Re: DD emulated by HHH cannot possibly terminate normally --- x86 code

On 3/1/25 7:36 PM, olcott wrote:

On 3/1/2025 4:08 PM, Fred. Zwarts wrote:

Op 01.mrt.2025 om 22:12 schreef olcott:

On 3/1/2025 2:22 PM, Fred. Zwarts wrote:

Op 01.mrt.2025 om 20:01 schreef olcott:

On 3/1/2025 10:05 AM, Richard Damon wrote:

On 3/1/25 9:41 AM, olcott wrote:

On 3/1/2025 6:49 AM, Richard Damon wrote:

On 2/28/25 7:47 PM, olcott wrote:

_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 we hypothesize that the code at machine address
0000213c is an x86 emulator then we know that DD
remains stuck in recursive emulation and cannot possibly
reach its own "ret" instruction and terminate normally.
>
When we add the additional complexity that HHH also
aborts this sequence at some point then every level
of recursive emulation immediately stops. This does
not enable any DD to ever reach its "ret" instruction.
>

>
But then you just negated your first assumption, as a partial emulator that aborts its emulation, then DD no longer gets stuck.
>

>
Cannot possibly reach its own "ret" instruction and terminate normally
proves non-termination whether aborted or not.
>
Cannot possibly reach its own "ret" instruction and terminate normally
proves non-termination whether aborted or not.
>
Cannot possibly reach its own "ret" instruction and terminate normally
proves non-termination whether aborted or not.
>
>

>
But it DOES terminate

>
DD emulated by HHH never terminates no matter how many
times you try to get away with the straw-man deception
of referring to anything at all besides
>
DD EMULATED BY HHH
DD EMULATED BY HHH
DD EMULATED BY HHH
DD EMULATED BY HHH
DD EMULATED BY HHH
DD EMULATED BY HHH

HHH emulating DD fails to reach the 'ret' instruction.

>
Because DD calls HHH(DD) in recursive emulation
thus non-termination is entirely the fault of DD.

No, DD calls HHH only once.

 When DD is correctly emulated by HHH and DD calls HHH(DD)
then DD calls HHH in recursive emulation.

ANd if HHH correctly emulates DD, then it, and all copies of it will never return, and thus HHH is not a decider.

 When HHH breaks this otherwise endless chain every level
of DD immediately stops with none of them ever reaching
their "ret" instruction.

It then fails to meet the above requirements, and thus your logic done above is incorrect.
Your logic is just based on a isomophism to the Liar's Paradox.

 It is not that hard when you understand these things
and impossible when you don't.
 

No, it isn't hard to see, and what is clear is that you are just an ignorant pathological lying idiot that doesn't care about the truth.