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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: Wed, 10 Jul 2024 07:24:27 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <a43d2f6552ebd7301bacd9096088d453353bebe7@i2pn2.org>
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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.