On 9/21/22 2:22 PM, olcott wrote:
> computation that halts … the Turing machine will halt whenever it enters
> a final state. (Linz:1990:234)
>
> When the copy of Linz H that is embedded within Linz Ĥ is a simulating
> halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated ⟨Ĥ⟩ would never
> reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞ steps of correct
> simulation.
WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL
reach the final state if H <H^> <H^> goes to H.Qn
If H <H^> <H^> doesn't go to H.Qn or H.Qy, then it fails to be a decider.
We see this at the path of the CORRECT (and thus complete) simulation if
the input to H will be: (here <H^> is wM from
The input to H <H^> <H^> represents H^ <H^>
that starts at q0 <H^> then goes to
H^.q0 <H^> <H^> after duplicating the input.
this is where the copy of H's q0 <H^> <H^> is
since we know that H <H^> <H^> is ending up at is y1 qn y2 we know the
copy will end up at H^ y1 qn y2
And this is defined as a final state.
Please point out the error here.
>
>
https://www.liarparadox.org/Linz_Proof.pdf
>
> I added this material from the Peter Linz text to address the Turing
> machine proofs. I paraphrased the Linz encoding because the Linz version
> has two start states.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy
> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final
> state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own
> final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.
>
> Ĥ.q0 copies its input then invokes its embedded copy of the original
> Linz H...
Which was STIPULATED to end up at H's qn in finite time.
Thus we KNOW that this H can't just unconditionally simulate its input,
or it does get stuck in this loop.
>
> computation that halts … the Turing machine will halt whenever it enters
> a final state. (Linz:1990:234)
>
> Because we can see that replacing H with a UTM would cause Ĥ to become
> stuck in infinitely recursive simulation we can see that the input
> correctly simulated by H would never reach the final state of this
> simulated input, thus never halts.
So, does your H have a condition in its simulation to stop its
simulaiton and return its answer to the machine it is embedded in, and
thus H^ doesn't get stuck in an infinite loop, or does H get stuck in
this loop and NEVER give an answer.
>
> Linz, Peter 1990. An Introduction to Formal Languages and Automata.
> Lexington/Toronto: D. C. Heath and Company. (317-320)
>
FAIL.