Re: A simulating halt decider applied to the The Peter Linz Turing Machine description ⟨Ĥ⟩

On 5/28/2024 6:34 AM, Richard Damon wrote:

On 5/27/24 11:24 PM, olcott wrote:

On 5/27/2024 7:17 PM, Richard Damon wrote:

>
He STARTS with a proof that one specific (but arbitrary) machine gets the wrong answer.
>

>
*Not exactly, you are misreading this*
>
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

   *** a single Turing Machine ***
 not singular
 

>
...
>
Proof: We assume the contrary, namely that there exists an algorithm,
and consequently some Turing machine H, that solves the halting problem
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

   *** some Turing Machine ***
 Note singular
 

>
Ordinary existential quantification looks for at least one
element not exactly one element:

 But you can look for at least one by looking for one without an assumption that it is unique.
 

The Peter Linz words refer to ordinary existential quantification
anchored in the infinite set of Turing machines.
∃H  ∈ Turing_Machines
∀x  ∈ Turing_Machines_Descriptions
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,x)

>
Does at least one Turing machine exist of the infinite set
of all Turing machines ...

 Right, so if we can prove that none of them are correct, you have shown that some is contradicted.
 

>
So like I have always said, the second ⊢* specifies
an infinite set of Turing machines.
>

 As a SPECIFICATION of the domain of selection, yes.
 As a SPECIFICATION of what THIS ONE machine does, no.
 

My formalization already handles this.
*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)
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer