Sujet : Re: DDD correctly emulated by HHH cannot possibly halt
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logicDate : 10. Jul 2024, 12:24:27
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <a43d2f6552ebd7301bacd9096088d453353bebe7@i2pn2.org>
References : 1 2 3
User-Agent : Mozilla Thunderbird
On 7/9/24 11:08 PM, olcott wrote:
On 7/9/2024 9:51 PM, Richard Damon wrote:
On 7/9/24 7:49 PM, olcott wrote:
>
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ steps of DDD can't make it
past the above line of code no matter what.
>
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
>
>
Nope, you have a problem with your definitons
>
No the problem is your ADD has again caused
you to not pay close enough attention. The
above version has always been airtight since
the first time that I wrote it.
As a submarine with a screen door.
You are just proving you don't know what you are talking about.
The emulation of DDD by HHH can't make it there, but the DDD that was emulated only a finite number of steps by HHH will, after the HHH aborts its emulation and returns to its caller (which was DDD).
You just don't understand the difference between Reality and the observation of it, which is why you confuse Truth with Knowledge.
Any HHH that only emulates a finite number of instructions and then stops does NOT do a fully correct emulation, since every instruction it emulated includes the property that the next instruction WILL run, and thus needs to be emulated, and thus doesn't get to see the full behavior of the input.
The part it misses is the difference.