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On 7/19/2024 2:30 AM, Mikko wrote:So, you ADMIT that there actually are some correct expression that are undecable, and only SOME are based on wrong statements.On 2024-07-18 13:36:53 +0000, olcott said:Some undecidable expressions are only undecidable because
>On 7/18/2024 2:55 AM, Mikko wrote:>On 2024-07-17 13:14:43 +0000, olcott said:>
>On 7/17/2024 2:08 AM, Mikko wrote:>On 2024-07-16 14:46:40 +0000, olcott said:>
>On 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:>
When the source of your disagreement is your own ignorance
then your disagreement has no actual basis.
>
*You can comprehend this is a truism or fail to*
*comprehend it disagreement is necessarily incorrect*
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.
If it is a logical impossibility then it places no
actual limit on computation otherwise we would have
"the CAD problem" of the logical impossibility of making
a CAD system that correctly draws a square circle.
A logical impossibility does place a limit on computation.
Otherwise it would be possible to build a CAD system that
can correctly draw a square circle.
Of the set of possible things TM's can do them all.
Depends on the meanings of "possible" and "thing". Of things other than
computation no TM can do any. A Turing machine can determine whether
a sentence of Presburger arithmetic is provable but no Turing machine
can determine whether a sentence of Peano arithmetic is provable.
>
they are self contradictory. In other words they are undecidable
because there is something wrong with them.
The Liar Paradox: "This sentence is not true"Says who? That is your lie.
(is neither true nor false) and the HP proof are that way,
yet, only when we expect a decider to return the halt status
of an input that does that opposite of whatever value the
decider returns.
typedef void (*ptr)();
int HHH(ptr P);
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
int main()
{
DD();
}
*When we understand that*
(a) The halt decider is not allowed to report on the computation
that it is contained within. Then the behavior of the directly
executed DD() is moot.
(b) The self-contradictory part of the input is unreachable fromNo, it is unrachable by the emulation of the decider, but not by the behavior of the machine described, this is your fundamental confusion because you can't seem to tell the difference between the full reality and a partial observation of it.
the emulated DD then a simulating partial halt decider does
correctly compute the mapping from the input finite string to
the non-halting behavior of this finite string.
int main { DD(); } calls HHH(DD) that must abort the emulationBut HHH is only the program that it actually is. And if HHH meets the requirements to be a decider, it WILL aborts is simulation and return and thus DDD will returm.
of its input or HHH, emulated DD and executed DD never stop
running.
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