Re: Defining a correct halt decider

On 2024-09-06 11:41:05 +0000, olcott said:

On 9/6/2024 6:12 AM, Mikko wrote:

On 2024-09-05 13:39:14 +0000, olcott said:
 

On 9/5/2024 2:39 AM, Mikko wrote:

On 2024-09-03 13:17:56 +0000, olcott said:
 

On 9/3/2024 3:44 AM, Mikko wrote:

On 2024-09-02 16:06:11 +0000, olcott said:
 

A correct halt decider is a Turing machine T with one accept state and one reject state such that:
 If T is executed with initial tape contents equal to an encoding of Turing machine X and its initial tape contents Y, and execution of a real machine X with initial tape contents Y eventually halts, the execution of T eventually ends up in the accept state and then stops.
 If T is executed with initial tape contents equal to an encoding of Turing machine X and its initial tape contents Y, and execution of a real machine X with initial tape contents Y does not eventually halt, the execution of T eventually ends up in the reject state and then stops.

 Your "definition" fails to specify "encoding". There is no standard
encoding of Turing machines and tape contents.

 That is why I made the isomorphic x86utm system.
By failing to have such a concrete system all kinds
of false assumptions cannot be refuted.

 If it were isnomorphic the same false assumtipns would apply to both.

 They do yet I cannot provide every single details of
the source-code of the Turing machine because these
details would be too overwhelming.
 So instead every author makes a false assumption that
is simply believed to be true with no sufficient basis
to show that it isn't true.
 Once I prove my point as the x86 level I show how the
same thing applies to the Peter Linz proof.

 Your recent presentations are so far from Linz' proof that they
look totally unrelated.

 I must begin where people are so far no one even understands
the concept of recursive emulation.

I don't know about you but most of the participants of this discussion
seem to understand recursive simulation and how it differs from
recursion.
--
Mikko