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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: DDD correctly emulated by H0 --- Why Lie?
Date: Wed, 26 Jun 2024 07:02:47 -0400
Organization: i2pn2 (i2pn.org)
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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.