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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 20:49:17 -0500
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On 3/14/2025 8:34 PM, dbush wrote:
> On 3/14/2025 9:27 PM, olcott wrote:
>> 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.
>>
> 
> But I want to know if any arbitrary X with input Y halts when executed 
> directly, 

Even when some inputs are BOGUS.

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