Sujet : Re: Who here understands that the last paragraph is Necessarily true?
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 15. Jul 2024, 14:49:46
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <027a58b80e88784ef5b3b6df9acbff84c0896fc0@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Mon, 15 Jul 2024 07:34:39 -0500 schrieb olcott:
On 7/15/2024 4:03 AM, joes wrote:
Am Sun, 14 Jul 2024 22:41:24 -0500 schrieb olcott:
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.
Your problem is you keep on ILEGALLY changing the input in your
argument because you have misdefined what the input is.
Specifically, the input HHH aborts.
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.
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*
Don't deflect. HHH as part of DDD (because it is called) needs to be
included in the input to the simulator.
Bedsides, TMs can be encoded as strings. Notwithstanding that HHH is
not and does not simulate a TM.
Richard insists that HHH report on the behavior of the TM that is *not*
encoded as finite string. TM's are not allowed to report on the behavior
of the computation that they are contained within.
What TM do you mean? They can all be encoded.
TMs CAN report on their input, if we arrange for it to be the same as
their environment.
The question is not whether or not HHH halts.
The question is does the finite string input to HHH mathematically map
to behavior that halts
And the answer is yes, as you have agreed.
when DDD is correctly emulated by HHH according
to the semantics of the x86 language?
If HHH can't map it, it is not simulating correctly.
-- Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:Objectively I am a genius.
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