On 3/7/2022 8:36 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 3/7/2022 4:13 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 3/7/2022 1:16 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 3/7/2022 10:53 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 3/7/2022 5:07 AM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 3/6/2022 8:46 PM, Ben Bacarisse wrote:
>>>>>>>>>>> I'm getting bored so I will skip to the chase at the end..
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>>>> As long as the Linz Ĥ relationship to its halt determiner is
>>>>>>>>>>>> maintained any damn thing that correctly determines that ⟨Ĥ⟩ ⟨Ĥ⟩ never
>>>>>>>>>>>> halts defeats the Linz proof.
>>>>>>>>>>>
>>>>>>>>>>> Not "any damn thing", no. It must be the supposed halt determiner, H,
>>>>>>>>>>> that correctly determines it. Now that might be what you are trying to
>>>>>>>>>>> so say with the mysterious "its", but then putting in "any damn thing"
>>>>>>>>>>> is just daft.
>>>>>>>>>>>
>>>>>>>>>>> More clearly, if you have an TM H such that either
>>>>>>>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qy when Ĥ halts on input ⟨Ĥ⟩,
>>>>>>>>>>> or
>>>>>>>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn when Ĥ does not halt on input ⟨Ĥ⟩,
>>>>>>>>>>> then you will have done the impossible thing what you claimed to have
>>>>>>>>>>> done more than three years ago.
>>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>
>>>>>>>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>>>>>>> Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
>>>>>>>>>>
>>>>>>>>>> Then these steps would keep repeating:
>>>>>>>>>> Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
>>>>>>>>>> Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
>>>>>>>>>> Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...
>>>>>>>>>
>>>>>>>>> So, just to be 100% clear, you are /not/ claiming to have an TM that
>>>>>>>>> does either
>>>>>>>>
>>>>>>>> What I am claiming is that the above shows that if embedded_H rejected
>>>>>>>> its input it would be correct.
>>>>>>>
>>>>>>> So what? We are interested in the halting problem and your claim to
>>>>>>> have addressed the key case: to have an H such that either
>>>>>>
>>>>>> The copy of Linz H embedded at Ĥ.qx would correctly decide that its
>>>>>> input ⟨Ĥ⟩ ⟨Ĥ⟩ never halts thus refuting the Linz proof that says this
>>>>>> is impossible.
>>>>> No. Linz's H is such that neither
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qy when Ĥ halts on input ⟨Ĥ⟩,
>>>>> nor
>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn when Ĥ does not halt on input ⟨Ĥ⟩,
>>>>> is possible.
>>>>
>>>> Like I said I am not talking about H.
>>>> I have not been talking about H for many months.
>>> There's no need to keep repeating that. I understand 100% that you are
>>> not talking about, and don't want to talk about, your H.
>>> If you really don't want to talk about your H, don't reply. I'll say
>>> what I like about it in peace. But you can't offer any response to my
>>> posts about it if you won't talk about it.
>>
>> I am only talking about the simulating halt decider that is embedded
>> in the middle of the Linz Ĥ at Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
>
> Oh, you are talking about Linz's H and Ĥ. OK. To "refute the proof"
> (as you put it) you'd need a TM that that does either
>
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qy when Ĥ halts on input ⟨Ĥ⟩,
> or
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn when Ĥ does not halt on input ⟨Ĥ⟩
>
> But you don't (though you once said you did). Unsurprisingly, the proof
> is solid. Such an H is impossible.
>
>>>> Also the key thing that you make sure to ignore is that it is never
>>>> the case that any decider reports on the behavior of the computation
>>>> that contains itself. Decider's are not capable of reporting on the
>>>> behavior of Turing machines they can only accept or reject finite
>>>> string inputs.
>>>
>>> I don't ignore it, I tell you that you are wrong -- every time you bring
>>> it up. What's more, I have, on multiple occasions, said that I can
>>> offer you a series of exercises that, if you were to take them
>>> seriously, might lead you to see why you are wrong.
>>> This is one of your favourite themes. You accuse people of ignoring
>>> something that they have answered every time you bring it up. But,
>>> unlike a crank, I will answer it every time you ask: you are wrong. Of
>>> course deciders are capable of reporting on the behaviour of Turing
>>> machines.
>
> So are you not interested in this fundamental error anymore? I suppose
> you need to wait a bit before saying I'm ignoring this point again!
>
