Re: HP counter-example INPUT cannot possibly exist

Am Fri, 20 Jun 2025 11:39:40 -0500 schrieb olcott:

On 6/19/2025 3:12 AM, Mikko wrote:

On 2025-06-18 14:39:02 +0000, olcott said:

On 6/16/2025 3:21 AM, Mikko wrote:

On 2025-06-15 14:44:47 +0000, olcott said:

On 6/15/2025 4:19 AM, Mikko wrote:

On 2025-06-13 15:33:45 +0000, olcott said:

On 6/13/2025 5:37 AM, Mikko wrote:

As you respond to my question without answering it it is obvious
that you don't see any proofs in your article.

>
It is a fact that there is no actual input D to any termination
analyzer H that does the opposite of whatever value that H
derives. The key element that all conventional HP proofs depend on
cannot possibly exist.

>
Nonsense is not a fact.

>
After studying these things for 22 years I found that every
conventional proof of the halting problem never provides an actual
input that would do the opposite of whatever value that its partial
halt decider (PHD) returns.

Only if that non-decider simulates.

The core part of those proofs is a constructive specification of that
test case.

Yet no one ever noticed that the counter-example input cannot even be
constructed thus the proof itself never actually existed.

How can't it be constructed? You are certainly passing that input to HHH.

Depending on the style of the proof one can ither prove that the
counter example exists or that if a halting decider exists then the
caunter example exists, too, and otherwise none is needed.
 

No this is counter-factual.
It has never been possible for *AN ACTUAL INPUT* to do the opposite of
whatever value that it decider decides.

Do you think that no programs halt when HHH is prepended? Do you think
that HHH doesn't halt?

int main()
{
   DD(); // IS NOT AN ACTUAL INPUT TO THE
}       // HHH(DD) THAT THIS DD() CALLS.

Then explain why you are not passing it to HHH for fuck's sake.

In the C programming language it has always been impossible for the
caller of a function to be an argument to this called function.

No, C does not forbid recursion.

The finite string of x86 machine language that is passed as an argument
to HHH *is not exactly one and the same thing as the directly executed
DD*

DD does have the same "string". How else would you pass a program?

Intuition would tell you that the behavior must be the same, yet
empirical proof proves they are not the same.

No, the definitions tell us the behaviour is the same; your intuition
tells you HHH is correct.

--
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.