Re: DDD correctly emulated by EEE --- Correct Emulation Defined

On 3/22/25 10:52 AM, 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 EEE(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 finite integer N instructions of the above x86
machine language DD are emulated by each x86 emulator
EEE[N] at machine address [000015c3] according to the
semantics of the x86 language no DD ever reaches its own
"ret" instruction at machine address [00002155] and
terminates normally.
 

Your can't emulate the above code for N > 4, as you get into undefine memory.
If EEE is defined to be a correct emulator, then yes, the correct emulation of this will not halt, but then EEE is not a correct halt decider either.
Note, the "proof program" that shows the decider is wrong is a specific program for each decider that we want to prove wrong, so that one shows that a correct emulator is not a correct halt decider.
Since none of your HHH's are that correct emulator (at least not if you claim they return an answer) this behavior is irrelevent, as you looked at the wrong program. DD needs to call the EXACT code of the decider you want to claim to be correct, and that code is consider part of the input.
Thus, you are just showing the fraud in your arguement, as you seem to be saying that EEE and HHH can both reside at the same address and be different programs, which is, of course, impossible.
Sorry, you are just proving your utter stupidity.