Re: HHH(DDD) computes the mapping from its input to HHH emulating itself emulating DDD --- anyone that says otherwise is a liar

Am Thu, 21 Nov 2024 15:19:43 -0600 schrieb olcott:

On 11/21/2024 3:11 PM, joes wrote:

Am Thu, 21 Nov 2024 09:19:03 -0600 schrieb olcott:

On 11/20/2024 10:00 PM, Richard Damon wrote:

On 11/20/24 9:57 PM, olcott wrote:

On 11/20/2024 5:51 PM, Richard Damon wrote:

On 11/20/24 5:03 PM, olcott wrote:

On 11/20/2024 3:53 AM, Mikko wrote:

On 2024-11-20 03:23:12 +0000, olcott said:

On 11/19/2024 4:12 AM, Mikko wrote:

On 2024-11-18 20:42:02 +0000, olcott said:

On 11/18/2024 3:41 AM, Mikko wrote:

The "the mapping" on the subject line is not correct. The
subject line does not specify which mapping and there is no
larger context that could specify that. Therefore it should
be "a mapping".
On 2024-11-17 18:36:17 +0000, olcott said:

 

My code is one example of the infinite set of every possible HHH
that emulates DDD according to the semantics of the x86 language.

Like all of them, it is unable to simulate DDD to its undeniable
halting state.

In your case you may simply not even understand what infinite recursion
is, thus cannot see the isomorphism.

But it gets the wrong answer for the halting problem, as DDD dpes
halt.

DDD emulated by HHH does not halt.

Whatever. DDD halts and HHH should return that.

IT IS NOT THE SAME INSTANCE OF DDD.

All instances of DDD behave the same (if it is a pure function and
the HHH called from it doesn't switch behaviour by a static variable).

HHH is reporting on the behavior of its INPUT.

And its input is (the encoding of) DDD.

HHH is not reporting on the behavior of some other different DDD
instance that behaves differently.

You mean the same DDD when it is not simulated?

I can make a "decider" this way: it simulates no steps at all and
returns that all inputs halt when simulated by it.

Then it is never wrong about its own result.

Are you a fake Christian?

I am a fake Scientologist.

That may be the correct Scientologist to be.

We can solve this conundrum by rejecting Scientologists as incorrect.

--
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.