On Thu, 12 May 2022 12:51:27 -0500
olcott <NoOne@NoWhere.com> wrote:

> On 5/12/2022 12:47 PM, Mr Flibble wrote:
> > On Thu, 12 May 2022 12:45:15 -0500
> > olcott <NoOne@NoWhere.com> wrote:
> >
> >> On 5/12/2022 12:43 PM, Mr Flibble wrote:
> >>> On Wed, 11 May 2022 19:23:45 -0500
> >>> olcott <NoOne@NoWhere.com> wrote:
> >>>
> >>>> On 5/11/2022 2:11 PM, Mr Flibble wrote:
> >>>>> On Wed, 11 May 2022 13:07:16 -0500
> >>>>> olcott <NoOne@NoWhere.com> wrote:
> >>>>>
> >>>>>> Proof that H(P,P)==0 is correct [ refuting the halting problem
> >>>>>> proofs ]
> >>>>>>
> >>>>>> The x86utm operating system was created so that every detail of
> >>>>>> the conventional halting problem counter example could be fully
> >>>>>> specified in C/x86.
> >>>>>>
> >>>>>> In computability theory, the halting problem is the
> >>>>>> problem of determining, from a description of an
> >>>>>> arbitrary computer program and an input, whether the
> >>>>>> program will finish running, or continue to run
> >>>>>> forever...
> >>>>>>
> >>>>>> For any program f that might determine if programs halt,
> >>>>>> a "pathological" program g, called with some input, can
> >>>>>> pass its own source and its input to f and then
> >>>>>> specifically do the opposite of what f predicts g will do. No f
> >>>>>> can exist that handles this case.
> >>>>>> https://en.wikipedia.org/wiki/Halting_problem
> >>>>>>
> >>>>>> This exact same relationship of f(g,g) was created as H(P,P),
> >>>>>> shown below.
> >>>>>>
> >>>>>> This is the overview of the method for proving that this
> >>>>>> analysis is correct:
> >>>>>> (a) Verify that the execution trace of P by H is correct by
> >>>>>> comparing this execution trace to the ax86 source-code of P.
> >>>>>>
> >>>>>> (b) Verify that this execution trace shows that P is stuck in
> >>>>>> infinitely nested simulation (a non-halting behavior).
> >>>>>>
> >>>>>> This proof can only be understood only by those having
> >>>>>> sufficient technical competence in:
> >>>>>> (a) software engineering (recognizing infinite recursion in C
> >>>>>> and x86 code) (b) the x86 programming language
> >>>>>> (c) the C programming language and
> >>>>>> (d) the details of how C is translated into x86 by the
> >>>>>> Microsoft C compilers.
> >>>>>>
> >>>>>> #include <stdint.h>
> >>>>>> #define u32 uint32_t
> >>>>>>
> >>>>>> void P(u32 x)
> >>>>>> {
> >>>>>> if (H(x, x))
> >>>>>> HERE: goto HERE;
> >>>>>> return;
> >>>>>> }
> >>>>>>
> >>>>>> int main()
> >>>>>> {
> >>>>>> Output("Input_Halts = ", H((u32)P, (u32)P));
> >>>>>> }
> >>>>>>
> >>>>>> _P()
> >>>>>> [00001352](01) 55 push ebp
> >>>>>> [00001353](02) 8bec mov ebp,esp
> >>>>>> [00001355](03) 8b4508 mov eax,[ebp+08]
> >>>>>> [00001358](01) 50 push eax
> >>>>>> [00001359](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>> [0000135c](01) 51 push ecx
> >>>>>> [0000135d](05) e840feffff call 000011a2 // call H
> >>>>>> [00001362](03) 83c408 add esp,+08
> >>>>>> [00001365](02) 85c0 test eax,eax
> >>>>>> [00001367](02) 7402 jz 0000136b
> >>>>>> [00001369](02) ebfe jmp 00001369
> >>>>>> [0000136b](01) 5d pop ebp
> >>>>>> [0000136c](01) c3 ret
> >>>>>> Size in bytes:(0027) [0000136c]
> >>>>>>
> >>>>>> _main()
> >>>>>> [00001372](01) 55 push ebp
> >>>>>> [00001373](02) 8bec mov ebp,esp
> >>>>>> [00001375](05) 6852130000 push 00001352 // push P
> >>>>>> [0000137a](05) 6852130000 push 00001352 // push P
> >>>>>> [0000137f](05) e81efeffff call 000011a2 // call H
> >>>>>> [00001384](03) 83c408 add esp,+08
> >>>>>> [00001387](01) 50 push eax
> >>>>>> [00001388](05) 6823040000 push 00000423 // "Input_Halts
> >>>>>> = " [0000138d](05) e8e0f0ffff call 00000472 // call
> >>>>>> Output [00001392](03) 83c408 add esp,+08
> >>>>>> [00001395](02) 33c0 xor eax,eax
> >>>>>> [00001397](01) 5d pop ebp
> >>>>>> [00001398](01) c3 ret
> >>>>>> Size in bytes:(0039) [00001398]
> >>>>>>
> >>>>>> machine stack stack machine assembly
> >>>>>> address address data code language
> >>>>>> ======== ======== ======== ========= =============
> >>>>>> ...[00001372][0010229e][00000000] 55 push ebp
