On 1/22/22 11:47 AM, olcott wrote:
On 1/22/2022 10:39 AM, Richard Damon wrote:
On 1/22/22 11:29 AM, olcott wrote:
On 1/22/2022 10:23 AM, Richard Damon wrote:
On 1/22/22 10:48 AM, olcott wrote:
Halting problem undecidability and infinitely nested simulation (V3) >>>>>Take FIFTY TWO, I think you are stuck in an apparent infinite loop.
You keep on repeating the same basic mistakes.
We define Linz H to base its halt status decision on the behavior
of its pure simulation of N steps of its input. N is either the
number of steps that it takes for its simulated input to reach its >>>>>> final state or the number of steps required for H to match an
infinite behavior pattern proving that the simulated input would
never reach its own final state. In this case H aborts the
simulation of this input and transitions to H.qn.
Such a pattern NOT existing for the <H^> <H^> pattern, thus your H
can't correctly abort and becomes non-answering and thus FAILS to
be a decider.
The non-existance has been proven and you have ignored that proof,
showing you have no counter for the proof.
If you want to claim such a pattern exists, you MUST provide it or
accept that your logic is just plain flawed as you are claiming the
existance of something that is impossible.
In effect, you are saying that if you have a halt decider for you
halt decider to use, you can write a halt decider.
FAIL.
The following simplifies the syntax for the definition of the Linz >>>>>> Turing machine Ĥ, it is now a single machine with a single start
state. A copy of Linz H is embedded at Ĥ.qx.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Because it is known that the UTM simulation of a machine is
computationally equivalent to the direct execution of this same
machine H can always form its halt status decision on the basis of >>>>>> what the behavior of the UTM simulation of its inputs would be.
When Ĥ applied to ⟨Ĥ⟩ has embedded_H simulate ⟨Ĥ⟩ ⟨Ĥ⟩ these steps
would keep repeating:
Ĥ copies its input ⟨Ĥ⟩ to ⟨Ĥ⟩ then embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩...
But only if H never aborts, if H does abort, then the pattern is
broken.
YOU ARE EITHER TOO IGNORANT OR DISHONEST TO ACKNOWLEDGE THE TRUTH OF
THIS:
The fact that there are no finite number of steps that the simulated
input to embedded_H would ever reach its final state conclusively
proves that embedded_H is correct to abort its simulation of this
input and transition to Ĥ.qn.
The problem is that any H that aborts after a finite number of steps,
gives the wrong answer because it only looked to see if the input
doesn't halt at some specific finite number.
OK IGNORANT it is. When there exists no finite (or infinite) number of
steps such that the simulated input to embedded_H reaches its final
state then we know that this simulated input does not halt.
And the only case when that happens is when H does not abort its
simulation,
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