Re: My reviewers think that halt deciders must report on the behavior of their caller

On 6/6/2025 2:30 PM, joes wrote:

Am Fri, 06 Jun 2025 14:21:07 -0500 schrieb olcott:

On 6/6/2025 2:18 PM, joes wrote:

Am Fri, 06 Jun 2025 12:32:56 -0500 schrieb olcott:

On 6/6/2025 3:19 AM, Fred. Zwarts wrote:

Op 05.jun.2025 om 18:03 schreef olcott:

On 6/5/2025 2:48 AM, Mikko wrote:

On 2025-06-04 15:50:25 +0000, olcott said:

 

The only possible way that HHH can report on the direct execution
of DDD() is for HHH to report on the behavior of its caller:

>
The relevant question is not what HHH can report but what it does
and what it is required. DDD() is known to halt so HHH(DDD) is
required to report that it halts. But HHH(DDD) does not report so.
>

The only DDD that is known to halt is the DDD that calls HHH(DDD).
HHH(DDD) IS NOT ACCOUNTABLE FOR THE BEHAVIOR OF ITS CALLER.
>

Counterfactual. It has nothing to do with the caller. World-class
simulators show that the exact same input halts.

>
You are incorrectly calling it an *INPUT* when it never was an actual
*INPUT* it was always a *NON-INPUT CALLER*
People have made this same stupid mistake for 90 years.

The thing is, DDD is both,

>
int main()
{
    DDD; // calls HHH(DDD)
}
>
*The input to HHH IS NOT ITS CALLER*

 DDD is DDD.
 

DDD correctly emulated by HHH
has a different sequence of steps than
DDD correctly emulated by HHH1
I have proven this many times yet it is simply
too difficult for anyone here to understand.

by construction. Data can be code. The contradiction is exactly that
HHH's return value makes the simulation
of DDD wrong. You can't sidestep that by saying they are different
DDD's.

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer