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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: How do simulating termination analyzers work? ---Truth Maker
 Maximalism
Date: Fri, 4 Jul 2025 19:09:01 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 04 Jul 2025 13:37:25 -0500 schrieb olcott:
> On 7/4/2025 1:23 PM, joes wrote:
>> Am Fri, 04 Jul 2025 12:30:43 -0500 schrieb olcott:
>>> On 7/4/2025 8:37 AM, joes wrote:
>>>> Am Fri, 04 Jul 2025 07:16:23 -0500 schrieb olcott:
>>>>> On 7/4/2025 3:55 AM, joes wrote:

>>>>>> You are effectively saying that all programs that start with a call
>>>>>> to HHH are the same.
> The nesting is too deep to see what you are responding to.
Lol, you could have responded immediately. You know how to look up posts.

>>>> Yes it is, HHH should compute whether the code of DD halts when run.
>>>> You can't be thinking that is uncomputable.
>>> Likewise we should also compute the area of a square circle with a
>>> radius of 2.
>> Are you seriously suggesting that you can't compute what the code of
>> DDD does when executed?
Don't complain later.

>>> Partial halt deciders have never been allowed to report on the
>>> behavior of any directly executed Turing machine. Instead of this they
>>> have used the behavior that their input machine description specifies
>>> as a proxy.
>> And you think that DDD's direct execution is not specified by its
>> description?
> I HAVE PROVEN THAT DDD CORRECTLY SIMULATED BY HHH DOES NOT HAVE THE SAME
> BEHAVIOR AS DDD() THOUSANDS OF TIMES IN THE LAST THREE YEARS
No disagreement; not my question.

>>> Now for the first time we see that DDD correctly simulated by HHH *IS
>>> NOT A PROXY* for the behavior of the directly executed DDD().
>> Indeed, HHH does not simulate it correctly. (You can't mean that DDD is
>> *executed* incorrectly.)
> You are using the wrong measure of correct.
So DDD specifies at least two different behaviours?

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.