On 10/25/2021 7:46 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 10/24/2021 6:30 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>>>>>> q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ
>>>>>>>>>>>>>> if M applied to Wm does not halt
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Linz cannot be correctly referring to q0 applied to Wm because there
>>>>>>>>>>>>>> is no halt decider at q0.
>>>>>>>>>>>>> This notation is hopeless in ASCII. Using the notation that has evolved
>>>>>>>>>>>>> in these threads (not my preferred one, but I don't want to complicate
>>>>>>>>>>>>> matters any more) Linz writes:
>>>>>>>>>>>>> Ĥ.q0 <M> ⊢* Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
>>>>>>>>>>>>> if M applied to <M> does not halt.
>>>>>>>>>>>>> And with these changes in notation you seem to be saying
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Linz cannot be correctly referring to Ĥ applied to <M> because there
>>>>>>>>>>>>>> is no halt decider at q0.
>>>>>>>>>>>>> Yes, there is no halt decider at Ĥ.q0, but he is correctly referring to
>>>>>>>>>>>>> what Ĥ applied to <M> should do: Ĥ applied to <M> should transition to
>>>>>>>>>>>>> Ĥ.qn if (and only if) M (the TM encoded in the string input) applied to
>>>>>>>>>>>>> <M> (that exactly same string input) does not halt.
>>>>>>>>>>>>> Your next point helps explain how he comes to say this about Ĥ.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Linz can only be referring to Ĥ.qx <M> <M> the machine of the first
>>>>>>>>>>>>>> <M> being applied to the machine description of the second <M>.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The annotation dose not specifically refer to what this sub-computation
>>>>>>>>>>>>> does. It simply gives the condition under which Ĥ will transition from
>>>>>>>>>>>>> Ĥ.q0 to Ĥ.qn.
>>>>>>>>>>>>
>>>>>>>>>>>> If a simulating halt decider H correctly determines that the
>>>>>>>>>>>> simulation of its input <M> applied to <M> never reaches its final
>>>>>>>>>>>> state whether or not this simulating halt decider aborts the
>>>>>>>>>>>> simulation of this input then this simulating halt decider does
>>>>>>>>>>>> correctly decide that this input never halts NO MATTER WHAT ELSE.
>>>>>>>>>>> This is not what Linz is saying.
>>>>>>>>>>
>>>>>>>>>> Of course this is not what Linz is saying.
>>>>>>>>> But until you understand what he is saying, you can't see why you are
>>>>>>>>> wrong.
>>>>>>>>
>>>>>>>> That Linz did not consider a simulating halt decider does not make me
>>>>>>>> wrong.
>>>>>>>
>>>>>>> Quite. You are wrong for the reasons I gave.
>>>>>>>
>>>>>>>> If any of this was wrong someone could point out an actual error:
>>>>>>>>
>>>>>>>> The machine at Ĥq0 transitions to Ĥqn if the simulation of ⟨Ĥ⟩ applied
>>>>>>>> to ⟨Ĥ⟩ would never reach its final state whether or not this
>>>>>>>> simulation is aborted.
>>>>>>> The error is that this is a statement about an sub-set of and empty set
>>>>>>> of Turing machines.
>>>>>>
>>>>>> You are simply making a false assumption. You are beginning with the
>>>>>> premise that Linz is correct then concluding that Linz is correct on
>>>>>> the basis of this premise.
>>>>>
>>>>> A proved theorem is not a "false assumption".
>>>>
>>>> Ah so you are back to your blasphemy. One can not rely on a proved
>>>> theorem as exactly equal to infallibility.
>>>
>>> You are being dishonest again. Setting aside the silly religious
>>> language, I am not asserting that any argument is infallible something
>>> that was obvious from what I wrote immediate after this.
>>> I am simply saying you have not shown any error in the argument. Until
>>> you show that, for example, 2+2 is not equal to 4, it is entirely
>>> reasonable to assert that it is. That is not blasphemy.
>>
>> I don't see how this is so difficult for you:
>> (a) If it is claimed that X cannot possibly be done
>> (b) I show X being correctly done
>> (c) then I have refuted X cannot possibly be done
>
> I don't think you honestly believe that the logic is difficult for me.

The alternative is that you are a liar and I prefer to give you the
benefit of the doubt.

> Neither do I think you have trouble knowing what part I am saying you
> have failed at (it's (b) by the way).

We have not gotten to B yet.
First we must have mutual agreement on the hypothetical.

Here is the long form of the hypothetical:

THIS STATEMENT IS NECESSARILY ALWAYS TRUE
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.

> But my objection to what you've been saying goes my deeper than a
> failure to show anything of value. You are being dishonest in that you
> claim to be talking about Linz's class of TMs, but you keep cutting key
> facts about them while trying to replace those facts with waffle of you
> your own invention.

THIS IS THE SHORT VERSION OF THE HYPOTHETICAL
If a halt decider correctly decides that its input never halts then the
halt decider did correct decide the halt status of its input.

You talk about me being dishonest when you continually refuse to agree
that when a halt decider is correct then it is not incorrect.

--
Copyright 2021 Pete Olcott

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