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From: joes <noreply@example.com>
Newsgroups: comp.theory,
Subject: Re: Simplified proof that DDD correctly simulated by HHH does not
 halt
Date: Sun, 9 Jun 2024 17:59:51 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sun, 09 Jun 2024 11:07:19 -0500 schrieb olcott:
> typedef void (*ptr)(); // pointer to void function
> 01   void HHH(ptr P, ptr I)
> 02   {
> 03     P(I);
> 04     return;
> 05   }
> 06 07   void DDD(int (*x)())
> 08   {
> 09     HHH(x, x);
> 10     return;
> 11   }
> 12 13   int main()
> 14   {
> 15     HHH(DDD,DDD);
> 16   }
> 17

> In the above Neither DDD nor HHH ever reach their own return statement
> thus never halt.
> Most of my reviewers incorrectly believe that when HH(DD,DD) aborts its
> simulated input that this simulated input halts.
If HH aborts the simulation of DD(DD), the call to itself also does, and
returns, making the whole chain terminate. Therefore the abortion was
premature.

> When HHH is a simulating halt decider just like HH then DDD correctly
> simulated by HHH cannot possibly terminate normally
HHH must be the same as HH. There is no distinction by simulation level.
That's the whole point of the argument.

>> When HHH is a simulating halt decider then HHH sees that DDD
>> cannot possibly reach its return statement, AKA
>>     simulating halt decider HHH simulates its input DDD
>>     until HHH determines that DDD would never
>>     stop running unless aborted
and aborts it.

-- 
joes