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On 7/17/2024 2:08 AM, Mikko wrote:It is impossible for a program to return the halting status of everyOn 2024-07-16 14:46:40 +0000, olcott said:If it is a logical impossibility then it places no actual limit onOn 7/16/2024 2:18 AM, Mikko wrote:On 2024-07-15 13:32:27 +0000, olcott said:>On 7/15/2024 2:57 AM, Mikko wrote:>On 2024-07-14 14:48:05 +0000, olcott said:>On 7/14/2024 3:49 AM, Mikko wrote:>On 2024-07-13 12:18:27 +0000, olcott said:>
Any input that must be aborted to prevent the non termination of
HHH necessarily specifies non-halting behavior or it would never
need to be aborted.
Disagreeing with the above is analogous to disagreeing with
arithmetic.
A lame analogy. A better one is: 2 + 3 = 5 is a proven theorem just
like the uncomputability of halting is.
The uncomputability of halting is only proven when the problem is
framed this way: HHH is required to report on the behavior of an
input that was defined to do exactly the opposite of whatever DDD
reports.
No, it is proven about the halting problem as that problem is.
Which is simply a logical impossibility
Yes, a halting decider is a logical impossibility, as can be and has
been proven.
computation otherwise we would have "the CAD problem" of the logical
impossibility of making a CAD system that correctly draws a square
circle.
In contrast to this example, the halting problem does have a welldefinedthus no actual limit to computation more that this logical
impossibility: What time is it (yes or no)?
You may not care about the halting decider counterexample. That hasAs construction of a halting decider is already known to be impossibleOnly because we have framed the problem as a logical impossibility. When
why would anyone care whether there is other limitations about it?
And of course the impossibility of halting decider prevents any
applicaions of it, for example as a tool to solve other problems.
we re-frame the problem so that it is not a logical impossibility then
the practical applications can still be derived.
And the answer is no, Carol cannot, but others can.*This is isomorphic the HP decider/input pair*
Can Carol correctly answer “no” to this (yes/no) question?
(Hehner:2018:2)
It is certainly a neat turn on itself. That makes it all the morePerhaps you can use the isomorphism to proove that Carol can't.Carol's question is isomorphic to the halting problem decider/input pair
But that should be faily easy anyway.
showing that the halting problem is simply a cheap trick.
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