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On 5/28/24 10:23 PM, olcott wrote:So it never was about any specific machine as Linz misleading wordsOn 5/28/2024 9:04 PM, Richard Damon wrote:But note, x, being a Turing Machine, is NOT a "template"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 }
>
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
*Formalizing the Linz Proof structure*
∃H ∈ Turing_Machines
∀x ∈ Turing_Machines_Descriptions
∀y ∈ Finite_Strings
such that H(x,y) = Halts(x,x)
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.
>
We have not gotten to that point yet this post is so that
you can fully understand what templates are and how they work.
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.
That is basically what Linz does.
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.
All the other H's might get this input right, but we don't care, we have shown that for every H we
>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.(And I think you have an error in your reference to Halts, I think you mean Halts(x,y) not Halts(x,x)>
>
Yes good catch. I was trying to model embedded_H / ⟨Ĥ⟩
and then changed my mind to make it more general.
>>>>
*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
>
as there are many C_Functions that are not the equivalent of Turing Machines.
>
The whole purpose here is to get you to understand what
templates are and how they reference infinite sets.
>
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.
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.>>>>
In both cases infinite sets are examined to see
if any H exists with the required properties.
>
Yes, but the logic of Turing Machines looks at them one at a time, and the input is a FULL INDEPENDENT PROGRAM.
>
∃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.
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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