Re: D correctly simulated by H cannot possibly halt --- templates and infinite sets --- deciders

On 5/30/2024 2:32 PM, Fred. Zwarts wrote:

Op 30.mei.2024 om 21:01 schreef olcott:

On 5/30/2024 1:50 PM, Fred. Zwarts wrote:

Op 30.mei.2024 om 19:00 schreef olcott:

On 5/30/2024 10:20 AM, Fred. Zwarts wrote:

Op 30.mei.2024 om 16:43 schreef olcott:

On 5/28/2024 11:16 AM, olcott wrote:

>
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
*Formalizing the Linz Proof structure*
∃H  ∈ Turing_Machines
∀x  ∈ Turing_Machines_Descriptions
∀y  ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
>

>
A decider computes the mapping from finite string inputs to
its own accept or reject state.
>
A decider does not and cannot compute the mapping from
Turing_Machine inputs to its own accept or reject state.
>
Halts(x,y) would report on the direct execution of x(y) thus ignores
the pathological behavior of x correctly simulated by pure function H.
This makes Halts(x,y) an incorrect measure of the correctness of H(x,y).

>
Why are you referring to the 'pathological behavior of x' if your claim is that the simulator does not even reach the part of DD (below) that contradicts the result of HH? This 'pathological behavior of x' is completely irrelevant.

>
It is totally relevant because it is the reason why D correctly
simulated by H cannot possibly halt.

>
Incorrect. Your own words are that lines 04, 05 and 06 are nor reachable for the simulator.

>
Because D correctly simulated by H remains stuck in recursive simulation
because D calls H(D,D) in recursive simulation D cannot possibly reach
past its own line 03.
>
You must must 100% complete attention to the exact words that I exactly say.
>

 If you ask to 100% attention, please, pay also attention to the replies and read more than the first few words. You remove most of my reply, probably because you did not read them.
 

I soon as yo show that you are starting with a fundamentally false assumption I stop reading.

The fact that D does not reach past line 03, means that lines 04, 05 and 06 do not play a role in the decision.

OK

Do you understand C?

I learned C back when K&R was the standard and have been a professional
C++ software engineer for two decades. I have been professional
programmer since 1984.

If line 04 cannot be reached, lines 05 and 06 do not cause any behaviour. In particular no 'pathological' behaviour.
 

The reason why these lines can't be reached is that D calls H(D,D)
in recursive simulation. This relationship between H and D is
typically called pathological.

It is H that keeps repeating the simulation of D

D calls H(D,D) in recursive simulation until H stops this.
THIS IS D'S FAULT!

and the next H, so the simulated H never reaches its abort, and therefore it does not reach its final state. D acts only as a quick parameter duplicator so that H simulates itself. Then simulated H gets stuck in an infinite recursion and never reaches the 'pathological' part of D.
Even a beginning C programmer will see that if the simulated H would really halt (as required),

It does yet you continue to fail to understand this.

then simulated D would continue to line 04.

No that is utterly impossible because the only reason
that H halts is that it totally stops simulating D.

But simulated H does not halt and must be aborted.
 

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