On 3/12/2017 6:25 AM, Ben Bacarisse wrote:
> peteolcott <peteolcott@gmail.com> writes:
>
>> http://LiarParadox.org/HP_Infinite_Recursion.pdf
>>
>> As this page 319 of An Introduction to Formal Languages and Automata
>> by Peter Linz 1990 indicates
>>
>> From H' we construct another Turing machine H-Hat. This new machine
>> takes as input Wm, copies it then behaves exactly like H'.
>>
>> q0 Wm |-* H-Hat q0 Wm Wm...
>>
>> Page 320 indicates that we apply H-Hat to itself as input.
>
> No, it shows H^ acting on w^ which is the representation of H^ in
> {0,1}*.
>
>> The problem is that every H-Hat needs a pair of inputs.
>
> No. H^ needs nothing. Like any TM, it acts on a tape and has no
> needs.
>
>> H-Hat takes an H-Hat as input
>
> No "an" H^. It takes *the* representation of H^ as input...
>
>> and copies it so that it
>> can analyze how its input H-hat would analyze the copy
>> of H-Hat that it just made.
>
> No, it copies it. It does this for no purpose at all. It might analyze
> the input or it may not. It might take the square root of the input, or
> it might immediately halt. H^ is simply H with an initial step. H is
> arbitrary. For every possible Turing machine M, M^ can be constructed.
>
>> The input H-Hat would have to copy its own input H-Hat
>> so that it can analyze what its own input H-Hat would
>> do on its own input, on and on forever...
>
> You write as H^ has some "intent" where in fact H^ just copies the input
> but is otherwise the same as H. If your humanising metaphors about what
> H^ "would have" to do is intended to explain why, in your mind, no H
> satisfying the requirements that were put on it can exist, then OK, but
> I doubt your explanation will help anyone else to understand the proof.
>
> <snip>
>

Do you agree that the definition of the Linz Ĥ on page 319
http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

specifies infinitely nested simulation to the internal halt
decider of Linz Ĥ @Ĥq0 wM wM (the second q0 start state of Ĥ)
when this internal halt decider is based on simulating its input?

--
Copyright 2021 Pete Olcott

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

olcott <NoOne@NoWhere.com> writes:

> Do you agree that the definition of the Linz Ĥ on page 319
> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
> specifies infinitely nested simulation to the internal halt
> decider of Linz Ĥ @Ĥq0 wM wM (the second q0 start state of Ĥ)
> when this internal halt decider is based on simulating its input?

No. H^ does not specify nested anything because it's not even a
computation.

Why do you feel you have to refer to Linz when you are not talking about
Linz's H? It makes what you write garbled.

Your TM, D, is a partial decider that naively uses simulation and gets
some halting problem instances wrong. Here are the facts:

(a) UTM^([UTM^]) is not a finite computation. The "hat" construction
guarantees it.

(b) Your partial decider, D, is a UTM that also scans for the
silly-non-halting pattern. It is not a UTM.

(c) D^([D^]) is a finite computation.

(d) D(<[D^], [D^]>) gives the wrong result for the halting problem but,
according to you, the correct result or the silly-non-halting
problem.

You declare D(<[D^], [D^]>) correct because UTM^([UTM^]) is not finite.

Notation: M(s) is the computation that results from Turing machine M
applied to string s. [M] is the string encoding of a TM M. <x, y> is
the encoding of a pair of strings. <[M], s> is therefore the string
representing M(s).

M^(s) is M'(<s, s>) where M' is a TM that does not halt on exactly those
inputs on which M halts.

--
Ben.

On 5/20/2021 7:23 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> Do you agree that the definition of the Linz Ĥ on page 319
>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>> specifies infinitely nested simulation to the internal halt
>> decider of Linz Ĥ @Ĥq0 wM wM (the second q0 start state of Ĥ)
>> when this internal halt decider is based on simulating its input?
>
> No. H^ does not specify nested anything because it's not even a
> computation.
>

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

Truism(1a)
The simulation of: ([Ĥ][Ĥ]) by the simulating halt decider @ Ĥ.qx would
never halt unless Ĥ.qx aborts this simulation.

Truism(1b)
(1a) is another way of saying that the simulation of: ([Ĥ][Ĥ]) by a UTM
@ Ĥ.qx would never halt.

Unless and until people understand that (1a) is a paraphrase of (1b) we
cannot move to truism(2).

The above does not specify two machines. In the purely hypothetical case
where Ĥ.qx did not stop simulating its input then in this purely
hypothetical case the behavior of its input would be the same as if the
partial halt decider @ Ĥ.qx was replaced by a UTM.

Any high school student would be able to analyze the basic logic of the
above and know that it is correct very easily.

If X is the only cause of Y and not X then not Y.

That you and others try to dance all around this very obviously correct
logic seems quite disingenuous.

X = Ĥ.qx stops simulating ([Ĥ][Ĥ])
Y = The simulation of ([Ĥ][Ĥ]) stops

~X = Ĥ.qx never stops simulating ([Ĥ][Ĥ])
~Y = The simulation of ([Ĥ][Ĥ]) never stops

> Why do you feel you have to refer to Linz when you are not talking about
> Linz's H? It makes what you write garbled.
>
> Your TM, D, is a partial decider that naively uses simulation and gets
> some halting problem instances wrong. Here are the facts:
>
> (a) UTM^([UTM^]) is not a finite computation. The "hat" construction
> guarantees it.
>
> (b) Your partial decider, D, is a UTM that also scans for the
> silly-non-halting pattern. It is not a UTM.
>
> (c) D^([D^]) is a finite computation.
>
> (d) D(<[D^], [D^]>) gives the wrong result for the halting problem but,
> according to you, the correct result or the silly-non-halting
> problem.
>
> You declare D(<[D^], [D^]>) correct because UTM^([UTM^]) is not finite.
>
> Notation: M(s) is the computation that results from Turing machine M
> applied to string s. [M] is the string encoding of a TM M. <x, y> is
> the encoding of a pair of strings. <[M], s> is therefore the string
> representing M(s).
>
> M^(s) is M'(<s, s>) where M' is a TM that does not halt on exactly those
> inputs on which M halts.
>

--
Copyright 2021 Pete Olcott

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