Re: Who knows that DDD correctly simulated by HHH cannot possibly reach its own return instruction final state? BUT ONLY that DDD

On 8/4/24 2:54 PM, olcott wrote:

On 8/4/2024 1:45 PM, Richard Damon wrote:

On 8/4/24 8:35 AM, olcott wrote:

On 8/4/2024 6:12 AM, Richard Damon wrote:

On 8/3/24 11:00 PM, olcott wrote:

On 8/3/2024 9:56 PM, Richard Damon wrote:

On 8/3/24 7:36 PM, olcott wrote:

On 8/3/2024 5:51 PM, Richard Damon wrote:

On 8/3/24 6:15 PM, olcott wrote:

On 8/3/2024 5:07 PM, Richard Damon wrote:

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The problem is that every one of those emulation is of a *DIFFERENT* input, so they don't prove anything together except that each one didn't go far enough.

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void DDD()
{
   HHH(DDD);
   return;
}
>
When each HHH correctly emulates 0 to infinity steps of
its corresponding DDD and none of them reach the "return"
halt state of DDD then even the one that emulated infinite
steps of DDD did not emulate enough steps?
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Just says lying YOU.
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You got any source for that other than yourself?
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It is self-evident and you know it. I do have four
people (two with masters in CS) that attest to that.
*It is as simple as I can possibly make it*

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Maybe to your mind filled with false facts, but it isn't true.
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I wonder how you think that you are not swearing your
allegiance to that father of lies?

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Because, I know I speak the truth.
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Why do you not think you are lying?
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Anyone that truly understands infinite recursion knows
that DDD correctly simulated by HHH cannot possibly reach
its own "return" final state.

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Right, but for every other HHH, which the ones that answer are, it isn't a fact.
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Surpisingly (to me) Jeff Barnett set the record straight
on exactly what halting means.
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No, there is one, and only one definition, it is a machine that reaches its final state.
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Note, *a machine*, not a (partial) emulation of the machine
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You already know that a complete emulation of a non-ending
sequence is impossible and you already acknowledged that
DDD emulated by HHH that never aborts is non-ending.
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WHy do you say it is impossible, it just takes forever,

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A complete emulation is after all of the instructions have been
emulated. That never happens with any infinite execution.
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But by never aborting, it never gives up, and thus is completely correct.
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 The bottom line here is that you are requiring the impossible
thus making you necessarily incorrect.

Nope. A correct emulation of a non-halting input just is non-halting itself.
There IS a correct answer about halting for EVERY possible input (and thus a correct answer for Each version of DDD for each version of HHH).
The fact that none of the HHH's give that answer doesn't mean the question doesn't have one, just that it turns out to be uncomputable, so for ever decider there exists an input that it doesn't get correct.
There is no issue with problems being uncomputable, as it can be shown that there MUST be uncomputable problems.

 Far worse than that you seem to be dishonest by never
admitting the truth that DDD correctly emulated by any
HHH cannot possibly reach its own "return" instruction.

But the HHH that answers DOESN'T correctly emulate its input (the DDD built on it)
The only HHH that correctly emulates the DDD built on it is the one that never aborts, and thus never answer, and thus fails to be a decider.

 

Yes, it never "finishes" but you can't finish an infinite task, but since all countable infinities are the same, it "reaches" "completion" at the same point the input does.
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Now, it can't ever give an answer after completing, since that completion is at infinity, but that isn't the problem for a correct emulator, only for a decider.
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All this shows is that correct (and complete) emulation can not be the method used to halt decide.

 Yet again being dishonest by requiring a complete simulation
of an infinite computation.