Re: Can D simulated by H terminate normally? --- Message_ID Provided

On 5/18/24 6:44 PM, olcott wrote:

On 5/18/2024 3:02 PM, Richard Damon wrote:

On 5/18/24 3:57 PM, olcott wrote:

On 5/1/2024 7:10 PM, Richard Damon wrote:

The second method uses the fact that you have not restricted what H is allowed to do, and thus H can remember that it is simulating, and if a call to H shows that it is currently doing a simulation, just immediately return 0.

>
Nice try but this has no effect on any D correctly simulated by H.
When the directly executed H aborts its simulation it only returns
to whatever directly executed it.

>
Why? My H does correctly simulate the D it was given.
>
You don't seem to understand how the C code actually works.
>

>
If the directly executed outermost H does not abort then none of
the inner simulated ones abort because they are the exact same code.
When the directly executed outermost H does abort it can only return
to its own caller.

>
WHAT inner simulatioin?
>
>
My H begins as:
>
int H(ptr x, ptr y) {
   static int flag = 0;
   if(flag) return 0;
   flag = 1;
>
followed by essentially your code for H, except that you need to disable the hack that doesn't simulate the call to H, but just let it continue into H where it will immediately return to D and D will then return.
>
>
Thus, your claim is shown to be wrong.
>

 We are talking about every element of an infinite set where
H correctly simulates 1 to ∞ steps of D thus including 0 to ∞
recursive simulations of H simulating itself simulating D.
 *At whatever point the directly executed H(D,D) stops simulating*
*its input it cannot possibly return to any simulated input*

And my H never stops simulating, so that doesn't apply. It will reach the final state.
You seem to not understand such simple things.
I guess this is because you are just an ignorant pathological liar that doesn't care avout the truth, so your rebuttal s don't need to have anythign to do with the statement you are claiming to rebute.