Re: Functions computed by Turing Machines MUST apply finite string transformations to inputs --- MT

On 5/4/25 9:23 PM, olcott wrote:

On 5/4/2025 8:04 PM, Ben Bacarisse wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:
...

As explained above, UTM(⟨Ĥ⟩ ⟨Ĥ⟩) simulates Ĥ run with input Ĥ (having the
same halting behaviour) and Ĥ run with input Ĥ HALTS.  So embedded_H does
not "gather enough information to deduce that UTM(⟨Ĥ⟩ ⟨Ĥ⟩) would never
halt".  THAT IS JUST A FANTASY THAT YOU HAVE.
>
UTM(⟨Ĥ⟩ ⟨Ĥ⟩) DOES halt, so embedded_H can't possibly gather information
that genuinely implies it DOESN'T halt.  The explanation is obvious:
embedded_H gathers information that *YOU* believe implies that UTM(⟨Ĥ⟩ ⟨Ĥ⟩)
would never halt, but *YOU ARE SIMPLY WRONG*.

>
He used to claim that false ("does not halt") was the correct answer,
/even though/ the computation in question halts!  Those were simpler
days.  Of course cranks will never admit to having been wrong about
anything other than a detail or two, so anyone who could be bothered
could try to get him to retract that old claim.
>

 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
      H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 In other words embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is correct to
reject its input if
 Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Would not halt.

Nope, because that isn't the input that it was given. Ĥ doesn't use a UTM (unless H is actually such a machine). H is allowed to abort its emulaiton if UTM (Ĥ) (Ĥ) would never halt, but UTM (Ĥ) (Ĥ) sees that
Ĥ (Ĥ) ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn since embedded_H is the same as H and your H does do that, and thus H never had the needed proof to do that, so it made an error.
You are just showing that you idea of "logic" is based on just lying about what words mean.

 

I know you'll not understand what I've just said, because it is all too
abstract and you don't understand the concepts involved, and consequently
you probably don't agree with my Sipser interpretation, and even if you did
I doubt you would be able to work out its consequences.  So I don't expect
to be posting any further.

>
Not you then!  I sympathise, though my reason for not talking to him is
his unacceptable insults.
>