> >>>>>> ...[00001373][0010229e][00000000] 8bec mov ebp,esp
> >>>>>> ...[00001375][0010229a][00001352] 6852130000 push 00001352 //
> >>>>>> push P ...[0000137a][00102296][00001352] 6852130000 push
> >>>>>> 00001352 // push P ...[0000137f][00102292][00001384] e81efeffff
> >>>>>> call 000011a2 // call H
> >>>>>>
> >>>>>> Begin Local Halt Decider Simulation Execution Trace Stored
> >>>>>> at:212352 ...[00001352][0021233e][00212342] 55 push ebp
> >>>>>> // enter P ...[00001353][0021233e][00212342] 8bec mov
> >>>>>> ebp,esp ...[00001355][0021233e][00212342] 8b4508 mov
> >>>>>> eax,[ebp+08] ...[00001358][0021233a][00001352] 50 push
> >>>>>> eax // push P ...[00001359][0021233a][00001352] 8b4d08 mov
> >>>>>> ecx,[ebp+08] ...[0000135c][00212336][00001352] 51 push
> >>>>>> ecx // push P ...[0000135d][00212332][00001362] e840feffff call
> >>>>>> 000011a2 // call H ...[00001352][0025cd66][0025cd6a] 55
> >>>>>> push ebp // enter P ...[00001353][0025cd66][0025cd6a] 8bec
> >>>>>> mov ebp,esp ...[00001355][0025cd66][0025cd6a] 8b4508
> >>>>>> mov eax,[ebp+08] ...[00001358][0025cd62][00001352] 50
> >>>>>> push eax // push P ...[00001359][0025cd62][00001352] 8b4d08
> >>>>>> mov ecx,[ebp+08] ...[0000135c][0025cd5e][00001352] 51
> >>>>>> push ecx // push P ...[0000135d][0025cd5a][00001362]
> >>>>>> e840feffff call 000011a2 // call H Local Halt Decider:
> >>>>>> Infinite Recursion Detected Simulation Stopped
> >>>>>>
> >>>>>> H sees that P is calling the same function from the same
> >>>>>> machine address with identical parameters, twice in sequence.
> >>>>>> This is the infinite recursion (infinitely nested simulation)
> >>>>>> non-halting behavior pattern.
> >>>>>>
> >>>>>> ...[00001384][0010229e][00000000] 83c408 add esp,+08
> >>>>>> ...[00001387][0010229a][00000000] 50 push eax
> >>>>>> ...[00001388][00102296][00000423] 6823040000 push 00000423 //
> >>>>>> "Input_Halts ="
> >>>>>> ---[0000138d][00102296][00000423] e8e0f0ffff call 00000472 //
> >>>>>> call Output Input_Halts = 0
> >>>>>> ...[00001392][0010229e][00000000] 83c408 add esp,+08
> >>>>>> ...[00001395][0010229e][00000000] 33c0 xor eax,eax
> >>>>>> ...[00001397][001022a2][00100000] 5d pop ebp
> >>>>>> ...[00001398][001022a6][00000004] c3 ret
> >>>>>> Number of Instructions Executed(15892) lines = 237 pages
> >>>>>>
> >>>>>>
> >>>>>> Halting problem undecidability and infinitely nested simulation
> >>>>>> (V5)
> >>>>>>
> >>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >>>>>>
> >>>>>
> >>>>> Your simulation approach is erroneous as there is no infinite
> >>>>> recursion; the key to understanding this is by realizing the
> >>>>> implication of the words "pass its own source" in the
> >>>>> following:
> >>>>
> >>>>
> >>>> YOU IGNORED THIS PART
> >>>>
> >>>> This proof can only be understood only by those having sufficient
> >>>> technical competence in:
> >>>> (a) software engineering - recognizing infinite recursion in
> >>>> C/x86 (b) the x86 programming language
> >>>> (c) the C programming language and
> >>>> (d) the details of how C is translated into x86 by the Microsoft
> >>>> C compilers.
> >>>
> >>> (a) I have been a software developer/engineer since 1993 and am
> >>> able to recognize infinite recursion and additionally, and more
> >>> importantly, a lack of infinite recursion.
> >>
> >> This has been disproven in that you did not see this:
> >>
> >> >>>> H sees that P is calling the same function from the same
> >> >>>> machine address with identical parameters, twice in
> >> >>>> sequence. This is the infinite recursion (infinitely nested
> >> >>>> simulation) non-halting behavior pattern.
> >
> > I am not talking about your braindead simulation, I am talking about
> > the halting problem proofs you are trying to refute: THEY DO NOT
> > HAVE AN INFINITE RECURSION.
> >
> > /Flibble
> >
>
> My proofs prove that they do.
> That you fail to comprehend this is no rebuttal at all.
Only your simulation contains an infinite recursion due to a category
error ON YOUR PART. Your category error is proof that your
simulation-based proof is in error.

/Flibble