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

On 2024-06-01 14:52:54 +0000, olcott said:

On 6/1/2024 3:20 AM, Mikko wrote:

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.
 

 ∃H  ∈ Turing_Machines
There exists at least one H
from the infinite set of all Turing_Machines
 ∃!H  ∈ Turing_Machines
There exists a single unique H
from the infinite set of all Turing_Machines

The symbol ∃! is non-standard and should be defined when used.
And there is no point to present some formulas without saying
anything about them.

    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".

 I know that he said that yet he meant this
∃H ∈ Turing_Machines *and didn't mean this* ∃!H ∈ Turing_Machines
or he would be contradicting every other HP proof.

He didn't mean either. Your false claim is merely an attempted
preparation of strawman deception.

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.

 In other words when there are two machines that solve the halting
problem then the halting problem IS NOT SOLVED?

If there are two then one of them is one and the other is irrelevant.
When Linz says "we are looking for a single Turing machine that ..."
he implies that when he (or someone) finds one he is not going to
look for another one. When he finally proves that it is not possible
to find one it is obvious that it is not possible to find more.
And if you need to ask that you don't need to as why you look stupid.
--
Mikko