>>>>>> https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>> if the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its own final state
>>>>>> in a finite number of steps.
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> if the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach its own final
>>>>>> state in any finite number of steps.
>>>>>
>>>>> Still don't care about this H of yours. You hooked people in with a lie
>>>>> that you kept repeating: your H is Linz's H. At least you've come clean
>>>>> now and are attempting a hand-waving definition of your H.
>>> So the situation is
>>
>> That you cannot pay attention to the fact that when the simulating
>> halt decider embedded within Linz Ĥ at Ĥ.qx would reject its input it
>> would be correct and refute the Linz proof.
>
> No. To "refute the proof" you need an H that does either
>
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qy when Ĥ halts on input ⟨Ĥ⟩,
> or
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn when Ĥ does not halt on input ⟨Ĥ⟩
>
> but you won't even talk about H!
>
> Anyway, with cranks, silence is often the only answer one can get, so I
> can do no more than take your silence on this point as consent: you have
> no such TM and never did.
>
> After all, you've spent years walking back this claim, to the extent
> that you were recently touting a decider for some made up problem of
> your own that is not the halting problem.
>
>>> that no one cares about your H and you dare not talk
>>> about it! There is nothing substantive to discuss if you won't talk
>>> about the big mistakes.
>>>
>>
>> This is verbatim Linz except it has been adapted to my clearer
>> notational conventions:
>>
>> <Linz quote>
>> Now Ĥ is a Turing machine, so that it will have some description
>> in Σ*, say ⟨Ĥ⟩. 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 ⟨Ĥ⟩.
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>
>> if Ĥ applied to ⟨Ĥ⟩ halts, and
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> if Ĥ applied to ⟨Ĥ⟩ does not halt. This is clearly nonsense. 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 quote>
>>
>> The copy of Linz H embedded at Ĥ.qx will be referred to as embedded_H
>>
>> Linz is confused into believing that embedded_H is reporting on the
>> behavior of the computation that contains itself Ĥ applied to ⟨Ĥ⟩,
>> rather than the behavior specified by its input: ⟨Ĥ⟩, ⟨Ĥ⟩.
>
> No he is not. No mention at all is made of what embedded_H reports.
> The nonsense is simply that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* ∞ if Ĥ applied to ⟨Ĥ⟩ halts, and
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
> Your only hope of removing this contradiction is to change the
> conditions on those lines. That's the ruse you've been trying to pull
> for months with your pompous "Halt status criterion measure", but I
> suppose you've realised that won't wash so, at least for today, we are
> back to Linz's H and his correct criterion: halting.
>
> But there was a time you made a bolder claim: to have an H that gave the
> correct answer for this one case -- a TM that does either
>
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qy when Ĥ halts on input ⟨Ĥ⟩,
> or
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn when Ĥ does not halt on input ⟨Ĥ⟩.
>
> But you no longer make that (impossible) claim.
>
>> THE KEY THING THAT YOU KEEP IGNORING IS
>> THE KEY THING THAT YOU KEEP IGNORING IS
>> THE KEY THING THAT YOU KEEP IGNORING IS
>> THE KEY THING THAT YOU KEEP IGNORING IS
>> THE KEY THING THAT YOU KEEP IGNORING IS
>> THE KEY THING THAT YOU KEEP IGNORING IS
>> THE KEY THING THAT YOU KEEP IGNORING IS
>> THE KEY THING THAT YOU KEEP IGNORING IS
>> THE KEY THING THAT YOU KEEP IGNORING IS
>
> That you act like an over-tired six year old? I do try to ignore that.
>
>> It is never the case that any decider reports on the behavior of the
>> computation that contains itself. Deciders are not capable of
>> reporting on the behavior of Turing machines they can only accept or
>> reject finite string inputs.
>
> I keep addressing this. Again: you are wrong, Deciders /are/ capable of
> reporting on the behavior of Turing machines.

Deciders only compute the mapping from input finite strings to accept or
reject states. If you disagree please provide a link that proves that
deciders compute mappings from non-input non-strings.

If you don't disagree then you know that embedded_H does not compute the
mapping from Ĥ ⟨Ĥ⟩ (not an input and not finite a string) and does
compute the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ (both an input and finite strings).

--
Copyright 2021 Pete Olcott

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