Deutsch   English   Français   Italiano  
<v4n7g7$61l9$7@i2pn2.org>

View for Bookmarking (what is this?)
Look up another Usenet article

Path: ...!weretis.net!feeder9.news.weretis.net!i2pn.org!i2pn2.org!.POSTED!not-for-mail
From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V3
 ---IGNORING ALL OTHER REPLIES
Date: Sun, 16 Jun 2024 13:30:47 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <v4n7g7$61l9$7@i2pn2.org>
References: <v4kf3h$3h3iu$7@dont-email.me> <v4kfoa$2218$19@i2pn2.org>
 <v4l2mr$3l6pa$1@dont-email.me> <v4l6gg$3n5d$1@i2pn2.org>
 <v4l87j$3m8b0$2@dont-email.me> <v4l8jn$3n5d$3@i2pn2.org>
 <v4la7d$3m8b0$4@dont-email.me> <v4lan7$3n5c$2@i2pn2.org>
 <v4lcoo$3n4dj$3@dont-email.me> <v4leiq$3n5d$8@i2pn2.org>
 <v4lfu5$3rfk3$2@dont-email.me> <v4m63j$3v7mm$1@dont-email.me>
 <v4mmvp$1qt6$4@dont-email.me>
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
Injection-Date: Sun, 16 Jun 2024 17:30:47 -0000 (UTC)
Injection-Info: i2pn2.org;
	logging-data="198313"; mail-complaints-to="usenet@i2pn2.org";
	posting-account="diqKR1lalukngNWEqoq9/uFtbkm5U+w3w6FQ0yesrXg";
User-Agent: Mozilla Thunderbird
X-Spam-Checker-Version: SpamAssassin 4.0.0
Content-Language: en-US
In-Reply-To: <v4mmvp$1qt6$4@dont-email.me>
Bytes: 6503
Lines: 124

