Eliminating the pathological self-reference error of the halting theorem (V8)

[olcott]

In epistemology (theory of knowledge), a self-evident proposition is a
proposition that is known to be true by understanding its meaning
without proof ... (Wikipedia: Self-evidence)

Self-evident(A) Every simulation of input P that never halts unless
simulating halt decider H aborts this simulation <is> a non-halting
computation. This remains true even after H stops simulating P.

Because we know that the only difference in the behavior of a simulating
halt decider and a simulator is that the simulating halt decider stops
simulating some of its inputs we can examine the behavior of these
inputs in a simulator to determine whether or not a simulating halt
decider would stop simulating these inputs.

If the simulation of input P to simulator S would never terminate then
we can know that simulating halt decider H must stop simulating input P.

∃H ∈ Simulating_Halt_Deciders
∀P ∈ Turing_Machine_Descriptions
∀I ∈ Finite_Strings
(UTM(P,I) = ∞) ⊢ (H(P,I) = 0)

In English it says that whenever the input (P,I) to UTM would never halt
then a simulating halt decider correctly decides not halting on this input.

The behavior of the proxy computation forms the halt deciding basis for
the inputs that would otherwise specify the pathological self-reference
error.

http://www.liarparadox.org/Halting_problem_undecidability_and_infinitely_nested_simulation.pdf

Wikipedia contributors. Self-evidence. Wikipedia, The Free Encyclopedia.
May 17, 2021, 19:44 UTC. Available at:
https://en.wikipedia.org/w/index.php?title=Self-evidence&oldid=1023688680.
Accessed May 22, 2021.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

[olcott]

On 5/22/2021 1:04 PM, André G. Isaak wrote:

> On 2021-05-22 11:46, olcott wrote:
>> In epistemology (theory of knowledge), a self-evident proposition is a
>> proposition that is known to be true by understanding its meaning
>> without proof ... (Wikipedia: Self-evidence)
>>
>> Self-evident(A) Every simulation of input P that never halts unless
>> simulating halt decider H aborts this simulation <is> a non-halting
>> computation. This remains true even after H stops simulating P.
>>
>> Because we know that the only difference in the behavior of a
>> simulating halt decider and a simulator is that the simulating halt
>> decider stops simulating some of its inputs we can examine the
>> behavior of these inputs in a simulator to determine whether or not a
>> simulating halt decider would stop simulating these inputs.
>

> Or, of course, we could just run the computation directly. Your
> fascination with simulation is odd to say the least...
>
> Either way, though,
>
> Your H(H_Hat, H_Hat) claims that H_Hat(H_Hat) doesn't halt.
>
> BUT
>
> If we run H_Hat(H_Hat) it *does* halt.
>
> Similarly, if we run
>
> S(H_Hat, H_Hat), it also halts.
>
> You are determined to instead run
>
> S(S_Hat, S_Hat), which admittedly does not halt, but S_Hat(S_Hat) is an
> *entirely* *different* computation from H_Hat(H_Hat).
>
> According to your own description "we can examine the behavior of THESE
> INPUTS [caps mine] in a simulator to determine whether or not a

> simulating halt decider would stop simulating these inputs."
>

> The input to H(H_Hat, H_Hat) is H_Hat, H_Hat. It isn't S_Hat, S_Hat. So
> by the very logic you give above you must give your simulator H_Hat,
> H_Hat as its input, not S_Hat, S_Hat.
>
> André
>

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn

You have to examine the rest of what I said before the point that I am
making is complete.

Until we have a mutual understanding on all the points in this thread
you will not be able to understand how it provides the correct basis for
the embedded halt decider at state Ĥ.qx to correctly decide not halting
on its input: ([Ĥ],[Ĥ]).