Re: Proof that H(D,D) meets its abort criteria --self-evident truth--

On 3/17/2024 7:28 PM, immibis wrote:

On 18/03/24 00:24, olcott wrote:

On 3/17/2024 12:35 PM, immibis wrote:

On 17/03/24 17:47, olcott wrote:

It is an empirically verified fact that either
Sipser_H(Sipser_D, Sipser_D) aborts its simulation or neither
Sipser_H(Sipser_D, Sipser_D) nor Sipser_D(Sipser_D) ever stops
running (not counting stack overflow).

>
Sipser_H and Sipser_D are the same code that you called H and D in x86utm, right?
>

Not quite.
int Sipser_D(int (*M)())
{
   if ( Sipser_H(M, M) )
     return 0;
   return 1;
}
>
Sipser_H is the same and Sipser_H(Sipser_D, Sipser_D)
is isomorphic to H(D,D)
>

It is an empirically verified fact that H(D,D) aborts its simulation.

That is not the question.
>
We are verifying that it is proven to be correct.
[Proof that H(D,D) meets its abort criteria --self-evident truth--]

 === DEFINITION OF CORRECT ===
 Output Q of a halt decider H(X,Y) is correct if and only if:
1. Q==1 and X(Y) halts, or
2. Q==0 and X(Y) does not halt
 === CONCLUSION ===
it is incorrect
 

∀H ∈ TM ∀D ∈ TMD | (H(D,D) simulates its input and D calls H(D,D))
(1) H(D,D) does not abort its simulation then simulated D(D) never stops running.
(2) H(D,D) aborts its simulation then simulated D(D) stops running.

>
For every possible way that H can be encoded and D(D)
calls H(D,D) either H(D,D) aborts its simulation or D(D)
never stops running.
>
>

 

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