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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: DDD correctly emulated by HHH cannot possibly halt
Date: Tue, 9 Jul 2024 22:51:21 -0400
Organization: i2pn2 (i2pn.org)
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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

If HHH "correctly" emulatates only a finite number of steps, and then 
stops, it did NOT ACTUALLY CORRECTLY emulate the input, as one part of 
the definition of all the instructions it saw was that the next 
instruction in sequence WILL be run.

Thus, an only PARTIAL emulation does not reveal the full behaior of the 
input, and says nothing about what happens in the behavior of the input 
after that point.

If that is what HHH does, as you claims make it clear is what actually 
happens, then the actual CORRECT behavior of DDD (that was partially 
emulated by HHH) will see that HHH return to it (but HHH will not see 
that happen) and thus it will return.

Yes, if your HHH actually does a COMPLETE correct emulation as you 
statement above seems to say, then DDD will not return, but HHH will 
never answer, and thus never reported on the behavior (as pure functions 
can only report by their return value)