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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows --- Truthmaker
 Maximalism
Date: Fri, 14 Mar 2025 21:00:13 -0400
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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


And they do so via proof-by-contradiction:

First they assume that such an H exists.  That is the ONLY assumption, 
so all possible implementations are included, including simulation.

Then they build a D such that H doesn't meet the above requirements for 
D.  This is a contradiction, so the assumption that such an H exists is 
proven false.