Re: D correctly simulated by H cannot possibly halt --- templates and infinite sets

On 5/29/24 9:24 AM, olcott wrote:

On 5/29/2024 4:14 AM, Mikko wrote:

On 2024-05-29 03:49:02 +0000, olcott said:
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On 5/28/2024 10:38 PM, Richard Damon wrote:

On 5/28/24 10:23 PM, olcott wrote:

On 5/28/2024 9:04 PM, Richard Damon wrote:

On 5/28/24 12:16 PM, olcott wrote:

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       }
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When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
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*Formalizing the Linz Proof structure*
∃H  ∈ Turing_Machines
∀x  ∈ Turing_Machines_Descriptions
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,x)

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But since for x being the description of the H^ built from that H and y being the same, it turns out that no matter what answer H gives, it will be wrong.
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We have not gotten to that point yet this post is so that
you can fully understand what templates are and how they work.

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But note, x, being a Turing Machine, is NOT a "template"
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And H, isn't a "set of Turing Machines", but an arbitrary member of that set, so all we need to do is find a single x, y, possible determined as a function of H (so, BUILT from a template, but not a template themselves) that shows that particular H was wrong.
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That is basically what Linz does.
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Given a SPECIFIC (but arbitary) H, we can construct a specific H^ built from a template from H, that that H can not get right.
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All the other H's might get this input right, but we don't care, we have shown that for every H we
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(And I think you have an error in your reference to Halts, I think you mean Halts(x,y) not Halts(x,x)
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Yes good catch. I was trying to model embedded_H / ⟨Ĥ⟩
and then changed my mind to make it more general.
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*Here is the same thing applied to H/D pairs*
∃H ∈ C_Functions
∀D ∈ x86_Machine_Code_of_C_Functions
such that H(D,D) = Halts(D,D)

>
Not the same thing.
∃H ∈ C_Functions
is not equivalent to
∃H  ∈ Turing_Machines
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as there are many C_Functions that are not the equivalent of Turing Machines.
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The whole purpose here is to get you to understand what
templates are and how they reference infinite sets.
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But the problem is that even in your formulation, H and D are, when doing the test, SPECIFIC PROGRAMS and not "templates" as Halts is defined on the domain of PROGRAMS.
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Similarly, a "Template" doesn't have a specific set of x86_Machine_Code_of_C_function, at least not one with defined behavior since if it tries to reference code outside of itself, then Halts of that just isn't defined, only Halts of that code + the specific machine deciding it.
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In both cases infinite sets are examined to see
if any H exists with the required properties.
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Yes, but the logic of Turing Machines looks at them one at a time, and the input is a FULL INDEPENDENT PROGRAM.
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∃H  ∈ Turing_Machines
That does not look at one machine it looks as an infinite set of
machines. I am very happy to find out that you were not playing head
games. Linz actually used the words that you referred to.

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while the ∃H part can create a set of machines, each element of that set is INDIVIDUALLY TESTED in the following conditions, so, when we get to your test  H(x,y) = Halts(x,x), each of H, x, y are individual members of the set, and we THEN collect the set of all of them.
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If we try to say
∃x ∈ Natural Numbers, such that  x+x = 3
we can't say that x is both 1 and 2 and thus as a set meet the requirement. For the conditions, each qualifier select a single prospective element, and those are tested to see if that meet the requirement.
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So it never was about any specific machine as Linz misleading words
seemed to indicate. It was always about examining each element of an
infinite set.
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Likewise: ∃H ∈ C_Functions is about examining each element
of an infinite set. A program template specifies a set of programs
the same way that an axiom schema specifies a set of axioms.
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I am very happy that the issue was the misleading words of Linz
and not you playing head games.

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In an inderect proof of an unversal claim the counter-hypothesis must
be about one example. Then the proof is about that specific example
until a contradiction is derived.
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 Does there exist at least one example of this when the
infinite set of Turing_Machines have been examined?
  Of the infinite set of Turing_Machines does there exist
at least one H that always gets this H(x,y) = Halts(x,y)
correctly for every {x,y} pair of the infinite set of {x,y} pairs?

Then why was Linz able to create, for any specific H, an H^ that it get wrong?

 *Formalizing the Linz Proof structure*
∃H  ∈ Turing_Machines
∀x  ∈ Turing_Machines_Descriptions
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
 

And since NO H, can get right the H^ built to contradict IT, that claim is proven false.
The statement can be proven to be incorrect if we can show:
∀H  ∈ Turing_Machines
∃x,y x  ∈ Turing_Machines_Descriptions
      y  ∈ Finite_Strings
such that H(x,y) != Halts(x,y)
And that is shown, since for ANY H that you can create, we can prove that your claim is just false.