Re: key error in all the proofs --- Correction of Fred

On 8/16/24 10:09 AM, olcott wrote:

On 8/16/2024 8:34 AM, Mikko wrote:

On 2024-08-16 12:02:00 +0000, olcott said:
>

>
I must go one step at a time.

>
That's reasonable in a discussion. The one thing you were discussing
above is what is the meaning of the output of HHH. Its OK to stay
at that step until we are sure it is understood.
>

 void DDD()
{
   HHH(DDD);
   return;
}
 Unless an unlimited emulation of DDD by HHH
can reach the "return" instruction of DDD it is
construed that this instance of DDD never halts.

But that also construes that HHH is a program that DOES an unlimited emulation of DDD, and thus isn't a decider

 For three years now at least most reviewers insisted
on disagreeing with the semantics of the x86 language.
 

No, the problem is HHH can't be two different things at once, and since DDD includes its HHH as part of its definition, the DDD that the HHH that DDD calls, can see any other DDD than the one that called IT, and not some other HHH.
The fact that you just don't understand that proper way to setup a correct emulator that can emulate any program given to it just wrecks your argument and renders your logic void.