Re: HHH(DDD)==0 is correct

On 7/4/25 6:24 PM, olcott wrote:

On 7/4/2025 3:40 PM, Richard Damon wrote:

On 7/4/25 2:25 PM, olcott wrote:

On 7/4/2025 1:14 PM, Richard Damon wrote:

On 7/4/25 10:43 AM, olcott wrote:

On 7/4/2025 8:23 AM, Richard Damon wrote:

On 7/4/25 8:50 AM, olcott wrote:

On 7/4/2025 2:35 AM, Mikko wrote:

On 2025-07-03 12:56:42 +0000, olcott said:
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On 7/3/2025 3:57 AM, Mikko wrote:

On 2025-07-03 02:50:40 +0000, olcott said:
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On 7/1/2025 11:37 AM, Mr Flibble wrote:

On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
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On 6/30/25 2:30 PM, Mr Flibble wrote:
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PO just works off the lie that a correct simulation of the input is
different than the direct execution, even though he can't show the
instruction actually correctly simulated where they differ, and thus
proves he is lying.
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The closest he comes is claiming that the simulation of the "Call HHH"
must be different when simulated then when executed, as for "some
reason" it must be just because otherwise HHH can't do the simulation.
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Sorry, not being able to do something doesn't mean you get to redefine
it,
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You ar4e just showing you are as stupid as he is.

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No. A simulator does not have to run a simulation to completion if it can
determine that the input, A PROGRAM, never halts.
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/Flibble

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The most direct way to analyze this is that
HHH(DDD)==0 and HHH1(DDD)==1 are both correct
because DDD calls HHH(DDD) in recursive simulation and
DDD does not call HHH1(DDD) in recursive simulation.

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Either "no" (encoded as 0) or "yes" (encoded as any other number) is the
wrong asnwer to the quesstion "does DDD specify a halting computation?".

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That is *not* the actual question.

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THe actual question is whatever someone asks.

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What is the area of a square circle with a radius of 2?
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However, if the question is
not "does DDD specify a halting computation?" or the same about some
other computation then it is not in the scope of the halting problem
or the termination problem.
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The halting problem has always been flatly incorrect
by making that the question. So I am reframing the
question the same way that ZFC reframed Russell's Paradox.

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Nope, just shows you are too stupid to understand it.
>

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Then tell me where I go wrong on this explanation:
ZFC conquered Russell's Paradox by redefining how
sets are defined such that a set that is a member
of itself can no longer be defined.
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"ZFC avoids this paradox by using axioms that restrict set formation."

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And what does that distraction have to do with halting problem?
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*Changing the definition of the problem is a way to solve it*
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But you aren't allowed to do that.
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Note, ZFC doesn't solve the problem of Russell's Paradox in Naive Set Theory, as it doesn't do anything to Naive Set Theory.
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 It replaced the erroneous naive set theory thus
conquering the misconception of Russell's Paradox.

Yes, by creating a totally new system.
Note, The don't say that Russell's Paradox no longer exists in Naive Set Theory, they

 Likewise I am conquering the misconception that
partial halt deciders must report on the behavior
of directly executed Turing machines.
 

But you aren't making a totally new system, just lying about the existing system. In Computability Theory, reporting on the behavior of the direct execution of a Turing Machine is a valid operation. To say it isn't is just a lie.
You are not making a new system as you say that in Computability Theory, the Halting Problem is wrong, but then here you imply a claim you are making a new theory (which you haven't actually worked to define) by your false comparison to ZFC and thus not doing anything in the classical Computability Theory,
You are just showing a fundamental confusion of how logic theories work, showing how stupid you are.