XPost: comp.theory, sci.logic, sci.math

On 3/18/2022 11:54 AM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

No BRAIN DEAD MORON this is not true, I keep calling you a BRAIN DEAD
MORON because after I have explain all the details you cannot remember
what I just said.

I see the petulant six-year-old is in residence today.

A decider maps its inputs to its own accept reject state.
A halt decider does not compute the halt status of itself.

A halt decider, let's call it H, maps the input <Ĥ> <Ĥ> to its accept
state if (and only if) Ĥ enter a final state on input <Ĥ>, and it maps
<Ĥ> <Ĥ> to its reject state if (and only if) Ĥ does not enter a final
state on input <Ĥ>.

No that is entirely incorrect. You have changed the subject away from
the Linz proof's conclusion that no matter what state Ĥ applied to <Ĥ> transitions to a contradiction is derived.

</Linz:1990:320>
Now Ĥ is a Turing machine, so that it will have some description in Σ*,
say ŵ . This string, in addition to being the description of Ĥ can also
be used as input string. We can therefore legitimately ask what would
happen if Ĥ is applied to ŵ .
…
The contradiction tells us that our assumption of the existence of H,
and hence the assumption of the decidability of the halting problem,
must be false.
</Linz:1990:320>

https://www.liarparadox.org/Linz_Proof.pdf

Here is the correct analysis of Ĥ applied to <Ĥ> (no contradiction):

embedded_H maps its input <Ĥ> <Ĥ> to its Ĥ.qn final reject state if the correctly simulated input <Ĥ> <Ĥ> would never reach its own final state
of <Ĥ>.qn in any finite number of steps of simulation.

No such H exists.

You implicitly accept this fact because you propose an H which does not
meet this specification but I can't see why you think anyone would care
about it.

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

XPost: comp.theory, sci.logic, sci.math

On 3/19/2022 8:10 AM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

On 3/18/2022 8:05 PM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

On 3/18/2022 11:54 AM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

No BRAIN DEAD MORON this is not true, I keep calling you a BRAIN DEAD >>>>>> MORON because after I have explain all the details you cannot remember >>>>>> what I just said.

I see the petulant six-year-old is in residence today.

A decider maps its inputs to its own accept reject state.
A halt decider does not compute the halt status of itself.

A halt decider, let's call it H, maps the input <Ĥ> <Ĥ> to its accept >>

You have changed the subject away from the Linz proof's conclusion
that deals with Ĥ applied to <Ĥ> and not H applied to <Ĥ> <Ĥ>.

You made an incorrect statement about a halt decider. I corrected it.
You will ignore the correction. The conversation remains exactly as it
was years ago.

You can't manage to stay on the topic at hand and can only do one
dishonest dodge or another because you cannot find any error in the
essence of my statements.

Perhaps in your case the concept of infinitely nested simulation is
beyond your capacity to understand so all that you can do is dodge this
subject to hide the fact that you don't even know the terminology.

No one can find any error in the following only because there is no
error to be found.

When Ĥ is applied to ⟨Ĥ⟩
Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then embedded_H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩

Then these steps would keep repeating:
Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H2 simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩...

(1) If embedded_H does not abort the simulation of its input the
simulation never stops and the simulated input never reaches its final
state.

(2) If embedded_H does abort the simulation of its input the simulation
is aborted at some point shown above and the simulated input never
reaches its final state.

This proves that ⟨Ĥ⟩ ⟨Ĥ⟩ presents embedded_H with a sequence of configurations that never reach a final state. This in turn makes the transition to Ĥ.qn by embedded_H correct.

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)