Re: Everyone on this forum besides Keith has been a damned liar about this point

On 6/9/25 10:43 AM, olcott wrote:

On 6/9/2025 5:31 AM, Fred. Zwarts wrote:

Op 09.jun.2025 om 06:15 schreef olcott:

On 6/8/2025 10:42 PM, dbush wrote:

On 6/8/2025 11:39 PM, olcott wrote:

On 6/8/2025 10:32 PM, dbush wrote:

On 6/8/2025 11:16 PM, olcott wrote:

On 6/8/2025 10:08 PM, dbush wrote:

On 6/8/2025 10:50 PM, olcott wrote:

void DDD()
{
   HHH(DDD);
   return;
}
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The *input* to simulating termination analyzer HHH(DDD)

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No it's not, as halt deciders / termination analyzers work with algorithms,

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That is stupidly counter-factual.
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That you think that shows that

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My understanding is deeper than yours.
No decider ever takes any algorithm as its input.

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But they take a description/specification of an algorithm,

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There you go.
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which is what is meant in this context.

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It turns out that this detail makes a big difference.
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And because your HHH does not work with the description/ specification of an algorithm, by your own admission, you're not working on the halting problem.
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HHH(DDD) takes a finite string of x86 instructions
that specify that HHH simulates itself simulating DDD.

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And HHH fails to see the specification of the x86 instructions. It aborts before it can see how the program ends.
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 This is merely a lack of sufficient technical competence
on your part. It is a verified fact that unless the outer
HHH aborts its simulation of DDD that DDD simulated by HHH
the directly executed DDD() and the directly executed HHH()
would never stop running. That you cannot directly see this
is merely your own lack of sufficient technical competence.

And it is a verified fact that you just ignore that if HHH does in fact abort its simulation of DDD and return 0, then the behavior of the input, PER THE ACTUAL DEFINITIONS, is to Halt, and thus HHH is just incorrect.
Because HHH must be a program to even start the process to ask the question, its behavior for this input is fixed and established.
Of course, your problem is you are too ignorant to understand this fact, and because of your reckless disregard for truth, you end you just spouting baseless lies out of your INTENTIONAL ignorance.
That you refuse to understand this just shows your pathological ignorance.
Your logic includes a statement the equivalent of "If we assume that 1 is equal to 2".

 

There are more steps, I don't want to overwhelm you.
>

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Yes. For some incorrect reason the programmer of HHH assumes that the simulation of HHH does not halt, when he knows that HHH does halt.
He does not see that these to assumptions contradict each other.