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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows --- Truthmaker
 Maximalism
Date: Sat, 15 Mar 2025 07:24:44 -0400
Organization: i2pn2 (i2pn.org)
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On 3/14/25 11:40 PM, olcott wrote:
> On 3/14/2025 8:56 PM, dbush wrote:
>> On 3/14/2025 9:49 PM, olcott wrote:
>>> 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.
>>>
>>
>> Did I stutter?
>>
>> I want to know if any arbitrary X with input Y halts when executed
> 
> If you reject "ls;dlfm skdofdfn 894&49.8244bewr" as a syntactically
> incorrect input then you are being inconsistent when you fail to reject
> semantically incorrect inputs.

And when you reject a valid input, that was PROVEN to be valid from the 
logic system.

All you have shown is that you don't know what you are talking about and 
your statements are just based on LIES and thus not semantically correct.

> 
>> directly.  If I had an H that could tell me that in *all* possible 
>> cases, I could solve the Goldbach conjecture, among many other 
>> unsolved problems.
>>
>> Does an H exist that can tell me that or not?
> 
> 
>