Re: How do simulating termination analyzers work? ---Truth Maker Maximalism

On 2025-07-05 15:46:21 +0000, olcott said:

On 7/5/2025 3:32 AM, Mikko wrote:

On 2025-07-04 12:34:39 +0000, olcott said:
 

On 7/4/2025 2:25 AM, Mikko wrote:

 What HHH correctly or otherwise simulates is merely an implementation
detail.

 It is a detail that defines a partial halt decider
that makes the "do the opposite" code unreachable.

 No, it does. The proof that a counter-example can be constructed
does not refer to any implementation details, so it applies to
every implementation that is does not violate the requirements
so obviously that the proof is not needed.
 

What matters is the beahviour DD specifies.

 The behavior that an input specifies is only correctly
measured by correctly simulating this input.

 Wrong. It is correctly measured by a direct execution.

 Since no Turing machine can possibly take another directly
executing Turing machine as an input this makes all directly
executed Turing machines outside of the domain of every Turing
machine based decider.

No, it does not. The domain is specified in the statement of the
halting problem which includes every Turing machine.

The requirement that a halt decider report on the behavior
of things outside of its domain has always been bogus.

The magic word "bogus" has no effect on the problem statement.

Instead of this deciders must report on the behavior that
their input actually specifies.

It does not work that way. Instead, the solution of the problem must
have, in addittion to the decider, a user's manual that specifies
what the input must be in order to specify the actual computation.
For example, if the computation asked about is DDD(), what must the
input be so that the decider says "it halts"?
--
Mikko