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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: embedded_H applied to =?UTF-8?B?4p+oxKTin6kg4p+oxKTin6k=?=
 computes the mapping from its input to =?UTF-8?B?xKQucW4=?=
Date: Thu, 1 Aug 2024 12:17:09 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 01 Aug 2024 06:49:13 -0500 schrieb olcott:
> On 8/1/2024 2:44 AM, Mikko wrote:
>> On 2024-07-31 17:27:33 +0000, olcott said:
>>> On 7/31/2024 2:32 AM, Mikko wrote:
>>>> On 2024-07-30 14:16:20 +0000, olcott said:
>>>>> On 7/30/2024 1:37 AM, Mikko wrote:
>>>>>> On 2024-07-29 16:16:13 +0000, olcott said:
>>>>>>> On 7/28/2024 3:02 AM, Mikko wrote:
>>>>>>>> On 2024-07-27 14:08:10 +0000, olcott said:
>>>>>>>>> On 7/27/2024 2:21 AM, Mikko wrote:
>>>>>>>>>> On 2024-07-26 14:08:11 +0000, olcott said:

>>>>>>> When we compute the mapping from the input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>>> to the behavior specified by this input we know that embedded_H is
>>>>>>> correct to transition to Ĥ.qn.
>>>>>>
>>>>>> The meaning of "correct" in this context is that if the transition
>>>>>> of embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn is correct if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions
>>>>>> to H.qn but incorrect otherwise.
>>>>>
>>>>> No you are wrong.
>>>> Which dictionary (or other authority) disagrees?
>>>
>>> The common knowledge that a decider computes the mapping from its
>>> input finite string...
>>> This is almost always the same as the direct execution of the machine
>>> represented by this finite string.
Not "almost". Otherwise it is doing something different.

>> None of above indicates any disagreement by any authority.
> Everyone (even Linz) has the wrong headed idea that a halt decider must
> report on the behavior of the computation that itself is contained
> within. This has always been wrong.
Dude. The halting problem /specifically/ asks about a machine simulating
itself.
> A halt decider must always report on the behavior that its finite string
> specifies. This is different only when an input invokes its own decider.
Um, no? Then it is making a mistake.

>>> The one rare exception is shown above where Ĥ ⟨Ĥ⟩ halts and the input
>>> to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach its own final state of
>>> ⟨Ĥ.qn⟩ when embedded_H acts as if it was a UTM.
H is not an UTM, though.

>> That is not supported by any anuthority.
>> 
> The authority says *given an input of the function domain it*
> *can return the corresponding output*
Which authority? Not that that would be a valid argument.

> In other words all deciders compute the mapping from their input (finite
> string) to an accept or reject state.
> This means that they do not compute the mapping of the executing process
> of themselves.
They do, if those happen to coincide.

> I am the first person in the world that noticed these two could be
> different. Everyone that has disagreed with me is disagreeing with the
> semantics of the x86 language.
What are the semantics that you disagree about?

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.