Sujet : Re: Formal systems that cannot possibly be incomplete except for unknowns and unknowable
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 08. May 2025, 02:20:34
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vvh0t2$1b939$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
User-Agent : Mozilla Thunderbird
On 5/7/2025 7:44 PM, dbush wrote:
On 5/7/2025 8:19 PM, olcott wrote:
On 5/7/2025 7:15 PM, dbush wrote:
On 5/7/2025 7:40 PM, olcott wrote:
On 5/7/2025 6:31 PM, dbush wrote:
On 5/7/2025 7:26 PM, olcott wrote:
>
When N instructions of DD are emulated by HHH
according to the rules of the x86 language then
>
The subject was "DD emulated by HHH", not "N instructions of DD emulated by HHH".
>
Changing the subject is the dishonest dodge of the strawman deception.
>
>
That you and Richard construe anything less than an
infinite number of steps of DD emulated by HHH
(according to the rules of the x86 language)
as an incorrect emulation IS MORONICALLY STUPID.
>
>
The fixed immutable code of HHH simulates a fixed number X of instructions of DD, the last of which was simulated incorrectly. Any number other than X is not what HHH simulates and is therefore irrelevant to HHH.
>
UTM simulates X+Y instruction of DD correctly and reaches a final state.
>
>
I will make it easier to understand.
>
void DDD()
{
HHH(DDD);
return;
}
>
Can DDD simulated by HHH reach its own "return" instruction?
>
Category error. There is no "can" as algorithm HHH is fixed and immutable, as is algorithm DDD.
Does there exist an HHH such that DDD emulated by
HHH according to the rules of the C programming language
where the DDD element of the infinite set of HHH/DDD
pairs reaches its own "return" instruction?
It is like I am asking you is there a positive
number that is less than zero? You don't have
to check the positive numbers one-at-a-time.
We can know with complete certainty that no DDD
simulated by any HHH can possibly reach its own
"return" instruction.
Algorithm HHH *does not* simulate algorithm DDD to the end but instead aborts in violation of the x86 language.
there is no end to reach.
Algorithm UTM *does* simulate algorithm DDD to the end.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
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