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On 4/28/2025 2:10 PM, dbush wrote:FALSE:On 4/28/2025 3:03 PM, olcott wrote:>On 4/28/2025 1:37 PM, dbush wrote:>On 4/28/2025 2:32 PM, olcott wrote:>On 4/28/2025 11:11 AM, dbush wrote:>On 4/28/2025 12:10 PM, olcott wrote:>On 4/28/2025 4:05 AM, Mikko wrote:>On 2025-04-27 18:23:03 +0000, olcott said:>
>On 4/27/2025 4:51 AM, Mikko wrote:>On 2025-04-26 16:15:44 +0000, olcott said:>
>_DD()>
[00002133] 55 push ebp ; housekeeping
[00002134] 8bec mov ebp,esp ; housekeeping
[00002136] 51 push ecx ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404 add esp,+04
[00002144] 8945fc mov [ebp-04],eax
[00002147] 837dfc00 cmp dword [ebp-04],+00
[0000214b] 7402 jz 0000214f
[0000214d] ebfe jmp 0000214d
[0000214f] 8b45fc mov eax,[ebp-04]
[00002152] 8be5 mov esp,ebp
[00002154] 5d pop ebp
[00002155] c3 ret
Size in bytes:(0035) [00002155]
>
When any HHH emulates DD according to the finite
string transformation rules specified by the x86
language (the line of demarcation between correct
and incorrect emulation) no emulated DD can possibly
reach its final halt state and halt.
There is a type error above. First DD is introduced as a proper name.
But later it is used in the phrase "no emulated DD" where the rules
of the language require a generic name.
>
*This of this as an axiom schema*
No DD correctly emulated by any HHH can possibly
reach its final halt state. This conclusively
proves that every HHH is correct to reject its
input DD as non-halting.
That cannot be used as a schema before you specify what symbols in it are
placeholders and what replacements can be used for the placeholders.
>
I have gone over this many hundreds of times
do you not remember anything that I already said?
>
int DD()
{
int Halt_Status = EEE(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
>
When each element of the set of x86 emulators
named EEE
Changing the input is not allowed.
Hypothetical possibilities numbskull.
>
So you're hypothesizing changing the input.
>
Changing the input, hypothetically or otherwise, is not allowed.
Examining the infinite set of HHH/DD pairs simultaneously
IS NOT CHANGING THE INPUT DUMBO.
>
When DD runs, HHH runs. So HHH is part of the input. So when you hypothesize changing HHH, you're hypothesizing changing the input.
>
Changing the input, hypothetically or otherwise, is not allowed.
HHH[0] emulates zero instructions of DD
HHH[1] emulates one instructions of DD
HHH[n] emulates n instructions of DD
HHH[∞] emulates ∞ instructions of DD
Each DD has the exactly same machine codeFalse, because the algorithm DD consists of the machine code of the function DD, the function HHH, and the machine code of everything that HHH calls down to the OS level.
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