Deutsch   English   Français   Italiano  
<v4k44j$3fmth$1@dont-email.me>

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

Path: ...!weretis.net!feeder9.news.weretis.net!feeder8.news.weretis.net!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V2
Date: Sat, 15 Jun 2024 08:14:57 -0500
Organization: A noiseless patient Spider
Lines: 79
Message-ID: <v4k44j$3fmth$1@dont-email.me>
References: <v4j0h2$39gh7$3@dont-email.me> <v4k0sr$3f4m3$1@dont-email.me>
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
Injection-Date: Sat, 15 Jun 2024 15:14:59 +0200 (CEST)
Injection-Info: dont-email.me; posting-host="020d44455d70acf7231aebb6a85d124b";
	logging-data="3660721"; mail-complaints-to="abuse@eternal-september.org";	posting-account="U2FsdGVkX18gZbQdRDjVBWJCi5nxZw2u"
User-Agent: Mozilla Thunderbird
Cancel-Lock: sha1:aFL7CHv/egfVXxxFYQFL21rAhzs=
Content-Language: en-US
In-Reply-To: <v4k0sr$3f4m3$1@dont-email.me>
Bytes: 4264

On 6/15/2024 7:19 AM, Mikko wrote:
> On 2024-06-15 03:07:14 +0000, olcott said:
> 
>> 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.
>>
>> _D()
>> [00000cfc](01) 55          push ebp
>> [00000cfd](02) 8bec        mov ebp,esp
>> [00000cff](03) 8b4508      mov eax,[ebp+08]
>> [00000d02](01) 50          push eax       ; push D
>> [00000d03](03) 8b4d08      mov ecx,[ebp+08]
>> [00000d06](01) 51          push ecx       ; push D
>> [00000d07](05) e800feffff  call 00000b0c  ; call H
>> [00000d0c](03) 83c408      add esp,+08
>> [00000d0f](02) 85c0        test eax,eax
>> [00000d11](02) 7404        jz 00000d17
>> [00000d13](02) 33c0        xor eax,eax
>> [00000d15](02) eb05        jmp 00000d1c
>> [00000d17](05) b801000000  mov eax,00000001
>> [00000d1c](01) 5d          pop ebp
>> [00000d1d](01) c3          ret
>> Size in bytes:(0034) [00000d1d]
>>
>> If there is no mapping from the input to H(D,D) to the behavior
>> of D(D) then H is not even being asked about the behavior of D(D).
>> H has no obligation to answer questions *THAT IT IS NOT BEING ASKED*
> 
> The halting problem specification does not say that a halt decider
> can be asked questions. 

*It assumes that you already know that*
In computability theory and computational complexity theory, a
decision problem is a computational problem that can be posed
as a yes–no question of the input values.
https://en.wikipedia.org/wiki/Decision_problem

Likewise algebra textbooks assume that you already know
arithmetic.

> It requires that a description of a Turing
> macine and a description of an input to that Turing machine can
> be given as an input.
> 

Yes and the above x86 machine code is the x86 equivalent
of a Turing Machine description.
This input DOES NOT MAP TO THE BEHAVIOR OF D(D)

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation

Two complete simulations show a pair of identical TMD's are
simulating a pair of identical inputs. We can see this thus
proving recursive simulation.


-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer