Re: ChatGPT totally understands exactly how I refuted the conventional halting problem proof technique

On 6/24/2025 3:05 PM, Alan Mackenzie wrote:

[ Followup-To: set. ]
 In comp.theory olcott <polcott333@gmail.com> wrote:

On 6/24/2025 1:14 PM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 6/24/2025 12:39 PM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 6/24/2025 11:43 AM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 6/24/2025 3:39 AM, Mikko wrote:

On 2025-06-23 16:37:53 +0000, olcott said:

 

I always interpret expressions of language according
to the literal base meaning of their words.

 

I interprete the above to mean that the author of those words is
stupid.

 

Counter factual, my IQ is in the top 3%

 

Pull the other one!

 

Given your demonstrated lack of understanding of abstraction, of
what a proof is, of so many other things, it is clear to all the
regulars in this group that your IQ is not "in the top 3%", or
anywhere near it.

 

It would seem to me you are, yet again, in the words of Sir Robert
Armstrong, being economical with the truth.

 

*I really did get that IQ on the Mensa entrance exam*

 

OK, let us be charitable, and suggest that that exam was a very long
time ago, and that your general intelligence has declined
substantially in the interval.

 

That I am unwilling to accept that textbooks on computer
science are inherently infallible is the broader minded
perspective of philosophy of computation.

 

That's an inaccurate summary.  You're clearly unable to understand these
textbooks.  If you were able, you'd see that the things they say are
necessarily correct, according to clear reasoning from obvious axioms.
Whether you'd accept these books if you could understand them is more
the question.

 

It is an easily verified fact that no *input* to any
partial halt decider (PHD) can possibly do the opposite
of what its corresponding PHD decides.

 

That's both a lie and a strawman.  The fact is, you're unable to
understand computer science textbooks.  If you could, you wouldn't simply
try and dodge the point.

 

.... In all of the years of all of these proofs no such *input* was
ever presented.

 

Of course not.  Such input can't exist.  What's happening here is that
you utterly fail to understand proof by contradiction, just as you fail
to understand so many abstractions.

 

*You are not paying close enough attention*
There cannot possibly be any *input* to any partial halt
decider that does the opposite of whatever this PHD decides
even when this (PHD) gets the wrong answer. All of the proofs
for all of these years have been bogus on this basis.

 All these proofs were valid and remain valid.  You're insufficiently
intelligent to understand them.  But you're right about me not paying
close attention.  Your continual repetitions of falsehoods got too dull
too long ago.
 

Of course you won't bother to try to actually refute
what I said by encoding any example of an *input*
that *is an actual input* that does the opposite of
whatever value that its termination analyzer determines.
All of the HP proofs are based on:
(a) The caller of the termination analyzer doing the opposite.
     int DD()
     {
       int Halt_Status = HHH(DD);
       if (Halt_Status)
         HERE: goto HERE;
       return Halt_Status;
     }
     int main()
     {
       DD();
     }
(b) The computation that the partial halt decider is embedded
within doing the opposite.
     *From the bottom of page 319 has been adapted to this*
     https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
     When Ĥ is applied to ⟨Ĥ⟩
     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞
       if Ĥ applied to ⟨Ĥ⟩ halts
     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
       if Ĥ applied to ⟨Ĥ⟩ does not halt
In none of these cases is *the actual input* doing the opposite.
When HHH is correctly simulated DD this simulated DD
cannot possibly reach its own simulated "return"
instruction final halt state.
When ⟨Ĥ⟩ ⟨Ĥ⟩ is correctly simulated Ĥ.embedded_H this
simulated ⟨Ĥ⟩ cannot possibly reach its own simulated
⟨Ĥ.qn⟩ final halt state.
Since Turing Machines cannot take directly executing
Turing Machines as inputs this means that the directly
executed Ĥ applied to ⟨Ĥ⟩ is not in the domain of
Ĥ.embedded_H so Ĥ.embedded_H does not get the wrong
answer after all.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer