating after some finite number
of simulated lines. The line count is stored in a local variable.
The pure function HH always returns the meaningless value of 56
after it stops simulating.
>

>
So, still no answer, to teh question.

>
You can pretend that you don't understand something that you do indeed
understand into perpetuity.
>
The key measure of dishonestly would be that you continue to say
that you don't understand yet never ever point out exactly what you
don't understand and why you don't understand it.
>

I giuess that Mean YOU don't even know what you are asking, though it seems that now you are admitting that your HH doesn't actually ANSWER the question, so it isn't ACTUALL a decider for any function except the "56" mapping.
>
I will repeat the question and until you answer the question of what that actually means, I will reply WHO CARES.
>
DO you mean the simulation of the TEMPLATE DD,

>
*Of course I don't mean that nonsense. I mean exactly what I specified*
>

which means that we CAN'T simulate the call HH as we have no code past point to simulate, and thus your claim is just a LIE.
>
Or, do you mean a given instance of HH simulating a given instance of DD, at which point we never have the 1 to infinte number of simulatons of THAT INPUT, so your claim is just a LIE.
>

>
Every element of the infinite set of every H/D pairs...
Every element of the infinite set of every H/D pairs...
Every element of the infinite set of every H/D pairs...
>
*Its not that hard when one refrains from dishonesty*
We can't even say that you forgot these details from one reply
to the next because the details are still in this same post.
>

>
And every one gives a meaningless answer,

>
*THEN TRY TO REFUTE THIS UNEQUIVOCAL STATEMENT*
DD correctly emulated by HH with an x86 emulator cannot possibly
reach past its own machine instruction [00001c2e] in any finite
number of steps of correct emulation.
>

>
Why? I don't care about it.
>
As I have said, the implication of your definition of "Correct SImulation" means that this says NOTHING about the halting behavior of DD. (only not halted yet)
>

>
*THEN TRY TO REFUTE THIS UNEQUIVOCAL STATEMENT*
DD correctly emulated by HH with an x86 emulator cannot possibly
reach past its own machine instruction [00001c2e] in any finite
*or infinite* number of steps of correct emulation.
>
When I say it that way you claim to be confused and what I do
not say it that way you claim what I say is incomplete proof.

>
WHy do I care? I won't spend the effort to even try to refute something that is clearly meaningless.
>
You seem to have a conflict of definitions, as a given DD will only ever be simulated by ONE given HH that only simuates for one number of steps.
>

>
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       }
08
09       int main()
10       {
11         HH(DD,DD);
12         return 0;
13       }
>
You continue to either fail to understand or seemingly more likely
simply lie about the fact that every DD correctly simulated by any
HH that can possibly exist cannot possibly reach past its own line 03.

>
Only if the simulation of HH simulated by HH does not reach HH's return, otherwise the simulation of DD would go to line 04.
>

>
*THIS MEANS THAT THE INPUT TO HH(DD,DD) DOES NOT HALT*
*THIS MEANS THAT THE INPUT TO HH(DD,DD) DOES NOT HALT*
*THIS MEANS THAT THE INPUT TO HH(DD,DD) DOES NOT HALT*
>

>
If true: The input to HH is both DD and HH called by DD, so both DD and HH do not halt, but keep starting new instances of each other.
However, HH is required to halt, but it doesn't. So, the HH that halts is phantasy.

>
I have fully operational code that proves otherwise.

>
But you are unable to show the ret instruction of HH simulated by itself. So, the proof is missing the crucial part.
>
>

Any expert in the C programming language knows the
same thing from the C source-code.
>

>

 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       }
08
09       int main()
10       {
11         HH(DD,DD);
12         return 0;
13       }
 

You don't need to be an expert in C to see that if the simulated HH would halt, then DD would continue to line 04.

 *Try and show how that can happen*

If you show how HH simulated by HH reaches its return and show the next 10 instructions. You can't and therefore we know the simulated HH does not halt and the non-halting HH is the reason that DD does not continue.

*You may simply lack the required prerequisite knowledge >
DD correctly emulated by HH with an x86 emulator cannot possibly
reach past its own machine instruction [00001c2e] in any finite
(or infinite) number of steps of correct emulation.
 _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 could if HH were a halting function as required. Then the call at [00001c2e] would return and the next instruction would be reachable. The reason that it is not, is that HH does not return.
(You don't need to show your x86 code. Is does not show more than the C code.)