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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Two dozen people were simply wrong
Date: Wed, 29 May 2024 13:31:52 -0500
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On 5/29/2024 1:14 PM, Ben Bacarisse wrote:
> Alan Mackenzie <acm@muc.de> writes:
> 
>> How about a bit of respect?  Mike specifically asked you not to cite his
>> name as a back up for your points.  Why do you keep doing it?
> 
> He does it to try to rope more people in.  It's the same ploy as
> insulting people by name.  It's hard to ignore being maligned in public
> by a fool.
> 

*Thanks for validating my simplified encoding of the Linz*

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

I really did believe that Ben Bacarisse was lying when I said it.

At the time I was talking about the easily verified fact of the actual
execution trace of fully operational code and everyone was denying the
easily verified facts.

typedef int (*ptr)();  // ptr is pointer to int function in C
00       int H(ptr p, ptr i);
01       int D(ptr p)
02       {
03         int Halt_Status = H(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         H(D,D);
12         return 0;
13       }

It turns out that two dozen people are easily proven wrong when
they claimed that the correct simulation of the input to H(D,D)
is the behavior of int main() { D(D); }

When D is correctly simulated by H using an x86 emulator the only
way that the emulated D can reach its own emulated final state
at line 06 and halt is
(a) The x86 machine code of D is emulated incorrectly
(b) The x86 machine code of D is emulated in the wrong order

*two dozen people were simply wrong*

It now turns out that Richard Damon was not lying when he referred
to the words of Peter Linz.

It did seem ridiculous that the Linz proof merely proved that
a single machine does not get the correct answer to a specific
input. Since Linz actually did use the term "single Turing machine"
I now see that was an honest mistake.

    The domain of this problem is to be taken as the set of all
    Turing machines and all w; that is, we are looking for a
    *single Turing machine* that, given the description of an arbitrary
    M and w, will predict whether or not the computation of M applied
    to w will halt

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer