Re: DDD correctly emulated by HHH is correctly rejected as non-halting. --- You are not paying attention

On 2024-07-19 13:48:49 +0000, olcott said:

On 7/19/2024 2:30 AM, Mikko wrote:

On 2024-07-18 13:36:53 +0000, olcott said:
 

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.

 Some undecidable expressions are only undecidable because
they are self contradictory. In other words they are undecidable
because there is something wrong with them.

Being self-contradictory is a semantic property. Being uncdecidable is
independent of any semantics. An arithmetic sentence is always about
numbers, not about sentences. A proof is about sentences, not about
numbers.

The Liar Paradox: "This sentence is not true"

cannot be said in the language of Peano arithmetic.
--
Mikko