Re: Every HHH(DDD) is correct to reject its input

On 6/5/25 11:56 AM, olcott wrote:

On 6/5/2025 2:21 AM, Fred. Zwarts wrote:

Op 04.jun.2025 om 17:00 schreef olcott:

On 6/4/2025 2:39 AM, Mikko wrote:

On 2025-06-02 05:12:26 +0000, olcott said:
>

On 6/1/2025 6:20 AM, Mikko wrote:

On 2025-05-31 19:21:10 +0000, olcott said:
>

On 5/31/2025 2:11 PM, Mr Flibble wrote:

Olcott is doing this:
>
int main()
{
DDD(); // DDD calls HHH
}
>
This is incorrect as it is a category (type) error in the form of
conflation of the EXECUTION of DDD with the SIMULATION of DDD: to
completely and correctly simulate/analyse DDD there must be no execution
of DDD prior to the simulation of DDD.
>
Olcott should be doing this:
>
int main()
{
HHH(DDD);
}

>
I would have left it there except that many dozens of
reviewers have pointed out that they believe that HHH
is supposed to report on the behavior of its caller.

>
A halt decider is required to report on the computation it is asked
about. There is no requirement that a halt decider knows or can find
out whether it is called by the program about which is required to
report. Consequently, whether the computaton asked about calls the
decider is irrelevant.

>
void DDD()
{
   HHH(DDD);
   return;
}
>
The *input* to simulating termination analyzer HHH(DDD)
specifies recursive simulation that can never reach its
*simulated "return" instruction final halt state*

>
If it does then the "input" is not DDD, which specifies a halting
behaviour if HHH is a decider.
>

>
You can say these things only by making
sure to ignore the verified facts.
>
HHH1(DDD) emulates DDD that calls HHH(DDD)
that emulates DDD that calls HHH(DDD)
that emulates DDD that calls HHH(DDD)

And a correct analysis of the code shows that this is only a finite recursion,

 It does not matter how many times that HHH emulates DDD.
HHH could even perform a static analysis and determine
that DDD simulated by HHH cannot possibly reach its own
"return" instruction final halt state. DDD can be rejected
on this basis.
 

No it can't as if HHH return 0 to HHH(DDD) then DDD() will halt, so no correct static analysis can say otherwise.
Your problem is you forget that HHH needs to be a program, and thus have fixed code, which doesn't change based on who is looking at it.
In other words, all you logic is based on LIES.

 

because the input for HHH includes the code of Halt7.c, where it specified that the program aborts and halts.
That HHH does not see that specification does not change the specification. It is childish to think that things that are not seen do not exist, in particular when there are other ways to observe them.