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

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.
(And I think you have an error in your reference to Halts, I think you mean Halts(x,y) not 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)

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.

 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.
I'm not sure what you can define your computation system to be actually based on, and what its supposed use is, since your 'decider' and 'input' are so intertwined.
And your supposed algorithm just doesn't work when you try to make you system "Turing Complete" by letting D have the ability to have a COPY of H, and being able to make copies of its input, like real Turing machines can.