Sujet : Re: Who here understands that the last paragraph is Necessarily true?
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 16. Jul 2024, 19:26:12
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v76e04$1cf96$4@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14
User-Agent : Mozilla Thunderbird
On 7/16/2024 3:16 AM, Mikko wrote:
On 2024-07-15 13:51:14 +0000, olcott said:
On 7/15/2024 3:37 AM, Mikko wrote:
On 2024-07-15 03:41:24 +0000, olcott said:
>
On 7/14/2024 9:04 PM, Richard Damon wrote:
On 7/14/24 9:27 PM, olcott wrote:
>
Any input that must be aborted to prevent the non termination
of simulating termination analyzer HHH necessarily specifies
non-halting behavior or it would never need to be aborted.
>
Excpet, as I have shown, it doesn't.
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Your problem is you keep on ILEGALLY changing the input in your argument because you have misdefined what the input is.
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The input to HHH is ALL of the memory that it would be accessed in a correct simulation of DDD, which includes all the codd of HHH, and thus, if you change HHH you get a different input.
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If you want to try to claim the input is just the bytes of the function DDD proper then you are just admitting that you are nothing more than a lying idiot that doesn't understand the problem,
Turing machines only operate on finite strings they do
not operate on other Turing machines *dumbo*
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That's right. But the finite string can be a description of a Turing machine.
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No that is wrong. The finite string must encode a Turing machine.
That you agree does not mean that it is wrong.
Your ignorance that a Turing machine cannot take its own
executing self as an input is no rebuttal what-so-ever.
-- 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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