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

On 2/19/2025 4:55 AM, Fred. Zwarts wrote:

Op 18.feb.2025 om 17:48 schreef olcott:

On 2/18/2025 8:11 AM, Fred. Zwarts wrote:

Op 18.feb.2025 om 14:37 schreef 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. Perhaps your technical skill is much more woefully
deficient than I ever imagined.
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.
>

The point Olcott misses is that if the non-terminating HHH is changed
to abort the simulation, the program is changed. He does not
understand that a modification of a program makes a change. Such a
change modifies the behaviour of the program. The non-termination
behaviour has disappeared with this change and only remains in his
dreams. After this change, the simulation would terminate normally
and HHH should no longer abort. But it does, because the code that
detects the 'special condition' has a bug, which makes that it does
not see that the program has been changed into a halting program.

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

It is not true that this point has never been addressed. Olcott ignores
it when it is addressed.
What is the point? Even if HHH fails to simulate the halting program DD
up to the end because it is logically impossible for it to complete the
simulation, it still fails.

 
It fails In the same way that every CAD system will never correctly
represent a geometric circle that has four equal length sides in the
same two dimensional plane.

If the logically impossible cannot be done,
we can admit that HHH's simulation fails to complete the impossible
task.
So, why is Olcott trying to fix the logically impossible? He could as
well try to draw a square circle.