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

On 6/1/2024 1:56 PM, joes wrote:

Am Sat, 01 Jun 2024 09:52:54 -0500 schrieb olcott:

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:
>

 

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 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.
>

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?

I misunderstood this, too, but we want a single machine that solves the
problem on its own, not multiple that each solve parts. There could be
many such machines.
 

∃H ∈ Turing_Machines
is fulfilled when there are one or more machines that independently
solve the halting problem.
∃!H  ∈ Turing_Machines
is ONLY fulfilled when there is exactly one machine that solves the halting problem.
I knew what Linz meant from all the other proofs that I read.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer