Re: not identical deciders

Am Thu, 25 Jul 2024 08:56:37 -0500 schrieb olcott:

On 7/24/2024 10:29 PM, Mike Terry wrote:

On 23/07/2024 14:31, olcott wrote:

On 7/23/2024 1:32 AM, 0 wrote:

On 2024-07-22 13:46:21 +0000, olcott said:

On 7/22/2024 2:57 AM, Mikko wrote:

On 2024-07-21 13:34:40 +0000, olcott said:

On 7/21/2024 4:34 AM, Mikko wrote:

On 2024-07-20 13:11:03 +0000, olcott said:

On 7/20/2024 3:21 AM, Mikko wrote:

On 2024-07-19 14:08:24 +0000, olcott said:
>

When we use your incorrect reasoning we would conclude that
Infinite_Loop() is not an infinite loop because it only
repeats until aborted and is aborted.

(b) We know that a decider is not allowed to report on the
behavior computation that itself is contained within.

No, we don't. There is no such prohibition.

If a Turing machine can take a description of a TM as its input or as
a part of its input it can also take its own description.
Every Turing machine can be given its own description as input but a
Turing machine may interprete it as something else.
>

In this case we have two x86utm machines that are identical except
that DDD calls HHH and DDD does not call HHH1.

So they are not identical.

It is empirically proven that this changes their behavior and the
behavior of DDD.

It is empirically proven according to the semantics of the x86 machine
code of DDD that DDD correctly emulated by HHH has different behavior
than DDD correctly emulated by HHH1.

So they can't be identical.

If you care study the code that I just provided
you can see that when DDD is correctly emulated by HHH that DDD does
correctly have different behavior than DDD correctly emulated by HHH1.

HHH1 and HHH are essentially identical and the only reason why DDD
correctly emulated by HHH has different behavior than DDD correctly
emulated by HHH1 is that DDD does call HHH in recursive emulation and
DDD does not call HHH1 in recursive emulation.

When DDD calls HHH in recursive emulation (as I have proven that it
does**) and DDD does not call HHH1 in recursive simulation (as I have
proven that it does not**)
then DDD will have different behavior.

I don't believe that someone of your intelligence and knowledge could
possibly actually fail to understand that the behavior of function DDD
emulated by a function that it calls in recursive emulation would not be
different than the behavior of this same function DDD emulated by a
function that it does not call in recursive emulation.

HHH does see recursive emulation that will never stop unless aborted.
HHH1 does not see this.

HHH does see recursive emulation that will never stop unless aborted.
HHH1 does not see this, because HHH has already aborted its DDD. When
any one call of infinite recursion has been aborted then it is infinite
recursion that abnormally terminates.

One function calls itself, one doesn't.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.