Re: How do simulating termination analyzers work? ---Truth Maker Maximalism

Am Thu, 03 Jul 2025 17:11:45 -0500 schrieb olcott:

On 7/2/2025 1:53 AM, Mikko wrote:

On 2025-07-01 11:46:11 +0000, olcott said:

On 7/1/2025 2:51 AM, Mikko wrote:

On 2025-06-30 17:49:20 +0000, olcott said:

On 6/30/2025 3:14 AM, Mikko wrote:

On 2025-06-29 14:04:43 +0000, olcott said:

On 6/29/2025 3:42 AM, Mikko wrote:

On 2025-06-28 13:28:04 +0000, olcott said:

On 6/28/2025 6:47 AM, Mikko wrote:

It is relevant to the halting problem because no input to a halt
decider can possibly do the opposite of whatever its halt decider
decides. The thing that does the opposite is not an input.

You are effectively saying that all programs that start with a call
to HHH are the same.

It is irrelevant because the halting problem clarly states that the
input is a description of a Turing machine and an input to that
machine. You may say that to decide halting of a directly executed
Turing machnie is not possible from the given input but the problem is
what it is.
 

Although it is called a description that term is inaccurate.
It leads people to believe that 98% of exactly what it does is close
enough. That DD() *DOES NOT DO* what DD correctly simulated by HHH does
is a key detail *THAT ALWAYS ESCAPES THEM*

WTH? It is rather obvious that HHH cannot simulate DDD or anything else
that calls HHH the same way as that input when run "directly".

It is actually has 100% of all of the details that the machine code of
DD has. The input to HHH(DD) *SPECIFIES*
100% of every detail of the exactly behavior *OF THIS INPUT*

Yeah, and you can also execute that code instead of simulating it.

--
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.