On 2/19/24 8:02 PM, polcot2 wrote:
> // Linz Turing machine H --- M applied to w
> // --- Does M halt on w?
> H.q0 ⟨M⟩ w ⊢* H.qy // M applied to w halts
> H.q0 ⟨M⟩ w ⊢* Hqn // M applied to w does not halt
>
> // Linz Turing machine H --- H applied to ⟨H⟩
> // --- Do you halt on your own Turing Machine description ?
Except that isn't what you are doing below!!!!
So, you are just lying.
Below H is applied to (H) (H)
> H.q0 ⟨H⟩ ⟨H⟩ ⊢* H.qy // H applied to ⟨H⟩ halts
> H.q0 ⟨H⟩ ⟨H⟩ ⊢* H.qn // H applied to ⟨H⟩ does not halt
> *Correctly transitions to H.qy*
Only if H applied to (H) will halt. Since that is asking it to determine
if H applied to -nothing- which is an invalid input, the question is
either invalid, or H must include the ability to determine if its input
is a valid computation.
If it does, the H.qy is correct for H (H) (H) as H (H) will also go to
qy as H - will go to qn and halt.
Except then the machine is no longer "H", and the input to the actual H
is no longer a description of itself. Note, Linz is careful to give his
modified machines new names and not think of them as just "version" of
the previous machine. They are a new machine.
Note, that the resultant H^, (when also adding the duplication of the
input needed to avoid the problem shown above) when applied to the
description of itself (H^) has a definite behavior, since it was built
on a SPECIFIC H, that has definite behavior.
IF H (H^) (H^) goes to qn as you have tried to claim is correct, then we
can show that H^ (H^) will also go to its qn and Halt, showing that H
was just wrong.
If we start with a DIFFERENT H (and thus get a DIFFERENT H^, and a
DIFFERENT question) that goes to qy when applied as H (H^) (H^) then we
can show that THIS H^ will also end at qy, and then to the infinite
loop, so it will never halt, and again, this different H is wrong.
In BOTH cases, there WAS a correct answer to the question "Does the
input represent a Halting Computation?", so the Halting question is
valid (as it always has a right answer).
Only your POOP question, which ignores that Computations have fixed
answers for given inputs becomes invalid.