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

On 6/9/2025 12:06 PM, olcott wrote:

On 6/9/2025 10:54 AM, dbush wrote:

On 6/9/2025 11:49 AM, olcott wrote:

On 6/9/2025 10:34 AM, dbush wrote:

On 6/9/2025 11:29 AM, olcott wrote:

On 6/9/2025 10:06 AM, dbush wrote:

On 6/9/2025 10:55 AM, olcott wrote:

On 6/9/2025 6:55 AM, dbush wrote:

On 6/9/2025 12:15 AM, olcott wrote:

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

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Which you stated only includes the instructions of the function DDD on multiple occasions (see below),

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It is proven that you are a liar by the part of
my reply that you erased.
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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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Then you admit that that finite string includes the machine code of the function DDD, the machine code of the function HHH, and the machine code of everything that HHH calls down to the OS level, and that address 000015c3 is part of DDD?

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I admit that:
(a) DDD correctly simulated by HHH,
(b) the directly executed DDD() and
(c) the directly executed HHH()
WOULD NEVER STOP RUNNING UNLESS
HHH ABORTS ITS SIMULATION OF DDD.
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Because this is true it derives conclusive proof
that the input to HHH(DDD) specifies a non-halting
sequence of configurations.
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That people here disagree with self-evident truth
seems to indicate that people here are liars.
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In epistemology (theory of knowledge), a self-evident
proposition is a proposition that is known to be true
by understanding its meaning without proof...
https://en.wikipedia.org/wiki/Self-evidence
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In other words, a non-answer.  I'll take that as a no.
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And since your HHH doesn't work with algorithms (or their description / specification) as you've admitted, you're not working on the halting problem.
>

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You are far too sloppy in your interpretation of the
meaning of words. Also when I do provide an answer
you simply ignore it.

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Replying with something other than "yes" or "no" to a yes or no question is not an answer.
>

 By replying to a yes or no question with the full
and complete justification forces the respondent
to look more deeply into these things than simply
dismissing a view out-of-hand without review.

But by not including the yes or no you dishonestly dodge the question.
An answer of "no" means you're not working with algorithms, and an answer of "yes" means you're changing the input.
Either way, you're wrong.

 That you are incapable of of analyzing such an answer
proves that you are not competent to review my work.

The only thing proven is that you'll dishonestly dodge direct questions that show you wrong, and that you're not actually working on the halting problem.
If you would just be honest about that fact that you're not working on the halting problem people would stop bothering you.

 

The input to HHH(DDD)

Which only consists of the the instructions of the function DDD, as you have admitted, you're not working with algorithms (or their description / specification) and therefore not working on the halting problem.
 If you would just quit lying about that people might take you seriously.