Re: DD specifies non-terminating behavior to HHH --- ONE POINT AT A TIME

On 2/18/25 11:34 AM, olcott wrote:

On 2/18/2025 7:48 AM, joes wrote:

Am Tue, 18 Feb 2025 07:37:54 -0600 schrieb olcott:

On 2/18/2025 6:25 AM, Richard Damon wrote:

On 2/18/25 6:26 AM, olcott wrote:

On 2/18/2025 3:24 AM, Mikko wrote:

On 2025-02-17 09:05:42 +0000, Fred. Zwarts said:

Op 16.feb.2025 om 23:51 schreef olcott:

On 2/16/2025 4:30 PM, joes wrote:

Am Sun, 16 Feb 2025 15:58:14 -0600 schrieb olcott:

On 2/16/2025 2:02 PM, joes wrote:

Am Sun, 16 Feb 2025 13:24:14 -0600 schrieb olcott:

On 2/16/2025 10:35 AM, joes wrote:

Am Sun, 16 Feb 2025 06:51:12 -0600 schrieb olcott:

On 2/15/2025 2:49 AM, Mikko wrote:

On 2025-02-14 12:40:04 +0000, olcott said:

On 2/14/2025 2:58 AM, Mikko wrote:

On 2025-02-14 00:07:23 +0000, olcott said:

On 2/13/2025 3:20 AM, Mikko wrote:

On 2025-02-13 04:21:34 +0000, olcott said:

On 2/12/2025 4:04 AM, Mikko wrote:

On 2025-02-11 14:41:38 +0000, olcott said:

>

DD  correctly simulated by HHH cannot possibly terminate
normally.

>
That claim has already shown to be false. Nothing above
shows that HHH does not return 0. If it does DD also returns
0.

When we are referring to the above DD simulated by HHH and
not trying to get away with changing the subject to some
other DD somewhere else

such as one that calls a non-aborting version of HHH
>

then anyone with sufficient knowledge of C programming knows
that no instance of DD shown above simulated by any
corresponding instance of HHH can possibly terminate
normally.

Well, then that corresponding (by what?) HHH isn’t a decider.

I am focusing on the isomorphic notion of a termination
analyzer.

(There are other deciders that are not termination analysers.)
>

A simulating termination analyzer correctly rejects any input
that must be aborted to prevent its own non-termination.

Yes, in particular itself is not such an input, because we
*know* that it halts, because it is a decider. You can’t have
your cake and eat it too.

I am not even using the confusing term "halts".
Instead I am using in its place "terminates normally".
DD correctly simulated by HHH cannot possibly terminate normally.

What’s confusing about „halts”? I find it clearer as it does not
imply an ambiguous „abnormal termination”. How does HHH simulate
DD terminating abnormally, then? Why doesn’t it terminate
abnormally itself?
You can substitute the term: the input DD to HHH does not need to
be aborted, because the simulated decider terminates.
>

Every simulated input that must be aborted to prevent the
non-termination of HHH is stipulated to be correctly rejected by
HHH as non-terminating.
>

A very strange and invalid stipulation.

>
It merely means that the words do not have their ordinary meaning.
>

Unless HHH(DD) aborts its simulation of DD itself cannot possibly
terminate normally. Every expert in the C programming language can see
this. People that are not experts get confused by the loop after the
"if" statement.
>

So? Since it does that, it needs to presume that the copy of itself it
sees called does that.
>

Not at all.

I mean, this is a deterministic program without any static variables,
amirite?
>

 When I focus on one single-point:
[D simulated by HHH cannot possibly terminate normally]
I get two years of dodging and this point is never addressed.

And you thus miss the point that what the partial simulation by HHH does is irerelvent, except to your strawman.

 Since there is about a 7% chance that my very drastic cancer
treatment will kill me in the next 100 days I must insist on
100% perfectly and completely addressing this point before
moving on to any other points.

And thus the world has about  7% chance of ridding itself of a lying fraud soon.
You NEED to insist on using CORRECT definitions so you can redeam your reputation from all your lies.

 

Here is the point that you just missed Unless the first HHH that sees
the non-terminating pattern aborts its simulation none of them do
because they all have the exact same code.