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

On 30/05/24 05:48, olcott wrote:

On 5/29/2024 9:55 PM, Richard Damon wrote:

On 5/29/24 10:36 PM, olcott wrote:

On 5/29/2024 9:25 PM, Richard Damon wrote:

On 5/29/24 9:55 PM, olcott wrote:

When the category is examined all at once then there is no need
to look at each individual element.

>
So, which one or ones gave the correct answer for their input?
>

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*Formalizing the Linz Proof structure*
∃H  ∈ Turing_Machines
∀x  ∈ *Turing_Machines_Descriptions*
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
>
When we formalize it that way then some simulating halt deciders
get the correct answer.
>
*Everyone else implicitly assumes this incorrect formalization*
∃H  ∈ Turing_Machines
∀x  ∈ *Turing_Machines*
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
>
>

>
Nope.
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You just don't understand the meaning of a "Description" in the problem.
>

 A deciders compute the mapping FROM ITS INPUTS
to it own accept or reject state
Deciders cannot take ACTUAL TURING MACHINES AS INPUTS
Deciders can only take FINITE STRINGS AS INPUTS
 

If you want to be pedantic, you made the mistake.
It's actually H(DescriptionOf(x),y) = Halts(x,y)
DescriptionOf is an injective function that converts Turing machines into finite strings.