Re: H(D,D) cannot even be asked about the behavior of D(D) V2

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