Re: Halt Deciders must be computable functions --- dbush was always wrong

Am Mon, 24 Mar 2025 09:49:34 -0500 schrieb olcott:

On 3/24/2025 6:23 AM, Richard Damon wrote:

On 3/23/25 8:59 PM, olcott wrote:

On 3/23/2025 7:01 PM, joes wrote:

Am Sun, 23 Mar 2025 15:08:25 -0500 schrieb olcott:
>

The behavior of a directly executing Turing Machine cannot be
computed because a directly executing Turing machine cannot be the
input to any computable function.

Lol. This is such a ridiculously silly objection. Of course a TM is
nothing but a finite string (or can be encoded as such). TMs are most
definitely computable - UTMs are possible.
>

It is enormously more nuanced than that.
https://www.liarparadox.org/Linz_Proof.pdf
>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

 
Incorrect, UTM was never there;, so you are just showing that your
logic was a lie.
 
 

When we define the Peter Linz Ĥ with a UTM that simulates a finite
number of "moves"
before transitioning to Ĥ.qn then Ĥ reaches Ĥ.qn and the simulated ⟨Ĥ⟩
cannot possibly reach ⟨Ĥ.qn⟩.
>

Which is impossible, as such a thing is not a UTM.
 

Saying that a UTM that simulates a finite number of states is not a UTM
is like saying that a red car is not a car.

The other way around: a limited TM is not universal.

You are just showing that your logic is based on the presumption of the
impossible and lies to fill the holes.

It is not impossible for a UTM to simulate a finite number of states.

Neither is it identical to the direct execution (which doesn't stop).

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.