Re: D correctly simulated by pure function H cannot possibly reach its, own line 06 --- Linz proof

On 5/26/24 1:54 PM, olcott wrote:

On 5/26/2024 12:48 PM, Richard Damon wrote:

On 5/26/24 1:26 PM, olcott wrote:

On 5/26/2024 12:16 PM, Richard Damon wrote:

On 5/26/24 1:01 PM, olcott wrote:

On 5/26/2024 11:31 AM, Richard Damon wrote:

On 5/26/24 10:13 AM, olcott wrote:

On 5/26/2024 6:43 AM, Richard Damon wrote:
>
typedef int (*ptr)();  // ptr is pointer to int function in C
00       int H(ptr p, ptr i);
01       int D(ptr p)
02       {
03         int Halt_Status = H(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         H(D,D);
12         return 0;
13       }
>
>

Because, as I have said, the answer and reasoning changes depending on what you acknowledged are the implications of your stipulations. For instance, if your actual understanding of being a "Pure Function" is that the program is the equivalent of a Turing Machine, then we need to add a strictness to the definition that isn't normally used for just "Pure Functions", like accessing value of registers like the program counter or stack pointer might not be allowed in some cases. (which breaks you H).
>

>
Since we can see (and you already agreed) that D correctly simulated
by pure simulator H remains stuck in infinite recursive simulation then
we also know that D never reaches its own line 06 and halts in less
than an infinite number of correctly simulated steps.

>
But it depends on what H actually does, which you refuse to agree to.
>

>
When I specify that D is correctly simulated by pure function H, that
means that H is a pure simulator that stops simulating D after some
finite number of correctly simulated steps. There is absolutely nothing
else that needs to be known about H besides this.
>

>
And, do you accept that this exact description means that H is NOT the computational equivalent of a Turing Machine, and that this simulation doesn't prove that its input is not a non-halting program?
>

>
In other words you are saying that it is 100% totally impossible to
adapt an actual UTM so that it stops simulating after a finite number of
steps.

>
Yep, at least and have it still be a UTM.
>
IMPOSSIBLE BY DEFINITION.

 *Yet not otherwise impossible at all*

In other words, you beleive that you can presume the impossible to happen.
You logic is DEAD.
YOu are worse than the election deniers
I guess you don't know the meaning of the word "Definition", which says a lot about you ideas of things being true by the meaning of the words, if words don't really have their meaning.

 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 Thus when we hypothesize that embedded_H is an exact UTM then we
see that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly
reach its own simulated final state of ⟨Ĥ.qn⟩ even in an infinite
number of simulated steps.

Except if Embedded) is an EXACT UTM, then so is H and as we have agreed, H fails to be a decide.

  From this we can definitively conclude that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated
by embedded_H cannot possibly reach its own simulated final state of
⟨Ĥ.qn⟩ in less than an infinite number of simulated steps.
 >

Well, if you H is a UTM, then yes, neither H nor embedded_H can return an answer for the H^ built on them, and thus fail to be a decider.
So, you are just showing your logic is based on the assumption of the impossible, and on you lying abou ti.