Sujet : Re: HHH maps its input to the behavior specified by it --- never reaches its halt state ---
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 10. Aug 2024, 13:48:37
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <03cabc6945effa1696e8ebfa691e6e661b896634@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
User-Agent : Mozilla Thunderbird
On 8/10/24 7:34 AM, olcott wrote:
On 8/10/2024 3:27 AM, Fred. Zwarts wrote:
Op 09.aug.2024 om 22:53 schreef olcott:
On 8/9/2024 2:35 PM, Fred. Zwarts wrote:
Op 09.aug.2024 om 18:19 schreef olcott:
>
void DDD()
{
HHH(DDD);
return;
}
>
Each HHH of every HHH that can possibly exist definitely
emulates zero to infinity instructions of DDD correctly.
Every expert in the C language sees that this emulated DDD
cannot possibly reaches its own "return" instruction halt state.
And you don't need to be an expert to see that this proves that all these simulations are incorrect.
In other words you are trying to get away with the lie that
Richard has been persistently pushing:
When N > 0 instructions of DDD are correctly emulated by HHH
then no instructions of DDD have been correctly emulated.
No, that is just a lie you say about me.
I never said that NO instructions of DDD have been correctly emulated, just tnat not all of the instructions of DDD have been correctly emulated.
The fact that you keep repeating the obviousl LIE, just shows that you have nothing to base you claims on.
That or you just don't understand the basic meaning of the words correct and incorrect.
For something to be correct, it must be TOTALLY correct, and incorrect doesn't mean nothing is right, just that it isn't ALL right.
You logic seems to think that if any part is correct, then it must be correct, and the only way something is incorrect is if nothing in it is correct, which is just the basis of so many lies, as that isn't what the words mean.