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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows --- Truthmaker
 Maximalism
Date: Fri, 14 Mar 2025 20:27:42 -0500
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On 3/14/2025 8:00 PM, dbush wrote:
> On 3/14/2025 8:45 PM, olcott wrote:
>> On 3/14/2025 12:54 PM, dbush wrote:
>>> On 3/14/2025 12:33 PM, olcott wrote:
>>>> On 3/14/2025 11:01 AM, wij wrote:
>>>>> On Fri, 2025-03-14 at 10:51 -0500, olcott wrote:
>>>>>> On 3/14/2025 10:04 AM, wij wrote:
>>>>>>> On Fri, 2025-03-14 at 09:35 -0500, olcott wrote:>>
>>>>>>>> void DDD()
>>>>>>>> {
>>>>>>>>      HHH(DDD);
>>>>>>>>      return;
>>>>>>>> }
>>>>>>>>
>>>>>>>> DDD correctly simulated by HHH cannot possibly reach
>>>>>>>> its own "return" instruction in any finite number of
>>>>>>>> correctly simulated steps.
>>>>>>>>
>>>>>>>> That you are clueless about the semantics of something
>>>>>>>> as simple as a tiny C function proves that you are not
>>>>>>>> competent to review my work.
>>>>>>>>
>>>>>>>
>>>>>>> https://en.wikipedia.org/wiki/Halting_problem
>>>>>>> In computability theory, the halting problem is the problem of 
>>>>>>> determining, from a description of
>>>>>>> an
>>>>>>> arbitrary computer program and an input, whether the program will 
>>>>>>> finish running, or continue to
>>>>>>> run
>>>>>>> forever.
>>>>>>>
>>>>>>> That means: H(D)=1 if D() halts and H(D)=0 if D() does not halt.
>>>>>>>
>>>>>>> But, it seems you don't understand English, as least as my 
>>>>>>> level, ....
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>     HHH(DDD);
>>>>>>     return;
>>>>>> }
>>>>>>
>>>>>> The only difference between HHH and HHH1 is that they are
>>>>>> at different locations in memory. DDD simulated by HHH1
>>>>>> has identical behavior to DDD() directly executed in main().
>>>>>>
>>>>>> The semantics of the finite string input DDD to HHH specifies
>>>>>> that it will continue to call HHH(DDD) in recursive simulation.
>>>>>>
>>>>>> The semantics of the finite string input DDD to HHH1 specifies
>>>>>> to simulate to DDD exactly once.
>>>>>>
>>>>>> When HHH(DDD) reports on the behavior that its input finite
>>>>>> string specifies it can only correctly report non-halting.
>>>>>>
>>>>>> When HHH(DDD) is required to report on behavior other than
>>>>>> the behavior that its finite string specifies HHH is not
>>>>>> a decider thus not a halt decider.
>>>>>>
>>>>>> All deciders are required to compute the mapping from
>>>>>> their input finite string to the semantic or syntactic property
>>>>>> that this string specifies. Deciders return true when this
>>>>>> string specifies this property otherwise they return false.
>>>>>>
>>>>>
>>>>> Are you solving The Halting Problem or not? Yes or No.
>>>>>
>>>>>
>>>>
>>>> I have only correctly refuted the conventional halting
>>>> problem proof. 
>>>
>>> And what exactly do you think this proof is proving?  More 
>>> specifically, what do you think the Linz proof is proving?
>>
>> All of the proofs merely show that there cannot
>> possibly exist any halt decider that returns a
>> value corresponding to the behavior of any input
>> that is actually able to do the opposite of whatever
>> value is returned.
>>
> Not exactly.  What they prove is that no H exists that satisfies these 
> requirements:
> 
> 
> Given any algorithm (i.e. a fixed immutable sequence of instructions) X 
> described as <X> with input Y:
> 
> A solution to the halting problem is an algorithm H that computes the 
> following mapping:
> 
> (<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
> 

The executed directly part is bogus as I have
shown and your indoctrination blindly ignores.

The easiest way to see that it is wrong that is not
over the head of everyone here (the behavior semantics
that a finite string specifies)

is that the above definition allows self-contradiction
to derive undecidability.


-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer