Re: Sufficient knowledge of C proves that DD specifies non-terminating behavior to HHH

On 2/7/25 8:12 PM, olcott wrote:

On 2/7/2025 5:56 PM, Richard Damon wrote:

On 2/7/25 11:26 AM, olcott wrote:

On 2/7/2025 6:20 AM, Richard Damon wrote:

On 2/6/25 10:02 PM, olcott wrote:

On 2/6/2025 8:21 PM, Richard Damon wrote:

On 2/6/25 5:18 PM, olcott wrote:

On 2/6/2025 1:51 PM, Richard Damon wrote:

On 2/6/25 1:26 PM, olcott wrote:

On 2/6/2025 10:52 AM, Bonita Montero wrote:

Am 05.02.2025 um 16:11 schrieb olcott:

On 2/5/2025 1:44 AM, Bonita Montero wrote:

Am 05.02.2025 um 04:38 schrieb olcott:

This treatment does not typically last very long and
will be immediately followed by a riskier fourth line
of treatment that has an initial success rate much higher
than its non progression mortality rate.
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Halting problem solved !
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The halting problem proof input does specify non-halting
behavior to its decider.
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https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D

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LOOOOOOOOL

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Anyone that understands the C programming language
sufficiently well (thus not confused by the unreachable
"if" statement) correctly understands that DD simulated
by HHH cannot possibly reach its own return instruction.
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And anyone that understand the halting problem knows that isn't the question being asked. The quesiton you NEED to ask is will the program described by the input halt when run?
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Since you start off with the wrong question, you logic is just faulty.
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Everyone that thinks my question is incorrect is wrong.
It has always been a mathematical mapping from finite
strings to behaviors. That people do not comprehend this
shows the shallowness of the depth of the learned-by-rote
(lack of) understanding.
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No, you are just incorreect as you don't know what you are talking about.
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Yes, it is a mapping of the string to the behavior, and that mapping is DEFINED to be the halting behavior of the program the string describes.
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No this is incorrect. The input finite string specifies
(not merely describes) non halting behavior to its decider.
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No, since the definition of "Halting Behavior" is the behavior of the progran being run.
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It may seem that way to people that have learned-by-rote
as their only basis. It is actually nothing like that.

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No, that *IS* the definition.
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 A termination analyzer computes the mapping from finite
strings to the actual behavior that these finite strings
specify. That this is not dead obvious to everyone here
merely proves that learned-by-rote does not involve any
actual comprehension.
 

And the behavior the finite string specifies is the behavior of running the program. WHich means the finite string has to fully define a program.
THose are DEFINITIONS.
You ignorance by never learning is just proving your stupidity.
Sorry, but everything you do just shows that you are nothing but a pathological liar that doesn't care about what is actually true, but what you want to be true, and are too stupid to understand the problem with that.
Your inability to understand that there ARE definitions are rules just futhers that observation.