Re: DDD correctly emulated by H0 --- Why Lie?

On 6/23/24 9:40 AM, olcott wrote:

_DDD()
[00002172] 55               push ebp
[00002173] 8bec             mov ebp,esp
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call HHH0
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
 According to the semantics of the x86 programming language
when DDD correctly emulated by H0 calls H0(DDD) this call
cannot possibly return.

According to the semantics of the x86 programming language, a "Correct emulation" of DDD must continue untul we reach the ret at 00002183, a fault occurs, or we hit a Halt instruction, as part of the behavior of every instruction is that the next instruction WILL be run.
Thus, the only HHH0 that can exist can not "abort" its simulation, and thus does not answer for this input.
If you want to alter the semantics to allow for PARTIAL simulation, then by the semantics of partial simulation, you can not talk about the future past the point of aborting (without otherwise proving the claim), and thus you can not say "can not possibly return", but "was not simulated to return", or "hasn't returned YET".
All such statements must allow for the fact that emulating further might occur by some bettert emulator.
Also, the very form of your question says that the code for HHH0 is INCLUDED in the input, or we can not actually emulate the call instruction itself.

 Likewise according to the semantics of arithmetic for
decimal integers: 2 + 3 = 5.

Exactly.

 Anyone disagreeing with these two statements is WRONG.
 

Right, so the ONLY HHH0 is the actual PURE EMULATOR which never answers.
Except of course for the HHH0 that begins as:
HHH0(ptr x) {
   static int flag = 0;
   if (flag) return 0;
...
That one (and similar) WILL be able to emulate its input to the final return.