Re: Proof that H(D,D) meets its abort criteria --Categorically Exhaustive Reasoning--

On 16/03/24 05:15, olcott wrote:

On 3/15/2024 10:57 PM, immibis wrote:

On 16/03/24 04:29, olcott wrote:

On 3/15/2024 10:17 PM, immibis wrote:

On 16/03/24 03:18, olcott wrote:

It means that every H(D,D) that correctly determines by a partial
simulation of its input: "H correctly simulates its input D until"
that "D would never stop running unless aborted" that this H(D,D)
<is> necessarily correct.

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And what does that mean?

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saying non-halting as it now is looking at a TOTALLY new set of input.
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You don't seem very "Exhaustive" in your reasoning.
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// The categorically exhaustive part
For every H(D,D) of the infinite set of encodings of H
that simulate their input

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And what does that mean?
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I am trying to make my words as clear as possible so reviews
like this one are very helpful.
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There cannot possibly exist any H(D,D) that is called by
D where H(D,D) simulates its input and D(D) stops running
and H never aborts its simulation.

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And what does THAT mean? What is "any H(D,D)"?
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 *software engineering and set theory*
Every C function named H(D,D) that simulates its input and
D(D) calls H(D,D) never stops running unless aborted by H.

"D(D) calls H(D,D) never stops running unless aborted by H." seems to be a nonsensical sentence.
C functions cannot be named H(D,D) because ( , and ) are illegal characters in C function names.

I think you are probably using words in a different way than everyone else so it would be nice to understand your words as well as I understand Turing machines.