On 6/16/24 8:48 AM, olcott wrote:
> On 6/16/2024 3:00 AM, Mikko wrote:
>> On 2024-06-16 01:42:29 +0000, olcott said:
>>
>>> On 6/15/2024 8:19 PM, Richard Damon wrote:
>>>> On 6/15/24 8:48 PM, olcott wrote:
>>>>> On 6/15/2024 7:13 PM, Richard Damon wrote:
>>>>>> On 6/15/24 8:05 PM, olcott wrote:
>>>>>>> On 6/15/2024 6:37 PM, Richard Damon wrote:
>>>>>>>> On 6/15/24 7:30 PM, olcott wrote:
>>>>>>>>> On 6/15/2024 6:01 PM, Richard Damon wrote:
>>>>>>>>>> On 6/15/24 5:56 PM, olcott wrote:
>>>>>>>>>>> On 6/15/2024 11:33 AM, Richard Damon wrote:
>>>>>>>>>>>> On 6/15/24 12:22 PM, olcott wrote:
>>>>>>>>>>>>> On 6/13/2024 8:24 PM, Richard Damon wrote:
>>>>>>>>>>>>>  > On 6/13/24 11:32 AM, olcott wrote:
>>>>>>>>>>>>>  >>
>>>>>>>>>>>>>  >> It is contingent upon you to show the exact steps of 
>>>>>>>>>>>>> how H computes
>>>>>>>>>>>>>  >> the mapping from the x86 machine language finite string 
>>>>>>>>>>>>> input to
>>>>>>>>>>>>>  >> H(D,D) using the finite string transformation rules 
>>>>>>>>>>>>> specified by
>>>>>>>>>>>>>  >> the semantics of the x86 programming language that 
>>>>>>>>>>>>> reaches the
>>>>>>>>>>>>>  >> behavior of the directly executed D(D)
>>>>>>>>>>>>>  >>
>>>>>>>>>>>>>  >
>>>>>>>>>>>>>  > Why? I don't claim it can.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The first six steps of this mapping are when instructions
>>>>>>>>>>>>> at the machine address range of [00000cfc] to [00000d06]
>>>>>>>>>>>>> are simulated/executed.
>>>>>>>>>>>>>
>>>>>>>>>>>>> After that the behavior of D correctly simulated by H diverges
>>>>>>>>>>>>> from the behavior of D(D) because the call to H(D,D) by D
>>>>>>>>>>>>> correctly simulated by H cannot possibly return to D.
>>>>>>>>>>>>
>>>>>>>>>>>> Nope, the steps of D correctly simulated by H will EXACTLY 
>>>>>>>>>>>> match the steps of D directly executed, until H just gives 
>>>>>>>>>>>> up and guesses.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> When we can see that D correctly simulated by H cannot possibly
>>>>>>>>>>> reach its simulated final state at machine address [00000d1d]
>>>>>>>>>>> after one recursive simulation and the same applies for 2,3,...N
>>>>>>>>>>> recursive simulations then we can abort the simulated input and
>>>>>>>>>>> correctly report that D correctly simulated by H DOES NOT HALT.
>>>>>>>>>>
>>>>>>>>>> Nope. Because an aborted simulation doesn't say anything about 
>>>>>>>>>> Halting,
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> It is the mathematical induction that says this.
>>>>>>>>>
>>>>>>>> WHAT "Mathematical Induction"?
>>>>>>>>
>>>>>>>
>>>>>>> A proof by induction consists of two cases. The first, the base
>>>>>>> case, proves the statement for n = 0 without assuming any knowledge
>>>>>>> of other cases. The second case, the induction step, proves that
>>>>>>> if the statement holds for any given case n = k then it must also
>>>>>>> hold for the next case n = k + 1 These two steps establish that the
>>>>>>> statement holds for every natural number n.
>>>>>>> https://en.wikipedia.org/wiki/Mathematical_induction
>>>>>>
>>>>>> Ok, so you can parrot to words.
>>>>>>
>>>>>>>
>>>>>>> It is true that after one recursive simulation of D correctly
>>>>>>> simulated by H that D does not reach its simulated final state
>>>>>>> at machine address [00000d1d].
>>>>>>
>>>>>> Which means you consider that D has been bound to that first H, so 
>>>>>> you have instruciton to simulate in the call H.
>>>>>>
>>>>>>>
>>>>>>> *We directly see this is true for every N thus no assumption needed*
>>>>>>> It is true that after N recursive simulations of D correctly
>>>>>>> simulated by H that D does not reach its simulated final state
>>>>>>> at machine address [00000d1d].
>>>>>>
>>>>>> Nope, because to do the first step, you had to bind the definition 
>>>>>> of the first H to D, and thus can not change it.
>>>>>
>>>>> So infinite sets are permanently beyond your grasp.
>>>>> The above D simulated by any H has the same property
>>>>> of never reaching its own simulated machine address
>>>>> at [00000d1d].
>>>>>
>>>>> What I mistook for dishonestly is simply a lack
>>>>> of comprehension.
>>>>>
>>>>
>>>>
>>>> But it isn't an infinite set.
>>>>
>>>
>>> Sure it is you are just clueless.
>>> I mistook your ignorance for deception.
>>>
>>>> We don't ask an infinite set a question, or give a decider an 
>>>> infinite set of inputs.
>>>>
>>>
>>> Yes we do and this is simply over your head.
>>>
>>> When Ĥ is applied to ⟨Ĥ⟩
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>> The second ⊢* wildcard specifies this infinite set.
>>
>> As you should already know, ⊢* as used by Linz is not a wildcard.
>> It is a repeated application of ⊢ without showing intermediate steps.
>>
> 
> It *is* a wild card such that the Linz template simultaneously
> specifies an infinite set of machines.
> 

of POTENTIAL machines, of which *ONE* is chosen (arbitrarily) to test.

After which, it is found that the potentially infinite set of machines 
is actually the empty set, when you apply the requirements.