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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1 ---
 STA
Date: Wed, 2 Apr 2025 22:35:10 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <912b0fe8dba03ab574ac0577f8eb754cabca4a5c@i2pn2.org>
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On 4/2/25 10:15 PM, olcott wrote:
> On 4/2/2025 8:52 AM, joes wrote:
>> Am Tue, 01 Apr 2025 21:37:40 -0500 schrieb olcott:
>>> On 4/1/2025 8:13 PM, Richard Damon wrote:
>>>> On 4/1/25 7:35 PM, olcott wrote:
>>>>> On 4/1/2025 5:36 AM, Richard Damon wrote:
>>>>>> On 3/31/25 10:19 PM, olcott wrote:
>>
>>>>>> But DDD doesn't prevent its own terminatation, as it calls an HHH
>>>>>> that WILL abort its emulation and return and answer.
>>>>>>
>>>>> You know that DDD stopping running and DDD reaching its final halt
>>>>> state are not the same thing you damned liar.
>>>>>
>>>> Right, the DDD who's simulation is stopped hasn't shown non-halting
>>>> behavior, just not-yet-halted.
>>>>
>>> You already admitted that you are lying about this.
>>> DDD emulated by HHH for an infinite number of steps never reaches its
>>> final halt state.
>> *finite
>>
>>> HHH sees this in one recursive emulation of DDD.
>>>
>>> *Simulating termination analyzer Principle*
>>> It is always correct for any simulating termination analyzer to stop
>>> simulating and reject any input that would otherwise prevent its own
>>> termination. The only rebuttal to this is rejecting the notion that
>>> deciders must always halt.
>> It must also return the right value.
>>
> 
> By process of elimination and by the above criteria
> we can determine that not stopping the emulation
> causes HHH to never halt. This only leaves stopping
> the emulation that semantically entails that the
> input must be rejected.
> 

No, because HHH must do what HHH does.

You don't seem to understand that basic property of programs, they do 
what they are programed, and the building blocks of programming (for 
this set of theories) are fixed, fully defined, deterministic operations.

Since the "input" calls its copy of the decider (even if as an 
optimization you reuse the memory) you can't change that code without 
changing the input. You can give FULL input to another version of HHH, 
which will need to be in some other memory locations, which is what HHH1 
actually is.

Sorry, you are just showing that you don't understand what you are 
talking about.