Re: V5 --- Professor Sipser

Am Wed, 21 Aug 2024 23:22:11 -0500 schrieb olcott:

On 8/21/2024 10:35 PM, Richard Damon wrote:

On 8/21/24 11:26 PM, olcott wrote:

On 8/21/2024 9:06 PM, Richard Damon wrote:

On 8/21/24 9:55 PM, olcott wrote:

On 8/21/2024 8:45 PM, Richard Damon wrote:

On 8/21/24 9:23 PM, olcott wrote:

On 8/21/2024 7:00 PM, Richard Damon wrote:

On 8/21/24 8:30 AM, olcott wrote:

On 8/21/2024 3:01 AM, Mikko wrote:

On 2024-08-21 03:01:38 +0000, olcott said:

It is crucial to the requirements in that it specifies that H is
required to predict (a) The behavior specified by the finite
string D

Which must include *ALL* of the code of the PROGRAM D, which
includes ALL the code of everything it calls, which includes H,
so with your system, changing H gives a DIFFERENT input, which is
not comparable in behavior to this input.

(d) This includes H simulating itself simulating D

Note, that is the emulation of this exact input, including D
calling the ORIGINAL H, not changing to the Hypothetical, since
by the rules of the field, the input is a fixed string, and fully
defines the behavior of the input.

The fact that you don't understand DOES make you stupid. I don't say
you are wrong because you are stupid, you are wrong because the words
you use don't mean what you think they do, and thus your conclusions
are just incorrect.
That you seem to NEVER LEARN is what makes you stupid.

Professor Sipser clearly agreed that an H that does a finite
simulation of D is to predict the behavior of an unlimited
simulation of D.

Right, H needs to predict in a finite number of steps, what an
unlimited simulation of this EXACT input, which means that it must
call the H that you claim to be getting the right answer, which is
the H that does abort and return non-halting.

OK then you seem to have this correctly, unless you interpret this as
a self-contradiction.

Why do you think it could be a self-contradiction?
It is an impossiblity for H to correctly do it, but that is why the
Halting Problem is non-computable.

THIS EXACTLY MATCHES THE SIPSER APPROVED CRITERIA The finite HHH(DDD)
emulates itself emulating DDD exactly once and this is sufficient for
this HHH to predict what a different HHH(DDD) do that never aborted its
emulation of its input.

That other HHH still has to simulate the HHH that aborts.

--
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.