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

On 2024-05-31 15:44:22 +0000, olcott said:

On 5/31/2024 8:10 AM, Mikko wrote:

On 2024-05-28 16:16:48 +0000, olcott said:
 

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)
 *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)
 In both cases infinite sets are examined to see
if any H exists with the required properties.

 That says nothing about correct simulation. It says
something abuout some D but not whether it is correctly
simulated. Also nothing is said about templates or
infinite sets. At the end is claimed that some
infinite sets are examined but not who examined, nor
how, nor what was found in the alleged examination.
 

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

The above is the counter hypothesis for the proof. Proof structore
is that a contradiction is derived from the counter hypthesis.

The above disavows Richard's claim based on a misinterpretation of
Linz that the Linz proof is about a single specific Turing machine.

Your ∃H declares H as a new symbol for a specific Turing machine.
Therefore everything that follows refers to that specific Turing machine.
There may be others that could be discussed the same way but they aren't.

    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.

Note the words "a single Turing machine".

Linz <IS NOT> looking for a single machine that gets the wrong answer.
Linz is looking for at least one Turing Machine that gets the right
answer: ∃H ∈ Turing_Machines

Not at least one but exactly one. The Halting Problem asks for one
or a proof that there is none.
--
Mikko