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:
>
>

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.

>
>

     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 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.