Re: How Partial Simulations correctly determine non-halting ---Ben's 10/2022 analysis

On 2024-06-09 13:59:57 +0000, olcott said:

On 6/9/2024 3:47 AM, Mikko wrote:

On 2024-06-08 13:04:14 +0000, olcott said:
 

On 6/8/2024 1:42 AM, Mikko wrote:

On 2024-06-07 22:26:05 +0000, olcott said:
 

On 6/7/2024 4:00 PM, joes wrote:

Am Fri, 07 Jun 2024 09:47:35 -0500 schrieb olcott:

On 6/7/2024 1:30 AM, Mikko wrote:

On 2024-06-06 15:31:36 +0000, olcott said:

On 6/6/2024 10:14 AM, Mikko wrote:

On 2024-06-06 13:53:58 +0000, olcott said:

On 6/6/2024 5:15 AM, Mikko wrote:

On 2024-06-05 13:29:28 +0000, olcott said:

On 6/5/2024 2:37 AM, Mikko wrote:

On 2024-06-04 18:02:03 +0000, olcott said:

 

A simulating halt decider cannot report on what the behavior of a
non-terminating input actually is because this would take forever.

Exactly. Didn't you say it is allowed to abort?
 

H is not allowed to report on any computation containing its actual self
because Turing machines can only take finite string inputs thus cannot
take Turing machines as inputs.

Bullshit. It can take other machines just fine. It doesn't know about
itself.
 

 No actual Turing machine can be the input to any other actual
Turing machine. Turing machines only take finite string inputs.

 Any finite string can be an input to some Turing machine.
Can you prove that a Turing machine is not a finite string?
 

 By definition Turing Machines are not finite strings in the
conventional model. In my x86utm model of computation x86
machine language <is> the input to another function written
in the x86 language.

 The definition does not say "Turing machine is not a finite string".

 The definition of puppy does not say it is not a fifteen
story office building.

Therefore you must use a different kind of argumentation if you want
to claim that a Turing machine is not a finite string.
--
Mikko