Re: Failure to meet this challenge proves that all of my reviewers are wrong

Am Mon, 03 Mar 2025 13:46:30 -0600 schrieb olcott:

On 3/3/2025 6:46 AM, Mikko wrote:

On 2025-03-02 19:06:52 +0000, olcott said:

On 3/2/2025 12:38 PM, Fred. Zwarts wrote:

Op 02.mrt.2025 om 15:42 schreef olcott:

HHH is an emulating termination analyzer that emulates the x86
machine code located at the address of a function using a fully
functional x86 emulator.
When HHH recognizes a non-terminating pattern in the execution trace
of its emulated input it aborts this emulation and returns 0.
I challenged everyone here to provide the machine address by machine
address (AKA line by line) execution trace of DD correctly emulated
by HHH that reaches its own "ret" instruction.

Olcott could as well challenge everyone to draw a correct square
circle.
>

No one made any attempt to do this because they know that this would
prove that they are stupidly wrong to say that my trace is
incorrect.

No one will attempt to draw a square circle. Does that imply that it
is wrong to say that another failed attempt to draw a square circle
is incorrect?
Similarly, HHH cannot possibly simulate itself correctly.

This C code conclusively proves that HHH does correctly emulate self
emulating DD correctly.
https://github.com/plolcott/x86utm/blob/master/Halt7.c

 
A C code does not prove. Only a proof would prove.
 

The C code proves exactly these things.
https://github.com/plolcott/x86utm/blob/master/Halt7.c The above code
proves that:
(a) HHH correctly emulates itself emulating DD.

No, it simulates HHH1 simulating DD.

(b) DD correctly emulated by HHH cannot possibly reach its own "ret"
instruction and terminate normally.
(c) The behavior of the input to HHH(DD) is different than the behavior
of the directly executed DD

Then DD is not the input.

because DD calls HHH(DD) in recursive
emulation and the directly executed DD does not call HHH(DD) in
recursive emulation.

The other way around.

This code proves that it reports integers > 5
void GT5(int X)
{
   if (X > 5)
     printf("X is greater than five\n");
}

This code proves nothing (although it does "report integers >5").

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.