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

On 11/10/2021 8:36 AM, Malcolm McLean wrote:

On Wednesday, 10 November 2021 at 13:37:01 UTC, Ben Bacarisse wrote:

I, for one, am fed up with trying to explain ever more silly
consequences of the initial assumption about H. The proof is simple and
the contradiction entailed is obvious.

There seems to be an oscillating between a pure simulator and a halt
decider. Then recently there's been a distinction between a decider which simulates its input and one which executes its input directly.

But the core problem remains that the halt decider returns "false" for a machine / program which halts. I haven't see any real advance on that, and
as you say, it become repetitive pointing it out.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

If the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts then the transition to ⊢* Ĥ.qn
is necessarily correct no matter what Ĥ ⟨Ĥ⟩ does. A halt decider is only accountable for correctly deciding the halt status of its actual input.

Indeed. So tell PO that there is no string <H^>. If there were, there
wold be a TM H^ halts if it does not and does not halt if it does.

There might be a assumption that H is a correct halt decider built into
PO's argument somewhere, but I haven't actually found it. However I
haven't been paying such close attention.

--
Copyright 2021 Pete Olcott

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

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

On 11/10/2021 9:44 AM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

In this Linz machine:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Remember to add that this must be the case if, and only if, Ĥ applied to ⟨Ĥ⟩ does not halt. Then it all becomes clear to the average reader.

If the actual input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ would never halt then the transition
to ⊢* Ĥ.qn is necessarily correct no matter what Ĥ ⟨Ĥ⟩ does. A halt decider is only accountable for correctly deciding the halt status of
its actual input.


--
Copyright 2021 Pete Olcott

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

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

On 11/10/2021 2:49 PM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

On 11/10/2021 1:20 PM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

On 11/10/2021 11:02 AM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

On 11/10/2021 9:44 AM, Ben Bacarisse wrote:

olcott <NoOne@NoWhere.com> writes:

In this Linz machine:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Remember to add that this must be the case if, and only if, Ĥ applied to
⟨Ĥ⟩ does not halt. Then it all becomes clear to the average reader.

If the actual input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ would never halt

Inputs don't halt or not halt. You've been told this many times. You >>>>> also know how to say what you are trying to say correctly, but I think >>>>> you want to avoid being clear.
Anyway, just make sure you keep the correct condition: if, and only if, >>>>> Ĥ applied to ⟨Ĥ⟩ does not halt. Any other "facts" care to add are >>>>> immaterial since the correct condition shows that there are not such >>>>> TMs.

then the transition to ⊢* Ĥ.qn is necessarily correct no matter what Ĥ
⟨Ĥ⟩ does.

What Ĥ ⟨Ĥ⟩ does is what makes the line above apply or not -- it's there
in the part you deliberately keep omitting. Ignoring (and not stating) >>>>> what Ĥ ⟨Ĥ⟩ does is central to why you are wrong.

A halt
decider is only accountable for correctly deciding the halt status of >>>>>> its actual input.

There is no halt decider present in the line you keep misquoting. There >>>>> is a "half-decider" at Ĥ.qx and we know what it does. You just keep >>>>> omitting the key statements so that you can add some waffle instead.

If the pure simulation of the actual input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ would never
reach a final state of Ĥ then the transition to ⊢* Ĥ.qn is necessarily >>>> correct no matter what Ĥ ⟨Ĥ⟩ does. A halt decider is only accountable
for correctly deciding the halt status of its actual input.

As I say, you can add any waffle you like provided you keep the correct
condition in place. You can prove that for every TM H that behaves as
Linz specifies, the string ⟨Ĥ⟩ is even. And you can prove that it's odd
as well. You can prove anything from a contradiction.

If it is necessarily true that input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts when
it is correctly simulated then it is necessarily correct for Ĥ.qx to
transition to Ĥ.qn on this input.

You don't seem to be listening. You certainly don't have anything
pertinent to say.

If an X is a Y then Z is always correct when it reports that an X is a Y.

If the correctly simulated input to any halt decider never halts then it
is always correct for this halt decider to report that this input never
halts.

You don't seem to be able to comprehend the concept of logical
necessity or that disagreeing with logical necessity is woefully
foolish.

I think it's rather foolish of you to think I might care about your
opinion of me. In case you have forgotten, I don't. I happy to stand
by my posts and let others decide who is being logical.

--
Copyright 2021 Pete Olcott

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

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